Chapter 1: Functions and Graphs

Problem 1

Step 1 — Read off the domain:

The domain is the set of all \(x\)-values that appear in the table.

$$\text{Domain} = \{-3,\, -2,\, -1,\, 0,\, 1,\, 2,\, 3\}$$

Step 2 — Read off the range:

The range is the set of distinct \(y\)-values that appear in the table. The values \(9,\, 4,\, 1,\, 0,\, 1,\, 4,\, 9\) reduce (after removing duplicates) to:

$$\text{Range} = \{0,\, 1,\, 4,\, 9\}$$

Step 3 — Check the function test:

A relation is a function exactly when every input \(x\) is paired with only one output \(y\). Scanning the table, each \(x\)-value appears exactly once, so each input has a unique output.

Answer: Domain \(= \{-3,-2,-1,0,1,2,3\}\), Range \(= \{0,1,4,9\}\). The relation is a function.

Problem 2

Step 1 — Read off the domain:

$$\text{Domain} = \{-3,\, -2,\, -1,\, 0,\, 1,\, 2,\, 3\}$$

Step 2 — Read off the range:

The \(y\)-values are \(-2,\, -8,\, -1,\, 0,\, 1,\, 8,\, -2\). Removing the duplicate \(-2\):

$$\text{Range} = \{-8,\, -2,\, -1,\, 0,\, 1,\, 8\}$$

Step 3 — Check the function test:

Every \(x\)-value in the table appears exactly once. Even though the output \(-2\) is reused (for \(x = -3\) and \(x = 3\)), the function test only forbids one input pairing with two different outputs — repeating outputs is allowed.

Answer: Domain \(= \{-3,-2,-1,0,1,2,3\}\), Range \(= \{-8,-2,-1,0,1,8\}\). The relation is a function.

Problem 3

Step 1 — Read off the domain:

The \(x\)-values listed are \(1,\, 2,\, 3,\, 0,\, 1,\, 2,\, 3\). The set of distinct inputs is:

$$\text{Domain} = \{0,\, 1,\, 2,\, 3\}$$

Step 2 — Read off the range:

The \(y\)-values listed are \(-3,\, -2,\, -1,\, 0,\, 1,\, 2,\, 3\). All are distinct:

$$\text{Range} = \{-3,\, -2,\, -1,\, 0,\, 1,\, 2,\, 3\}$$

Step 3 — Check the function test:

Look for any \(x\)-value that pairs with two different \(y\)-values. The input \(x = 1\) appears twice: once with \(y = -3\) and once with \(y = 1\). One input is matched to two different outputs, which violates the definition of a function. (The inputs \(x = 2\) and \(x = 3\) also fail in the same way.)

Answer: Domain \(= \{0,1,2,3\}\), Range \(= \{-3,-2,-1,0,1,2,3\}\). The relation is not a function.

Problem 4

Step 1 — Read off the domain:

$$\text{Domain} = \{1,\, 2,\, 3,\, 4,\, 5,\, 6,\, 7\}$$

Step 2 — Read off the range:

Every \(y\)-value in the table equals \(1\), so the range is the single-element set:

$$\text{Range} = \{1\}$$

Step 3 — Check the function test:

Each \(x\)-value appears exactly once, paired with exactly one output (the constant value \(1\)). The function test allows different inputs to share the same output; it only forbids one input pairing with two outputs.

Answer: Domain \(= \{1,2,3,4,5,6,7\}\), Range \(= \{1\}\). The relation is a function (a constant function).

Problem 5

Step 1 — Read off the domain:

$$\text{Domain} = \{3,\, 5,\, 8,\, 10,\, 15,\, 21,\, 33\}$$

Step 2 — Read off the range:

The \(y\)-values are \(3,\, 2,\, 1,\, 0,\, 1,\, 2,\, 3\). Removing duplicates:

$$\text{Range} = \{0,\, 1,\, 2,\, 3\}$$

Step 3 — Check the function test:

Every \(x\)-value listed is distinct, so each input has exactly one output. The reused \(y\)-values (\(1,\, 2,\, 3\)) do not break the function rule — different inputs are allowed to share an output.

Answer: Domain \(= \{3,5,8,10,15,21,33\}\), Range \(= \{0,1,2,3\}\). The relation is a function.

Problem 6

Step 1 — Read off the domain:

The \(x\)-values are \(-7,\, -2,\, -2,\, 0,\, 1,\, 3,\, 6\). Removing the duplicate:

$$\text{Domain} = \{-7,\, -2,\, 0,\, 1,\, 3,\, 6\}$$

Step 2 — Read off the range:

The \(y\)-values are \(11,\, 5,\, 1,\, -1,\, -2,\, 4,\, 11\). Removing the duplicate \(11\):

$$\text{Range} = \{-2,\, -1,\, 1,\, 4,\, 5,\, 11\}$$

Step 3 — Check the function test:

The input \(x = -2\) is paired with two different outputs: \(y = 5\) and \(y = 1\). A single input cannot map to two outputs in a function, so this relation fails the test.

Answer: Domain \(= \{-7,-2,0,1,3,6\}\), Range \(= \{-2,-1,1,4,5,11\}\). The relation is not a function.

Problem 7

Step 1 — Substitute each input into \(f(x) = 5x - 2\):

We evaluate \(f\) at each of the six requested inputs.

$$\begin{aligned}

f(0) &= 5(0) - 2 = -2 \\

f(1) &= 5(1) - 2 = 3 \\

f(3) &= 5(3) - 2 = 13 \\

f(-x) &= 5(-x) - 2 = -5x - 2 \\

f(a) &= 5a - 2 \\

f(a+h) &= 5(a + h) - 2 = 5a + 5h - 2

\end{aligned}$$

Step 2 — Simplify each result:

All expressions above are already fully simplified.

Answer: \(f(0) = -2,\ f(1) = 3,\ f(3) = 13,\ f(-x) = -5x - 2,\ f(a) = 5a - 2,\ f(a+h) = 5a + 5h - 2\).

Problem 8

Step 1 — Substitute each input into \(f(x) = 4x^2 - 3x + 1\):

$$\begin{aligned}

f(0) &= 4(0)^2 - 3(0) + 1 = 1 \\

f(1) &= 4(1)^2 - 3(1) + 1 = 4 - 3 + 1 = 2 \\

f(3) &= 4(3)^2 - 3(3) + 1 = 36 - 9 + 1 = 28

\end{aligned}$$

Step 2 — Evaluate \(f(-x)\):

Replace every \(x\) with \(-x\). Note \((-x)^2 = x^2\).

$$f(-x) = 4(-x)^2 - 3(-x) + 1 = 4x^2 + 3x + 1$$

Step 3 — Evaluate \(f(a)\):

$$f(a) = 4a^2 - 3a + 1$$

Step 4 — Evaluate \(f(a + h)\):

$$\begin{aligned}

f(a+h) &= 4(a+h)^2 - 3(a+h) + 1 \\

&= 4(a^2 + 2ah + h^2) - 3a - 3h + 1 \\

&= 4a^2 + 8ah + 4h^2 - 3a - 3h + 1

\end{aligned}$$

Answer: \(f(0) = 1,\ f(1) = 2,\ f(3) = 28,\ f(-x) = 4x^2 + 3x + 1,\ f(a) = 4a^2 - 3a + 1,\ f(a+h) = 4a^2 + 8ah + 4h^2 - 3a - 3h + 1\).

Problem 9

Step 1 — Note the domain restriction:

Because \(f(x) = \dfrac{2}{x}\) divides by \(x\), the value \(f(0)\) is undefined.

Step 2 — Substitute the remaining inputs:

$$\begin{aligned}

f(0) &\text{ does not exist} \\

f(1) &= \frac{2}{1} = 2 \\

f(3) &= \frac{2}{3} \\

f(-x) &= \frac{2}{-x} = -\frac{2}{x} \quad (x \neq 0) \\

f(a) &= \frac{2}{a} \quad (a \neq 0) \\

f(a+h) &= \frac{2}{a + h} \quad (a + h \neq 0)

\end{aligned}$$

Answer: \(f(0)\) does not exist; \(f(1) = 2,\ f(3) = \tfrac{2}{3},\ f(-x) = -\tfrac{2}{x},\ f(a) = \tfrac{2}{a},\ f(a+h) = \tfrac{2}{a+h}\).

Problem 10

Step 1 — Substitute each input into \(f(x) = |x - 7| + 8\):

Use \(|y| = y\) when \(y \ge 0\) and \(|y| = -y\) when \(y < 0\).

$$\begin{aligned}

f(0) &= |0 - 7| + 8 = 7 + 8 = 15 \\

f(1) &= |1 - 7| + 8 = 6 + 8 = 14 \\

f(3) &= |3 - 7| + 8 = 4 + 8 = 12

\end{aligned}$$

Step 2 — Evaluate \(f(-x)\):

$$f(-x) = |-x - 7| + 8 = |-(x + 7)| + 8 = |x + 7| + 8$$

Step 3 — Evaluate \(f(a)\) and \(f(a + h)\):

These cannot be simplified further without knowing the sign of the inside expression.

$$f(a) = |a - 7| + 8, \qquad f(a + h) = |a + h - 7| + 8$$

Answer: \(f(0) = 15,\ f(1) = 14,\ f(3) = 12,\ f(-x) = |x + 7| + 8,\ f(a) = |a - 7| + 8,\ f(a+h) = |a + h - 7| + 8\).

Problem 11

Step 1 — Substitute each input into \(f(x) = \sqrt{6x + 5}\):

The square root is defined only when \(6x + 5 \ge 0\), i.e. \(x \ge -\tfrac{5}{6}\).

$$\begin{aligned}

f(0) &= \sqrt{6(0) + 5} = \sqrt{5} \\

f(1) &= \sqrt{6(1) + 5} = \sqrt{11} \\

f(3) &= \sqrt{6(3) + 5} = \sqrt{23}

\end{aligned}$$

Step 2 — Evaluate \(f(-x)\):

$$f(-x) = \sqrt{6(-x) + 5} = \sqrt{-6x + 5} = \sqrt{5 - 6x}$$

This requires \(5 - 6x \ge 0\), i.e. \(x \le \tfrac{5}{6}\).

Step 3 — Evaluate \(f(a)\) and \(f(a + h)\):

$$f(a) = \sqrt{6a + 5}, \qquad f(a + h) = \sqrt{6(a + h) + 5} = \sqrt{6a + 6h + 5}$$

(Each requires the radicand to be non-negative.)

Answer: \(f(0) = \sqrt{5},\ f(1) = \sqrt{11},\ f(3) = \sqrt{23},\ f(-x) = \sqrt{5 - 6x},\ f(a) = \sqrt{6a + 5},\ f(a+h) = \sqrt{6a + 6h + 5}\).

Problem 12

Step 1 — Substitute each input into \(f(x) = \dfrac{x - 2}{3x + 7}\):

The denominator vanishes when \(3x + 7 = 0\), i.e. \(x = -\tfrac{7}{3}\), so that input is excluded.

$$\begin{aligned}

f(0) &= \frac{0 - 2}{3(0) + 7} = -\frac{2}{7} \\

f(1) &= \frac{1 - 2}{3(1) + 7} = \frac{-1}{10} = -\frac{1}{10} \\

f(3) &= \frac{3 - 2}{3(3) + 7} = \frac{1}{16}

\end{aligned}$$

Step 2 — Evaluate \(f(-x)\):

$$f(-x) = \frac{-x - 2}{3(-x) + 7} = \frac{-(x + 2)}{-3x + 7} = \frac{-(x+2)}{7 - 3x}$$

Step 3 — Evaluate \(f(a)\) and \(f(a + h)\):

$$f(a) = \frac{a - 2}{3a + 7}, \qquad f(a + h) = \frac{(a + h) - 2}{3(a + h) + 7} = \frac{a + h - 2}{3a + 3h + 7}$$

(In each case the denominator must be non-zero.)

Answer: \(f(0) = -\tfrac{2}{7},\ f(1) = -\tfrac{1}{10},\ f(3) = \tfrac{1}{16},\ f(-x) = \dfrac{-(x+2)}{7 - 3x},\ f(a) = \dfrac{a - 2}{3a + 7},\ f(a+h) = \dfrac{a + h - 2}{3a + 3h + 7}\).

Problem 13

Step 1 — Recognize the constant function:

\(f(x) = 9\) returns \(9\) regardless of the input.

Step 2 — Evaluate at every requested input:

$$f(0) = f(1) = f(3) = f(-x) = f(a) = f(a + h) = 9$$

Answer: All six values equal \(9\).

Problem 14

Step 1 — Find the domain of \(f(x) = \dfrac{x}{x^2 - 16}\):

The function is undefined where the denominator is zero. Solve \(x^2 - 16 = 0\):

$$x^2 = 16 \implies x = \pm 4$$

So we must exclude \(x = 4\) and \(x = -4\).

$$\text{Domain} = (-\infty,\, -4) \cup (-4,\, 4) \cup (4,\, \infty)$$

Step 2 — Find the zeros:

Set the numerator to zero (and confirm the denominator is non-zero there):

$$x = 0, \quad \text{denominator} = 0^2 - 16 = -16 \neq 0$$

So \(x = 0\) is a zero, giving \(x\)-intercept \((0, 0)\). Because \(f(0) = 0\), the \(y\)-intercept is also \((0, 0)\).

Step 3 — Determine the range:

Solve \(y = \dfrac{x}{x^2 - 16}\) for \(x\). Multiplying through, \(y(x^2 - 16) = x\), i.e.

$$y x^2 - x - 16 y = 0.$$

If \(y = 0\), then \(x = 0\), which lies in the domain — so \(y = 0\) is attainable. Otherwise treat it as a quadratic in \(x\) with discriminant \(1 + 64 y^2 \ge 1 > 0\); the discriminant is always positive, so real \(x\) exists for every non-zero \(y\) as well. Therefore every real number is in the range.

$$\text{Range} = (-\infty,\, \infty)$$

Answer: Domain \(= (-\infty, -4) \cup (-4, 4) \cup (4, \infty)\); Range \(= (-\infty, \infty)\); only intercept is the origin \((0, 0)\).

Problem 15

Step 1 — Find the domain of \(g(x) = \sqrt{8x - 1}\):

The radicand must be non-negative.

$$8x - 1 \ge 0 \implies x \ge \tfrac{1}{8}$$

$$\text{Domain} = \left[\tfrac{1}{8},\, \infty\right)$$

Step 2 — Find the range:

A principal square root is always \(\ge 0\), and as \(x \to \infty\) the expression grows without bound.

$$\text{Range} = [0,\, \infty)$$

Step 3 — Find the zeros and intercepts:

Set \(g(x) = 0\): \(\sqrt{8x - 1} = 0 \implies 8x - 1 = 0 \implies x = \tfrac{1}{8}\). So the \(x\)-intercept is \(\left(\tfrac{1}{8},\, 0\right)\).

For the \(y\)-intercept, evaluate \(g(0)\). But \(x = 0\) is not in the domain, so there is no \(y\)-intercept.

Answer: Domain \(= \left[\tfrac{1}{8}, \infty\right)\); Range \(= [0, \infty)\); \(x\)-intercept \(\left(\tfrac{1}{8}, 0\right)\); no \(y\)-intercept.

Problem 16

Step 1 — Find the domain of \(h(x) = \dfrac{3}{x^2 + 4}\):

The denominator \(x^2 + 4\) is always positive (\(\ge 4\)), so the function is defined for every real \(x\).

$$\text{Domain} = (-\infty,\, \infty)$$

Step 2 — Find the zeros:

\(h(x) = 0\) would require \(3 = 0\), which is impossible. There are no zeros, so there is no \(x\)-intercept.

Step 3 — Find the \(y\)-intercept:

$$h(0) = \frac{3}{0 + 4} = \frac{3}{4}$$

So the \(y\)-intercept is \(\left(0,\, \tfrac{3}{4}\right)\).

Step 4 — Find the range:

\(h(x)\) is positive everywhere. It is maximized when the denominator is smallest, at \(x = 0\), giving \(h(0) = \tfrac{3}{4}\). As \(|x| \to \infty\), \(h(x) \to 0^+\) but never reaches \(0\).

$$\text{Range} = \left(0,\, \tfrac{3}{4}\right]$$

Answer: Domain \(= (-\infty, \infty)\); Range \(= \left(0, \tfrac{3}{4}\right]\); no \(x\)-intercept; \(y\)-intercept \(\left(0, \tfrac{3}{4}\right)\).

Problem 17

Step 1 — Find the domain of \(f(x) = -1 + \sqrt{x + 2}\):

Require \(x + 2 \ge 0\):

$$x \ge -2 \implies \text{Domain} = [-2,\, \infty)$$

Step 2 — Find the range:

The square root \(\sqrt{x + 2}\) ranges over \([0, \infty)\). Subtracting \(1\) shifts the range down by \(1\):

$$\text{Range} = [-1,\, \infty)$$

Step 3 — Find the \(x\)-intercept (zeros):

Set \(f(x) = 0\):

$$\sqrt{x + 2} = 1 \implies x + 2 = 1 \implies x = -1$$

So the \(x\)-intercept is \((-1, 0)\).

Step 4 — Find the \(y\)-intercept:

$$f(0) = -1 + \sqrt{2}$$

So the \(y\)-intercept is \(\left(0,\, -1 + \sqrt{2}\right) \approx (0,\, 0.414)\).

Answer: Domain \(= [-2, \infty)\); Range \(= [-1, \infty)\); \(x\)-intercept \((-1, 0)\); \(y\)-intercept \(\left(0,\, \sqrt{2} - 1\right)\).

Problem 18

Step 1 — Find the domain of \(f(x) = \dfrac{1}{\sqrt{x - 9}}\):

The radicand must be strictly positive (the square root sits in a denominator, so it cannot be zero either).

$$x - 9 > 0 \implies x > 9 \implies \text{Domain} = (9,\, \infty)$$

Step 2 — Find the range:

For \(x > 9\), \(\sqrt{x - 9}\) takes every positive value, so its reciprocal also takes every positive value.

$$\text{Range} = (0,\, \infty)$$

Step 3 — Find the zeros and intercepts:

\(f(x) = 0\) would require the numerator to be zero, but the numerator is the constant \(1\). No zeros, so no \(x\)-intercept.

Since \(0\) is not in the domain, there is no \(y\)-intercept.

Answer: Domain \(= (9, \infty)\); Range \(= (0, \infty)\); no \(x\)- or \(y\)-intercepts.

Problem 19

Step 1 — Find the domain of \(g(x) = \dfrac{3}{x - 4}\):

Exclude where the denominator is zero: \(x - 4 = 0 \implies x = 4\).

$$\text{Domain} = (-\infty,\, 4) \cup (4,\, \infty)$$

Step 2 — Find the range:

Solving \(y = \dfrac{3}{x - 4}\) for \(x\) gives \(x = 4 + \dfrac{3}{y}\), which is defined for every \(y \neq 0\). So every non-zero \(y\) is attainable.

$$\text{Range} = (-\infty,\, 0) \cup (0,\, \infty)$$

Step 3 — Find the zeros and intercepts:

\(g(x) = 0\) would require \(3 = 0\), impossible — no \(x\)-intercept.

For the \(y\)-intercept, \(g(0) = \dfrac{3}{0 - 4} = -\dfrac{3}{4}\), so the \(y\)-intercept is \(\left(0,\, -\tfrac{3}{4}\right)\).

Answer: Domain \(= (-\infty, 4) \cup (4, \infty)\); Range \(= (-\infty, 0) \cup (0, \infty)\); no \(x\)-intercept; \(y\)-intercept \(\left(0,\, -\tfrac{3}{4}\right)\).

Problem 20

Step 1 — Find the domain of \(f(x) = 4|x + 5|\):

Absolute value is defined for every real input.

$$\text{Domain} = (-\infty,\, \infty)$$

Step 2 — Find the range:

\(|x + 5| \ge 0\), so \(4|x + 5| \ge 0\). The minimum value \(0\) occurs at \(x = -5\), and the expression grows without bound as \(|x + 5| \to \infty\).

$$\text{Range} = [0,\, \infty)$$

Step 3 — Find the zeros:

Set \(4|x + 5| = 0\): \(|x + 5| = 0 \implies x = -5\). So the \(x\)-intercept is \((-5, 0)\).

Step 4 — Find the \(y\)-intercept:

$$f(0) = 4|0 + 5| = 4(5) = 20$$

\(y\)-intercept: \((0, 20)\).

Answer: Domain \(= (-\infty, \infty)\); Range \(= [0, \infty)\); \(x\)-intercept \((-5, 0)\); \(y\)-intercept \((0, 20)\).

Problem 21

Step 1 — Find the domain of \(g(x) = \sqrt{\dfrac{7}{x - 5}}\):

The expression inside the square root must be \(\ge 0\), and the denominator \(x - 5\) cannot be zero.

Because \(7 > 0\), the sign of \(\dfrac{7}{x - 5}\) matches the sign of \(x - 5\). So we need \(x - 5 > 0\) (we cannot include \(x = 5\) — it makes the denominator zero, and the expression would not be \(\ge 0\) below \(5\) either).

$$x > 5 \implies \text{Domain} = (5,\, \infty)$$

Step 2 — Find the range:

As \(x \to 5^+\), \(\dfrac{7}{x - 5} \to +\infty\), so \(g(x) \to \infty\). As \(x \to \infty\), \(\dfrac{7}{x - 5} \to 0^+\), so \(g(x) \to 0^+\). The function is continuous and strictly decreasing on \((5, \infty)\), taking every value in between.

$$\text{Range} = (0,\, \infty)$$

Step 3 — Find the zeros and intercepts:

\(g(x) = 0\) would need \(\dfrac{7}{x - 5} = 0\), which is impossible. So no \(x\)-intercept. Since \(0 \notin\) domain, no \(y\)-intercept.

Answer: Domain \(= (5, \infty)\); Range \(= (0, \infty)\); no \(x\)- or \(y\)-intercepts.

Problem 22

Step 1 — Plot the seven listed points:

From the table, plot \((-3, 10),\ (-2, 5),\ (-1, 2),\ (0, 1),\ (1, 2),\ (2, 5),\ (3, 10)\).

Step 2 — Identify the shape:

\(f(x) = x^2 + 1\) is a parabola. Adding \(1\) to \(x^2\) shifts the standard parabola up by one unit, so the vertex is at \((0, 1)\) and the curve opens upward.

Step 3 — Connect the points with a smooth curve:

Draw a smooth, symmetric U-shape through the plotted points. The graph is symmetric about the \(y\)-axis because \(f(-x) = (-x)^2 + 1 = x^2 + 1 = f(x)\) (the function is even).

Step 4 — Describe key features:

- Vertex (minimum) at \((0, 1)\). - \(y\)-intercept \((0, 1)\). - No \(x\)-intercepts (since \(x^2 + 1 \ge 1 > 0\)). - Symmetric about the \(y\)-axis.

Answer: The graph is an upward-opening parabola with vertex \((0, 1)\), passing through the seven tabulated points and symmetric about the \(y\)-axis.

Problem 23

Step 1 — Plot the seven points from the table:

Plot \((-3, -15),\ (-2, -12),\ (-1, -9),\ (0, -6),\ (1, -3),\ (2, 0),\ (3, 3)\).

Note: the table appears to have used the rule \(y = 3x - 6\) with an unrelated tabulation error at \(x = -3, -2, -1, 0\); however the problem text fixes the rule as \(f(x) = 3x - 6\). Using the rule directly: \(f(-3) = -15,\ f(-2) = -12,\ f(-1) = -9,\ f(0) = -6,\ f(1) = -3,\ f(2) = 0,\ f(3) = 3\). These match the table (the table is consistent with the rule \(f(x) = 3x - 6\)).

Step 2 — Identify the shape:

The function is linear with slope \(3\) and \(y\)-intercept \(-6\).

Step 3 — Connect the points with a straight line:

Draw a straight line through the plotted points, extending in both directions.

Step 4 — Describe key features:

- Slope \(= 3\), \(y\)-intercept \((0, -6)\). - \(x\)-intercept: set \(3x - 6 = 0 \implies x = 2\), so \((2, 0)\). - The line is increasing everywhere; no symmetry about either axis.

Answer: The graph is a straight line through \((0, -6)\) with slope \(3\), passing through every point in the table.

Problem 24

Step 1 — Plot the seven points from the table:

Plot \(\left(-3, -\tfrac{1}{2}\right),\ (-2, 0),\ \left(-1, \tfrac{1}{2}\right),\ (0, 1),\ \left(1, \tfrac{3}{2}\right),\ (2, 2),\ \left(3, \tfrac{5}{2}\right)\).

Step 2 — Identify the shape:

\(f(x) = \tfrac{1}{2}x + 1\) is linear with slope \(\tfrac{1}{2}\) and \(y\)-intercept \(1\).

Step 3 — Connect the points with a straight line:

The line rises by \(\tfrac{1}{2}\) for every increase of \(1\) in \(x\); each plotted \(y\)-value matches \(\tfrac{x}{2} + 1\). Draw the line through the points and extend in both directions.

Step 4 — Describe key features:

- Slope \(\tfrac{1}{2}\), \(y\)-intercept \((0, 1)\). - \(x\)-intercept: \(\tfrac{1}{2}x + 1 = 0 \implies x = -2\), so \((-2, 0)\). - Strictly increasing; no symmetry.

Answer: The graph is a straight line through \((0, 1)\) with slope \(\tfrac{1}{2}\), passing through each tabulated point.

Problem 25

Step 1 — Plot the seven points from the table:

Plot \((-3, 6),\ (-2, 4),\ (-1, 2),\ (0, 0),\ (1, 2),\ (2, 4),\ (3, 6)\).

Step 2 — Identify the shape:

\(f(x) = 2|x|\) is a V-shaped absolute-value graph. The factor \(2\) makes the V twice as steep as \(y = |x|\). For \(x \ge 0\) the graph is the line \(y = 2x\); for \(x < 0\) it is the line \(y = -2x\).

Step 3 — Connect the points:

Draw a V whose vertex is at the origin, with both arms having slope \(\pm 2\).

Step 4 — Describe key features:

- Vertex (minimum) at \((0, 0)\). - \(x\)- and \(y\)-intercepts both at the origin. - Symmetric about the \(y\)-axis (since \(f(-x) = 2|-x| = 2|x| = f(x)\)); the function is even. - Decreasing on \((-\infty, 0]\), increasing on \([0, \infty)\).

Answer: The graph is a V opening upward with vertex \((0, 0)\) and arms of slope \(\pm 2\), passing through every tabulated point.

Problem 26

Step 1 — Plot the seven points from the table:

Plot \((-3, -9),\ (-2, -4),\ (-1, -1),\ (0, 0),\ (1, -1),\ (2, -4),\ (3, -9)\).

Step 2 — Identify the shape:

\(f(x) = -x^2\) is a downward-opening parabola — the reflection of \(y = x^2\) across the \(x\)-axis.

Step 3 — Connect the points with a smooth curve:

Draw a smooth, symmetric arch through the plotted points, opening downward.

Step 4 — Describe key features:

- Vertex (maximum) at \((0, 0)\). - \(x\)- and \(y\)-intercepts both at the origin. - Symmetric about the \(y\)-axis; the function is even because \(f(-x) = -(-x)^2 = -x^2 = f(x)\). - Increasing on \((-\infty, 0]\), decreasing on \([0, \infty)\).

Answer: The graph is a downward-opening parabola with vertex at the origin, passing through every tabulated point.

Problem 27

Step 1 — Plot the seven points from the table:

Plot \((-3, -27),\ (-2, -8),\ (-1, -1),\ (0, 0),\ (1, 1),\ (2, 8),\ (3, 27)\).

Step 2 — Identify the shape:

\(f(x) = x^3\) is the standard cubic curve — passing through the origin, increasing everywhere, with an inflection point at \((0, 0)\).

Step 3 — Connect the points with a smooth curve:

Draw a smooth S-shaped curve that flattens near the origin and steepens away from it.

Step 4 — Describe key features:

- \(x\)- and \(y\)-intercepts at the origin. - Symmetric about the origin (rotational symmetry): \(f(-x) = (-x)^3 = -x^3 = -f(x)\), so the function is odd. - Strictly increasing on all of \((-\infty, \infty)\). - Inflection point at \((0, 0)\).

Answer: The graph is the standard cubic — a smooth, strictly increasing S-curve through the origin, passing through every tabulated point and symmetric about the origin.

Problem 28

Note on missing graph: The problem refers to "graph 28," which is a textbook figure not reproduced in the markdown source available here. The full eight-part analysis (domain/range, intercepts, intervals of monotonicity, symmetry, even/odd) cannot be completed without the image. The following solution lays out the procedure the student should follow with the actual graph.

Step 1 — Apply the vertical line test:

Slide a vertical line across the graph from left to right. If every vertical line meets the graph in at most one point, the relation is a function; otherwise it is not. If it fails the vertical line test, stop — none of the remaining items (b)–(h) apply.

Step 2 — (a) Domain and range:

- Domain: the set of \(x\)-values for which the graph has any point. Read this off the horizontal extent (and follow the instruction to assume the graph continues beyond the grid when it appears to extend to the edges). - Range: the corresponding set of \(y\)-values.

Step 3 — (b)–(c) Intercepts:

- \(x\)-intercepts: points where the curve crosses the \(x\)-axis. - \(y\)-intercept: the point where the curve crosses the \(y\)-axis (a function has at most one).

Step 4 — (d)–(f) Monotonicity intervals:

- Increasing: \(x\)-intervals where the graph rises left-to-right. - Decreasing: \(x\)-intervals where it falls. - Constant: \(x\)-intervals where it is horizontal.

Step 5 — (g) Symmetry; (h) even/odd classification:

- If the graph is mirrored across the \(y\)-axis, the function is even (\(f(-x) = f(x)\)). - If the graph has \(180^\circ\) rotational symmetry about the origin, the function is odd (\(f(-x) = -f(x)\)). - Otherwise, neither.

Answer: Solution procedure stated above; numerical values for graph 28 cannot be supplied without the figure. (Stub solution — figure missing from source.)

Problem 29

Note on missing graph: As with problem 1.1.28, "graph 29" is a textbook figure not included in the markdown source. Without the image we cannot enumerate domain, range, intercepts, monotonicity, or symmetry numerically. The procedure below is identical to 1.1.28.

Step 1 — Apply the vertical line test:

If every vertical line crosses the graph at most once, the relation is a function; otherwise it is not, and the remaining items do not apply.

Step 2 — (a) Read domain and range:

Domain: the projection of the graph onto the \(x\)-axis; range: the projection onto the \(y\)-axis. Treat segments that run off the grid as continuing.

Step 3 — (b)–(c) Intercepts:

Estimate \(x\)-intercepts (where the curve meets the \(x\)-axis) and the \(y\)-intercept (where it meets the \(y\)-axis).

Step 4 — (d)–(f) Intervals of monotonicity:

Identify the \(x\)-intervals on which the curve is rising (increasing), falling (decreasing), or horizontal (constant).

Step 5 — (g)–(h) Symmetry and parity:

- Mirror symmetry about the \(y\)-axis \(\to\) even. - Rotational symmetry about the origin \(\to\) odd. - No such symmetry \(\to\) neither.

Answer: Procedure outlined above; specific values for graph 29 cannot be provided without the figure. (Stub solution — figure missing from source.)

Problem 30

Step 1 — Recognize what is needed: This problem refers to "graph 30," a figure that is not embedded in the source text provided to me, so I cannot read the graph directly. The solution below is a stub that explains exactly which procedure to follow once the graph is visible.

Step 2 — Apply the vertical line test: Mentally (or with a ruler) sweep a vertical line from left to right across the graph. If every vertical line crosses the curve in at most one point, the relation is a function of \(x\); otherwise it is not.

Step 3 — If the relation is a function, extract (a)–(h):

- (a) Domain and range. Read the smallest and largest \(x\)-values where the curve exists; do the same for \(y\). Use interval notation, e.g. \([-5, 5]\) or \((-\infty, \infty)\). - (b) \(x\)-intercept(s). Points where the curve crosses the \(x\)-axis (i.e. \(y = 0\)); estimate to the nearest grid mark if needed. - (c) \(y\)-intercept. The single point where the curve crosses the \(y\)-axis (\(x = 0\)). - (d) Increasing intervals. Intervals where, as \(x\) moves right, \(y\) moves up. - (e) Decreasing intervals. Intervals where, as \(x\) moves right, \(y\) moves down. - (f) Constant intervals. Intervals where \(y\) does not change (flat segment). - (g) Symmetry. Check whether the picture is unchanged by reflection across the \(y\)-axis (\(y\)-axis symmetry), across the \(x\)-axis (not allowed for a function unless it is the zero function), or through the origin (\(180^\circ\) rotation). - (h) Even / odd / neither. Even iff symmetric about the \(y\)-axis; odd iff symmetric about the origin; otherwise neither.

Answer: A worked answer requires the actual figure for graph 30. Once the figure is available, run steps 2–3 above to fill in items (a)–(h).

Problem 31

Step 1 — Recognize what is needed: The figure for "graph 31" is not embedded in the supplied source text, so a numeric answer cannot be produced. Follow the procedure below once the figure is available.

Step 2 — Apply the vertical line test: If any vertical line meets the graph at two or more points, the relation is not a function and items (a)–(h) do not apply. Otherwise continue.

Step 3 — Extract (a)–(h): Use the same reading rules listed in problem 1.1.30 above: read domain/range from the horizontal/vertical extent of the curve, locate intercepts on the axes, identify rising/falling/flat intervals, and test for \(y\)-axis or origin symmetry to classify the function as even, odd, or neither.

Answer: A complete numeric answer requires the actual figure for graph 31; the procedure above produces items (a)–(h) once the graph is visible.

Problem 32

Step 1 — Recognize what is needed: The figure for "graph 32" is not embedded in the supplied source text. The solution below explains the procedure rather than reading values from a missing image.

Step 2 — Apply the vertical line test: Sweep a vertical line across the picture; if it ever intersects the curve in two or more points, the relation fails the vertical line test and is not a function.

Step 3 — Extract (a)–(h): If the test is passed, read the domain (horizontal extent of the curve), range (vertical extent), \(x\)- and \(y\)-intercepts, increasing/decreasing/constant intervals, and check for \(y\)-axis or origin symmetry to label the function even, odd, or neither.

Answer: Final answers require the figure for graph 32; the steps above show how to obtain (a)–(h) once it is visible.

Problem 33

Step 1 — Recognize what is needed: The image for "graph 33" is not embedded in the source provided here. The solution gives the procedure to follow when the figure is available.

Step 2 — Apply the vertical line test: A vertical line that intersects the curve more than once means the relation is not a function. Otherwise proceed.

Step 3 — Extract (a)–(h): Read domain and range from the horizontal/vertical extent, locate the intercepts on the axes, identify the intervals on which the graph rises, falls, or is flat, and test for symmetry across the \(y\)-axis (even) or through the origin (odd).

Answer: A definitive answer requires the figure for graph 33; apply the procedure above once the image is in hand.

Problem 34

Step 1 — Recognize what is needed: The image for "graph 34" is missing from the supplied source. The solution outlines the procedure.

Step 2 — Apply the vertical line test: Determine whether each vertical line meets the curve at most once. If so, the relation is a function; otherwise it is not, and (a)–(h) are not applicable.

Step 3 — Extract (a)–(h): Reading from the graph, record domain (horizontal extent), range (vertical extent), intercepts, increasing/decreasing/constant intervals, and check whether \(f(-x) = f(x)\) (even, \(y\)-axis symmetry) or \(f(-x) = -f(x)\) (odd, origin symmetry), or neither.

Answer: A complete numeric answer requires the actual figure for graph 34; the procedure above produces items (a)–(h) once it is visible.

Problem 35

Step 1 — Recognize what is needed: The image for "graph 35" is not embedded in the source given to me. The solution below provides the reading procedure to use with the figure.

Step 2 — Apply the vertical line test: If every vertical line meets the curve in at most one point, the relation is a function and items (a)–(h) follow. Otherwise the relation is not a function and (a)–(h) do not apply.

Step 3 — Extract (a)–(h): Read the domain and range from the curve's horizontal and vertical extent, locate any \(x\)- and \(y\)-intercepts, find intervals of increase/decrease/constancy, and check for \(y\)-axis or origin symmetry to classify the function as even, odd, or neither.

Answer: A worked numeric answer for graph 35 requires the image; the procedure above produces items (a)–(h) once the figure is available.

Problem 36

We are given \(f(x) = 3x + 4\) and \(g(x) = x - 2.\) Both \(f\) and \(g\) are polynomials with domain \((-\infty, \infty),\) so the domains of \(f+g,\) \(f-g,\) and \(fg\) are all \((-\infty, \infty),\) and the domain of \(f/g\) excludes the zero of \(g.\)

Step 1 — Sum \(f + g\):

$$(f + g)(x) = (3x + 4) + (x - 2) = 4x + 2.$$

Domain: \((-\infty, \infty).\)

Step 2 — Difference \(f - g\):

$$(f - g)(x) = (3x + 4) - (x - 2) = 2x + 6.$$

Domain: \((-\infty, \infty).\)

Step 3 — Product \(f \cdot g\):

$$(f \cdot g)(x) = (3x + 4)(x - 2) = 3x^2 - 6x + 4x - 8 = 3x^2 - 2x - 8.$$

Domain: \((-\infty, \infty).\)

Step 4 — Quotient \(f / g\):

$$\left(\frac{f}{g}\right)(x) = \frac{3x + 4}{x - 2}.$$

Exclude \(x\) where \(g(x) = 0,\) i.e. \(x = 2.\) Domain: \((-\infty, 2) \cup (2, \infty).\)

Answer: (a) \(4x + 2,\) domain \((-\infty, \infty);\) (b) \(2x + 6,\) domain \((-\infty, \infty);\) (c) \(3x^2 - 2x - 8,\) domain \((-\infty, \infty);\) (d) \(\dfrac{3x + 4}{x - 2},\) domain \((-\infty, 2) \cup (2, \infty).\)

Problem 37

Given \(f(x) = x - 8\) and \(g(x) = 5x^2.\) Both are polynomials with domain \((-\infty, \infty),\) so \(f \pm g\) and \(fg\) have domain \((-\infty, \infty),\) and \(f/g\) excludes the zero of \(g.\)

Step 1 — Sum:

$$(f + g)(x) = (x - 8) + 5x^2 = 5x^2 + x - 8.$$

Domain: \((-\infty, \infty).\)

Step 2 — Difference:

$$(f - g)(x) = (x - 8) - 5x^2 = -5x^2 + x - 8.$$

Domain: \((-\infty, \infty).\)

Step 3 — Product:

$$(f \cdot g)(x) = (x - 8)(5x^2) = 5x^3 - 40x^2.$$

Domain: \((-\infty, \infty).\)

Step 4 — Quotient:

$$\left(\frac{f}{g}\right)(x) = \frac{x - 8}{5x^2}.$$

\(g(x) = 5x^2 = 0\) iff \(x = 0,\) so exclude \(x = 0.\) Domain: \((-\infty, 0) \cup (0, \infty).\)

Answer: (a) \(5x^2 + x - 8;\) (b) \(-5x^2 + x - 8;\) (c) \(5x^3 - 40x^2;\) (d) \(\dfrac{x - 8}{5x^2},\) domain \((-\infty, 0) \cup (0, \infty).\) All others have domain \((-\infty, \infty).\)

Problem 38

Given \(f(x) = 3x^2 + 4x + 1\) and \(g(x) = x + 1.\) Both are polynomials, so \(f \pm g\) and \(fg\) have domain \((-\infty, \infty);\) for \(f/g\) we must exclude \(x = -1.\)

Step 1 — Sum:

$$(f + g)(x) = (3x^2 + 4x + 1) + (x + 1) = 3x^2 + 5x + 2.$$

Step 2 — Difference:

$$(f - g)(x) = (3x^2 + 4x + 1) - (x + 1) = 3x^2 + 3x.$$

Step 3 — Product:

$$(f \cdot g)(x) = (3x^2 + 4x + 1)(x + 1).$$

Expand:

$$= 3x^3 + 3x^2 + 4x^2 + 4x + x + 1 = 3x^3 + 7x^2 + 5x + 1.$$

Step 4 — Quotient:

Notice \(3x^2 + 4x + 1 = (3x + 1)(x + 1),\) so

$$\frac{f(x)}{g(x)} = \frac{(3x + 1)(x + 1)}{x + 1} = 3x + 1, \quad x \neq -1.$$

Even though the simplified form is the polynomial \(3x + 1,\) the domain still excludes \(x = -1\) because that point makes the original denominator zero.

Answer: (a) \(3x^2 + 5x + 2;\) (b) \(3x^2 + 3x;\) (c) \(3x^3 + 7x^2 + 5x + 1;\) (d) \(3x + 1,\ x \neq -1,\) domain \((-\infty, -1) \cup (-1, \infty).\) All others have domain \((-\infty, \infty).\)

Problem 39

Given \(f(x) = 9 - x^2\) and \(g(x) = x^2 - 2x - 3.\) Both are polynomials; the quotient excludes the zeros of \(g.\)

Step 1 — Sum:

$$(f + g)(x) = (9 - x^2) + (x^2 - 2x - 3) = -2x + 6.$$

Step 2 — Difference:

$$(f - g)(x) = (9 - x^2) - (x^2 - 2x - 3) = 9 - x^2 - x^2 + 2x + 3 = -2x^2 + 2x + 12.$$

Step 3 — Product:

$$(f \cdot g)(x) = (9 - x^2)(x^2 - 2x - 3).$$

Expand:

$$= 9x^2 - 18x - 27 - x^4 + 2x^3 + 3x^2 = -x^4 + 2x^3 + 12x^2 - 18x - 27.$$

Step 4 — Quotient:

Factor: \(9 - x^2 = (3-x)(3+x)\) and \(x^2 - 2x - 3 = (x-3)(x+1).\) Then

$$\frac{f(x)}{g(x)} = \frac{(3-x)(3+x)}{(x-3)(x+1)} = \frac{-(x-3)(x+3)}{(x-3)(x+1)} = \frac{-(x+3)}{x+1}, \quad x \neq 3.$$

The denominator \(g(x) = 0\) when \(x = 3\) or \(x = -1,\) so both must be excluded from the domain. Domain: \((-\infty, -1) \cup (-1, 3) \cup (3, \infty).\)

Answer: (a) \(-2x + 6,\) domain \((-\infty, \infty);\) (b) \(-2x^2 + 2x + 12,\) domain \((-\infty, \infty);\) (c) \(-x^4 + 2x^3 + 12x^2 - 18x - 27,\) domain \((-\infty, \infty);\) (d) \(\dfrac{-(x+3)}{x+1},\) domain \((-\infty, -1) \cup (-1, 3) \cup (3, \infty).\)

Problem 40

Given \(f(x) = \sqrt{x}\) (domain \([0, \infty)\)) and \(g(x) = x - 2\) (domain \((-\infty, \infty)\)). The domain of each combination is the intersection of the individual domains, further restricted by removing zeros of the denominator for the quotient.

Step 1 — Intersection of domains:

\([0, \infty) \cap (-\infty, \infty) = [0, \infty).\)

Step 2 — Sum, difference, product:

$$(f + g)(x) = \sqrt{x} + x - 2,$$

$$(f - g)(x) = \sqrt{x} - x + 2,$$

$$(f \cdot g)(x) = \sqrt{x}\,(x - 2) = x\sqrt{x} - 2\sqrt{x}.$$

All three have domain \([0, \infty).\)

Step 3 — Quotient:

$$\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{x - 2}.$$

We must remove \(x = 2\) (denominator zero). Domain: \([0, 2) \cup (2, \infty).\)

Answer: (a) \(\sqrt{x} + x - 2;\) (b) \(\sqrt{x} - x + 2;\) (c) \(x\sqrt{x} - 2\sqrt{x};\) each with domain \([0, \infty);\) (d) \(\dfrac{\sqrt{x}}{x - 2},\) domain \([0, 2) \cup (2, \infty).\)

Problem 41

Given \(f(x) = 6 + \dfrac{1}{x}\) (domain \(x \neq 0\)) and \(g(x) = \dfrac{1}{x}\) (domain \(x \neq 0\)). The intersection of domains is \(\{x : x \neq 0\},\) i.e. \((-\infty, 0) \cup (0, \infty).\)

Step 1 — Sum:

$$(f + g)(x) = 6 + \frac{1}{x} + \frac{1}{x} = 6 + \frac{2}{x}.$$

Step 2 — Difference:

$$(f - g)(x) = 6 + \frac{1}{x} - \frac{1}{x} = 6.$$

Step 3 — Product:

$$(f \cdot g)(x) = \left(6 + \frac{1}{x}\right)\cdot\frac{1}{x} = \frac{6}{x} + \frac{1}{x^2}.$$

Step 4 — Quotient:

$$\left(\frac{f}{g}\right)(x) = \frac{6 + \dfrac{1}{x}}{\dfrac{1}{x}} = \left(6 + \frac{1}{x}\right) \cdot x = 6x + 1, \quad x \neq 0.$$

The simplified result is a polynomial, but the original quotient was undefined at \(x = 0,\) so the domain still excludes \(0.\)

Answer: (a) \(6 + \dfrac{2}{x};\) (b) \(6;\) (c) \(\dfrac{6}{x} + \dfrac{1}{x^2};\) (d) \(6x + 1,\ x \neq 0.\) All four have domain \((-\infty, 0) \cup (0, \infty).\)

Problem 42

Given \(f(x) = 3x\) and \(g(x) = x + 5.\) Both are polynomials with domain \((-\infty, \infty).\)

Step 1 — Compose \((f \circ g)(x) = f(g(x)):\)

$$f(g(x)) = 3 \cdot g(x) = 3(x + 5) = 3x + 15.$$

Since \(g\) is defined for all real \(x\) and \(f\) accepts every real input, the domain is \((-\infty, \infty).\)

Step 2 — Compose \((g \circ f)(x) = g(f(x)):\)

$$g(f(x)) = f(x) + 5 = 3x + 5.$$

Same reasoning: domain \((-\infty, \infty).\)

Answer: (a) \((f \circ g)(x) = 3x + 15,\) domain \((-\infty, \infty);\) (b) \((g \circ f)(x) = 3x + 5,\) domain \((-\infty, \infty).\)

Problem 43

Given \(f(x) = x + 4\) and \(g(x) = 4x - 1.\) Both are polynomials with domain \((-\infty, \infty).\)

Step 1 — Compute \((f \circ g)(x):\)

$$f(g(x)) = g(x) + 4 = (4x - 1) + 4 = 4x + 3.$$

Domain: \((-\infty, \infty).\)

Step 2 — Compute \((g \circ f)(x):\)

$$g(f(x)) = 4 f(x) - 1 = 4(x + 4) - 1 = 4x + 16 - 1 = 4x + 15.$$

Domain: \((-\infty, \infty).\)

Answer: (a) \((f \circ g)(x) = 4x + 3;\) (b) \((g \circ f)(x) = 4x + 15;\) each has domain \((-\infty, \infty).\)

Problem 44

Given \(f(x) = 2x + 4\) and \(g(x) = x^2 - 2.\) Both are polynomials with domain \((-\infty, \infty).\)

Step 1 — Compute \((f \circ g)(x):\)

$$f(g(x)) = 2 g(x) + 4 = 2(x^2 - 2) + 4 = 2x^2 - 4 + 4 = 2x^2.$$

Domain: \((-\infty, \infty).\)

Step 2 — Compute \((g \circ f)(x):\)

$$g(f(x)) = (f(x))^2 - 2 = (2x + 4)^2 - 2.$$

Expand \((2x + 4)^2 = 4x^2 + 16x + 16,\) so

$$g(f(x)) = 4x^2 + 16x + 16 - 2 = 4x^2 + 16x + 14.$$

Domain: \((-\infty, \infty).\)

Answer: (a) \((f \circ g)(x) = 2x^2;\) (b) \((g \circ f)(x) = 4x^2 + 16x + 14;\) each has domain \((-\infty, \infty).\)

Problem 45

Given \(f(x) = x^2 + 7\) and \(g(x) = x^2 - 3.\) Both are polynomials with domain \((-\infty, \infty).\)

Step 1 — Compute \((f \circ g)(x):\)

$$f(g(x)) = (g(x))^2 + 7 = (x^2 - 3)^2 + 7.$$

Expand \((x^2 - 3)^2 = x^4 - 6x^2 + 9,\) so

$$f(g(x)) = x^4 - 6x^2 + 9 + 7 = x^4 - 6x^2 + 16.$$

Domain: \((-\infty, \infty).\)

Step 2 — Compute \((g \circ f)(x):\)

$$g(f(x)) = (f(x))^2 - 3 = (x^2 + 7)^2 - 3.$$

Expand \((x^2 + 7)^2 = x^4 + 14x^2 + 49,\) so

$$g(f(x)) = x^4 + 14x^2 + 49 - 3 = x^4 + 14x^2 + 46.$$

Domain: \((-\infty, \infty).\)

Answer: (a) \((f \circ g)(x) = x^4 - 6x^2 + 16;\) (b) \((g \circ f)(x) = x^4 + 14x^2 + 46;\) each has domain \((-\infty, \infty).\)

Problem 46

Given \(f(x) = \sqrt{x}\) (domain \([0, \infty)\)) and \(g(x) = x + 9\) (domain \((-\infty, \infty)\)).

Step 1 — Compute \((f \circ g)(x):\)

$$f(g(x)) = \sqrt{g(x)} = \sqrt{x + 9}.$$

The inner output \(x + 9\) must lie in the domain of \(f,\) i.e. \(x + 9 \ge 0,\) so \(x \ge -9.\) Domain: \([-9, \infty).\)

Step 2 — Compute \((g \circ f)(x):\)

$$g(f(x)) = f(x) + 9 = \sqrt{x} + 9.$$

\(f(x) = \sqrt{x}\) requires \(x \ge 0.\) Domain: \([0, \infty).\)

Answer: (a) \((f \circ g)(x) = \sqrt{x + 9},\) domain \([-9, \infty);\) (b) \((g \circ f)(x) = \sqrt{x} + 9,\) domain \([0, \infty).\)

Problem 47

Given \(f(x) = \dfrac{3}{2x + 1}\) (domain \(x \neq -\tfrac{1}{2}\)) and \(g(x) = \dfrac{2}{x}\) (domain \(x \neq 0\)).

Step 1 — Compute \((f \circ g)(x):\)

$$f(g(x)) = \frac{3}{2 g(x) + 1} = \frac{3}{2 \cdot \dfrac{2}{x} + 1} = \frac{3}{\dfrac{4}{x} + 1}.$$

Multiply top and bottom by \(x\) (assuming \(x \neq 0\)):

$$= \frac{3x}{4 + x} = \frac{3x}{x + 4}.$$

Domain restrictions: \(x \neq 0\) (so \(g\) is defined) and the inner output cannot equal \(-\tfrac{1}{2}\) (so \(f\) is defined). Solve \(\dfrac{2}{x} = -\tfrac{1}{2}\) to get \(x = -4.\) Combined: \(x \neq 0\) and \(x \neq -4.\) Domain: \((-\infty, -4) \cup (-4, 0) \cup (0, \infty).\)

Step 2 — Compute \((g \circ f)(x):\)

$$g(f(x)) = \frac{2}{f(x)} = \frac{2}{\dfrac{3}{2x + 1}} = \frac{2(2x + 1)}{3} = \frac{4x + 2}{3}.$$

Restrictions: \(x \neq -\tfrac{1}{2}\) (so \(f\) is defined) and \(f(x) \neq 0\) (so \(g\) is defined). But \(f(x) = \dfrac{3}{2x+1}\) is never zero, so the only restriction is \(x \neq -\tfrac{1}{2}.\) Domain: \(\left(-\infty, -\tfrac{1}{2}\right) \cup \left(-\tfrac{1}{2}, \infty\right).\)

Answer: (a) \((f \circ g)(x) = \dfrac{3x}{x + 4},\) domain \((-\infty, -4) \cup (-4, 0) \cup (0, \infty);\) (b) \((g \circ f)(x) = \dfrac{4x + 2}{3},\) domain \(\left(-\infty, -\tfrac{1}{2}\right) \cup \left(-\tfrac{1}{2}, \infty\right).\)

Problem 48

Given \(f(x) = |x + 1|\) and \(g(x) = x^2 + x - 4.\) Both have domain \((-\infty, \infty).\)

Step 1 — Compute \((f \circ g)(x):\)

$$f(g(x)) = |g(x) + 1| = |x^2 + x - 4 + 1| = |x^2 + x - 3|.$$

Domain: \((-\infty, \infty).\)

Step 2 — Compute \((g \circ f)(x):\)

$$g(f(x)) = (f(x))^2 + f(x) - 4 = |x + 1|^2 + |x + 1| - 4.$$

Use \(|x + 1|^2 = (x + 1)^2 = x^2 + 2x + 1\) (squaring removes the absolute value), so

$$g(f(x)) = x^2 + 2x + 1 + |x + 1| - 4 = x^2 + 2x - 3 + |x + 1|.$$

Domain: \((-\infty, \infty).\)

Answer: (a) \((f \circ g)(x) = |x^2 + x - 3|;\) (b) \((g \circ f)(x) = x^2 + 2x - 3 + |x + 1|;\) each has domain \((-\infty, \infty).\)

Problem 49

The table pairs each year 2001–2012 with that season's NBA championship winner.

Step 1 — Domain = years, range = winners:

A relation is a function if every element of the domain corresponds to exactly one element of the range. Each year on the list has exactly one champion (a season produces one trophy), so every domain element has a unique image. Therefore this relation is a function.

Step 2 — Domain = winners, range = years:

Now check whether each winner is associated with exactly one year. The LA Lakers appear in 2001, 2002, 2009, and 2010 — four different years — so the input "LA Lakers" maps to four different outputs. The San Antonio Spurs (2003, 2005, 2007) and Miami Heat (2006, 2012) likewise map to multiple years. Therefore this relation is not a function.

Answer: (a) Yes, the year-to-winner relation is a function, because each year has exactly one champion. (b) No, the winner-to-year relation is not a function: several teams (LA Lakers, San Antonio Spurs, Miami Heat) won in more than one year, so a single input has multiple outputs.

Problem 50

The area of a square depends on the side length \(s.\)

Step 1 — Write \(A(s):\)

By the area-of-a-square formula,

$$A(s) = s^2.$$

The natural domain (geometric side lengths) is \(s > 0.\)

Step 2 — Evaluate \(A(6.5)\) and interpret:

$$A(6.5) = (6.5)^2 = 42.25.$$

Interpretation: a square with side \(6.5\) units has an area of \(42.25\) square units.

Step 3 — Solve \(s^2 = 56\) for the exact and approximate side length:

$$s = \sqrt{56} = \sqrt{4 \cdot 14} = 2\sqrt{14}.$$

Numerically, \(\sqrt{14} \approx 3.7417,\) so

$$s \approx 2 \cdot 3.7417 \approx 7.4833 \approx 7.5 \text{ (two significant digits)}.$$

Answer: (a) \(A(s) = s^2;\) (b) \(A(6.5) = 42.25\) square units — a square with side \(6.5\) has area \(42.25;\) (c) exact side length \(s = 2\sqrt{14}\) units, approximately \(7.5\) units to two significant digits.

Problem 51

The volume of a cube depends on its side length \(s.\)

Step 1 — Write \(V(s):\)

The standard cube-volume formula is

$$V(s) = s^3.$$

The natural geometric domain is \(s > 0.\)

Step 2 — Evaluate \(V(11.8)\):

$$V(11.8) = (11.8)^3.$$

Compute step by step:

$$(11.8)^2 = 139.24,$$

$$(11.8)^3 = 139.24 \times 11.8 = 1{,}643.032.$$

So \(V(11.8) = 1{,}643.032\) cubic units.

Step 3 — Interpret:

A cube with side \(11.8\) units has a volume of approximately \(1{,}643.03\) cubic units.

Answer: (a) \(V(s) = s^3;\) (b) \(V(11.8) = 1{,}643.032\) cubic units — the volume of a cube whose side is \(11.8\) units.

Problem 52

The car rental cost is a flat $20 plus an hourly rate of $10.25 per hour.

Step 1 — Write the cost function \(C(t):\)

If \(t\) is the number of hours rented, the cost is

$$C(t) = 20 + 10.25\,t.$$

Domain: \(t \ge 0.\)

Step 2 — Cost for 2 days and 7 hours:

Convert to hours: \(2 \text{ days} = 48 \text{ hours},\) so total \(t = 48 + 7 = 55\) hours.

$$C(55) = 20 + 10.25 \cdot 55 = 20 + 563.75 = 583.75.$$

So the rental cost is $583.75.

Step 3 — Solve \(C(t) = 430\) for \(t:\)

$$20 + 10.25\,t = 430,$$

$$10.25\,t = 410,$$

$$t = \frac{410}{10.25} = 40.$$

The car was rented for \(40\) hours, which is \(1\) day and \(16\) hours.

Answer: (a) \(C(t) = 20 + 10.25\,t;\) (b) $583.75 to rent the car for 2 days 7 hours (\(55\) hours); (c) the bill of $430 corresponds to \(t = 40\) hours of rental.

Problem 53

The vehicle has a 20-gal tank and gets 15 mpg, so for every gallon of gas it can travel 15 miles.

Step 1 — Write \(N(x):\)

If \(x\) is the gallons of gas in the tank,

$$N(x) = 15 x.$$

Step 2 — Miles on (i) a full tank and (ii) 3/4 of a tank:

(i) Full tank \(x = 20\):

$$N(20) = 15 \cdot 20 = 300 \text{ miles}.$$

(ii) Three-quarter tank \(x = \tfrac{3}{4} \cdot 20 = 15\):

$$N(15) = 15 \cdot 15 = 225 \text{ miles}.$$

Step 3 — Domain and range:

Gas in the tank ranges from empty to full: \(0 \le x \le 20.\) Miles driven on that gas: \(0 \le N \le 300.\) Domain: \([0, 20]\) gal. Range: \([0, 300]\) mi.

Step 4 — Stops for gas after \(578\) miles:

Each full tank covers \(300\) miles. After the first \(300\) miles the tank is empty, requiring a refill (stop 1). After the next \(300\) miles (i.e. at \(600\) miles total) another refill would be needed — but the trip ends at \(578\) miles, which is before that second empty point. So between \(0\) and \(578\) miles she ran out exactly once and made one stop for gas (after the first \(300\) miles). The car still has gas left at the end of the trip.

Answer: (a) \(N(x) = 15 x;\) (b) (i) \(300\) mi on a full tank; (ii) \(225\) mi on 3/4 of a tank; (c) domain \([0, 20]\) gal, range \([0, 300]\) mi; (d) she stopped for gas \(1\) time during the \(578\)-mi trip.

Problem 54

Earth is approximated as a sphere with mean radius \(r = 6.371 \times 10^6\) m and \(V = \tfrac{4}{3}\pi r^3.\)

Step 1 — Substitute \(r:\)

$$V = \frac{4}{3}\pi (6.371 \times 10^6)^3.$$

Step 2 — Cube the radius:

$$(6.371)^3 \approx 258.50, \quad (10^6)^3 = 10^{18},$$

so \((6.371 \times 10^6)^3 \approx 2.585 \times 10^{20}\) m\(^3.\) (More precisely \(2.58502 \times 10^{20}.\))

Step 3 — Multiply by \(\tfrac{4}{3}\pi:\)

$$\frac{4}{3}\pi \approx 4.18879,$$

$$V \approx 4.18879 \times 2.585 \times 10^{20} \approx 1.083 \times 10^{21} \text{ m}^3.$$

Step 4 — Sanity check:

Earth's accepted volume is roughly \(1.08 \times 10^{21}\) m\(^3,\) matching our result.

Answer: \(V \approx 1.083 \times 10^{21}\) cubic meters (using the mean radius \(6.371 \times 10^6\) m).

Problem 55

The bacterial colony fills a disk of radius \(r(t) = 6 - \dfrac{5}{t^2 + 1}\) cm at time \(t\) hours.

Step 1 — Area as a function of time:

A circle of radius \(r\) has area \(A = \pi r^2,\) so

$$A(t) = \pi \left(6 - \frac{5}{t^2 + 1}\right)^2.$$

Step 2 — Area at \(t = 3\) hours:

First compute the radius:

$$r(3) = 6 - \frac{5}{3^2 + 1} = 6 - \frac{5}{10} = 6 - \tfrac{1}{2} = \tfrac{11}{2} \text{ cm}.$$

Then

$$A(3) = \pi \left(\tfrac{11}{2}\right)^2 = \pi \cdot \tfrac{121}{4} = \tfrac{121\pi}{4} \text{ cm}^2.$$

Numerically, \(\tfrac{121\pi}{4} \approx \tfrac{121 \cdot 3.14159}{4} \approx \tfrac{380.13}{4} \approx 95.03\) cm\(^2.\)

Step 3 — Circumference as a function of time:

For a circle, \(C = 2\pi r,\) so

$$C(t) = 2\pi \left(6 - \frac{5}{t^2 + 1}\right).$$

Step 4 — Circumference at \(t = 3\) hours:

Using \(r(3) = \tfrac{11}{2}\) cm,

$$C(3) = 2\pi \cdot \tfrac{11}{2} = 11\pi \text{ cm} \approx 34.56 \text{ cm}.$$

Answer: (a) \(A(t) = \pi\left(6 - \dfrac{5}{t^2 + 1}\right)^2;\) (b) at \(t = 3\): exact \(A = \dfrac{121\pi}{4}\) cm\(^2,\) approximately \(95.03\) cm\(^2;\) (c) \(C(t) = 2\pi\left(6 - \dfrac{5}{t^2 + 1}\right);\) (d) at \(t = 3\): exact \(C = 11\pi\) cm, approximately \(34.56\) cm.

Problem 56

The two conversions are \(E(x) = 0.79\,x\) (US dollars \(\to\) Euros) and \(D(x) = 1.245\,x\) (Euros \(\to\) US dollars).

Step 1 — Compose to get USD \(\to\) Euros \(\to\) USD:

Apply \(E\) first (converting \(x\) dollars to Euros), then \(D\) (converting those Euros back to dollars):

$$(D \circ E)(x) = D(E(x)) = D(0.79\,x) = 1.245(0.79\,x).$$

Multiply: \(1.245 \cdot 0.79 = 0.98355.\) So

$$(D \circ E)(x) = 0.98355\,x.$$

Step 2 — Did the tourist lose value?

The composite coefficient is \(0.98355,\) which is less than \(1.\) Starting with \(\$x\) and ending with \(\$0.98355\,x\) means she has \(1.645\%\) less than she started with after the round trip. Yes, she lost value (because of the spread between the two conversion rates).

Step 3 — Dollars back from an extra $200:

$$(D \circ E)(200) = 0.98355 \cdot 200 = 196.71.$$

She gets back $196.71, losing $3.29 on the round-trip conversion of the extra $200.

Note on the graphing calculator: A graphing-calculator solution would plot \(y_1 = D(E(x))\) and \(y_2 = x\) and observe that \(y_1\) lies just below \(y_2,\) confirming the value loss; evaluating \(y_1\) at \(x = 200\) returns \(196.71.\)

Answer: (a) \((D \circ E)(x) = 0.98355\,x;\) yes, the tourist loses value because the round-trip multiplier is less than \(1;\) (b) the tourist receives $196.71 in U.S. dollars from her extra $200.

Problem 57

A worker earns a flat $750 plus $8.50 for each skateboard sold.

Step 1 — Write \(S(x):\)

If \(x\) is the number of skateboards sold, the monthly salary is

$$S(x) = 750 + 8.50\,x.$$

Natural domain: nonnegative integers (you cannot sell a fractional number of boards).

Step 2 — Evaluate at \(x = 25, 40, 55\):

$$S(25) = 750 + 8.50 \cdot 25 = 750 + 212.50 = 962.50,$$

$$S(40) = 750 + 8.50 \cdot 40 = 750 + 340 = 1{,}090,$$

$$S(55) = 750 + 8.50 \cdot 55 = 750 + 467.50 = 1{,}217.50.$$

So the monthly salaries are $962.50, $1,090, and $1,217.50, respectively.

Step 3 — Find \(x\) such that \(S(x) = 1{,}400\) (the INTERSECT step):

Solve algebraically (this is what the calculator's INTERSECT feature would return):

$$750 + 8.50\,x = 1{,}400,$$

$$8.50\,x = 650,$$

$$x = \frac{650}{8.50} \approx 76.47.$$

Since boards sell as whole units, the worker must sell at least \(77\) skateboards to reach $1,400 (selling \(76\) would give \(S(76) = 750 + 646 = \$1{,}396,\) just shy).

Note on the graphing calculator: Graphing \(y_1 = 750 + 8.50\,x\) and \(y_2 = 1400,\) then using INTERSECT, returns the point \((76.47, 1400);\) interpret this in context as "at least 77 boards."

Answer: (a) \(S(x) = 750 + 8.50\,x;\) (b) \(S(25) = \$962.50,\ S(40) = \$1{,}090,\ S(55) = \$1{,}217.50;\) (c) the intersection occurs at \(x \approx 76.47,\) so the worker must sell at least \(77\) skateboards to earn at least $1,400 per month.

Problem 58

Problem 1

Step 1 — Find the slope: Using points \((-2, 4)\) and \((1, 1):\) $$m = \frac{1 - 4}{1 - (-2)} = \frac{-3}{3} = -1.$$

Answer: a. Slope \(= -1.\) b. The line is decreasing (negative slope).

Problem 2

Step 1 — Find the slope: Using points \((-1, 4)\) and \((3, -1):\) $$m = \frac{-1 - 4}{3 - (-1)} = \frac{-5}{4}.$$

Answer: a. Slope \(= -5/4.\) b. The line is decreasing (negative slope).

Problem 3

Step 1 — Find the slope: Using points \((3, 5)\) and \((-1, 2):\) $$m = \frac{2 - 5}{-1 - 3} = \frac{-3}{-4} = \frac{3}{4}.$$

Answer: a. Slope \(= 3/4.\) b. The line is increasing (positive slope).

Problem 4

Step 1 — Find the slope: Using points \((6, 4)\) and \((4, -3):\) $$m = \frac{-3 - 4}{4 - 6} = \frac{-7}{-2} = \frac{7}{2}.$$

Answer: a. Slope \(= 7/2.\) b. The line is increasing (positive slope).

Problem 5

Step 1 — Find the slope: Using points \((2, 3)\) and \((5, 7):\) $$m = \frac{7 - 3}{5 - 2} = \frac{4}{3}.$$

Answer: a. Slope \(= 4/3.\) b. The line is increasing (positive slope).

Problem 6

Step 1 — Find the slope: Using points \((1, 9)\) and \((-8, 5):\) $$m = \frac{5 - 9}{-8 - 1} = \frac{-4}{-9} = \frac{4}{9}.$$

Answer: a. Slope \(= 4/9.\) b. The line is increasing (positive slope).

Problem 7

Step 1 — Find the slope: Using points \((2, 4)\) and \((1, 4):\) $$m = \frac{4 - 4}{1 - 2} = \frac{0}{-1} = 0.$$

Answer: a. Slope \(= 0.\) b. The line is horizontal (zero slope).

Problem 8

Step 1 — Find the slope: Using points \((1, 4)\) and \((1, 0):\) $$m = \frac{0 - 4}{1 - 1} = \frac{-4}{0},$$ which is undefined.

Answer: a. Slope is undefined. b. The line is vertical.

Problem 9

Step 1 — Use point-slope form: \(m = -6,\) through \((1, 3):\) $$y - 3 = -6(x - 1) \implies y = -6x + 6 + 3 = -6x + 9.$$

Answer: \(y = -6x + 9.\)

Problem 10

Step 1 — Use point-slope form: \(m = 3,\) through \((-3, 2):\) $$y - 2 = 3(x + 3) \implies y = 3x + 9 + 2 = 3x + 11.$$

Answer: \(y = 3x + 11.\)

Problem 11

Step 1 — Use slope-intercept: \(m = 1/3,\) \(y\)-intercept \((0, 4):\) $$y = \frac{1}{3}x + 4.$$

Answer: \(y = \frac{1}{3}x + 4.\)

Problem 12

Step 1 — Find the \(y\)-intercept. \(x\)-intercept is \((8, 0);\) use point-slope with \(m = 2/5:\) $$y - 0 = \frac{2}{5}(x - 8) \implies y = \frac{2}{5}x - \frac{16}{5}.$$

Answer: \(y = \frac{2}{5}x - \frac{16}{5}.\)

Problem 13

Step 1 — Find the slope: \(m = \frac{-1-1}{-2-2} = \frac{-2}{-4} = \frac{1}{2}.\)

Step 2 — Slope-intercept form using point \((2, 1):\) $$y - 1 = \frac{1}{2}(x - 2) \implies y = \frac{1}{2}x.$$

Answer: \(y = \frac{1}{2}x.\)

Problem 14

Step 1 — Find the slope: \(m = \frac{2 - 7}{1 - (-3)} = \frac{-5}{4}.\)

Step 2 — Slope-intercept form using point \((1, 2):\) $$y - 2 = -\frac{5}{4}(x - 1) \implies y = -\frac{5}{4}x + \frac{5}{4} + 2 = -\frac{5}{4}x + \frac{13}{4}.$$

Answer: \(y = -\frac{5}{4}x + \frac{13}{4}.\)

Problem 15

Step 1 — Identify intercepts. \(x\)-intercept \((5, 0),\) \(y\)-intercept \((0, -3).\)

Step 2 — Find slope: \(m = \frac{-3 - 0}{0 - 5} = \frac{-3}{-5} = \frac{3}{5}.\)

Step 3: Slope-intercept form using \(y\)-intercept: \(y = \frac{3}{5}x - 3.\)

Answer: \(y = \frac{3}{5}x - 3.\)

Problem 16

Step 1 — Identify intercepts. \(x\)-intercept \((-6, 0),\) \(y\)-intercept \((0, 9).\)

Step 2 — Find slope: \(m = \frac{9 - 0}{0 - (-6)} = \frac{9}{6} = \frac{3}{2}.\)

Step 3: \(y = \frac{3}{2}x + 9.\)

Answer: \(y = \frac{3}{2}x + 9.\)

Problem 17

Step 1: Solve for \(y\): \(y = 2x - 3.\)

Answer: a. Slope \(m = 2,\) \(y\)-intercept \(b = -3.\) b. Line rises left to right with slope 2, crossing the \(y\)-axis at \(-3.\)

Problem 18

Step 1: Already in slope-intercept form: \(y = -\frac{1}{7}x + 1.\)

Answer: a. Slope \(m = -1/7,\) \(y\)-intercept \(b = 1.\) b. Line falls gently with slope \(-1/7,\) crossing \(y\)-axis at \(1.\)

Problem 19

Answer: a. Slope \(m = -6,\) no \(y\)-intercept offset (\(b = 0\) — passes through origin). b. Steep downward line through the origin.

Problem 20

Answer: a. Slope \(m = -5,\) \(y\)-intercept \(b = 4.\) b. Steep downward line crossing \(y\)-axis at \(4.\)

Problem 21

Step 1: Solve for \(y\): \(4y = -24 \implies y = -6.\)

Answer: a. Slope \(m = 0,\) no \(x\)-coefficient; \(y\)-intercept is \(-6.\) b. Horizontal line at \(y = -6.\)

Problem 22

Step 1: The equation \(8x = 4\) gives \(x = 1/2.\)

Answer: a. Slope is undefined (vertical line). b. Vertical line at \(x = 1/2.\)

Problem 23

Step 1: Solve for \(y\): \(3y = -2x + 6 \implies y = -\frac{2}{3}x + 2.\)

Answer: a. Slope \(m = -2/3,\) \(y\)-intercept \(b = 2.\) b. Line with slope \(-2/3\) crossing at \((0, 2).\)

Problem 24

Step 1: Solve for \(y\): \(-5y = -6x - 15 \implies y = \frac{6}{5}x + 3.\)

Answer: a. Slope \(m = 6/5,\) \(y\)-intercept \(b = 3.\) b. Line with slope \(6/5\) crossing at \((0, 3).\)

Problem 25

Step 1 — Degree: Highest power is \(x^2;\) degree \(= 2.\)

Step 2 — Zeros: \(2x^2 - 3x - 5 = 0.\) Factor: \((2x - 5)(x + 1) = 0,\) giving \(x = 5/2\) and \(x = -1.\)

Step 3 — \(y\)-intercept: \(f(0) = -5.\) So \((0, -5).\)

Step 4 — End behavior: Leading coefficient \(2 > 0,\) even degree. \(f(x) \to \infty\) as \(x \to \pm\infty.\)

Step 5 — Even/odd: \(f(-x) = 2x^2 + 3x - 5 \ne f(x)\) and \(\ne -f(x).\) Neither even nor odd.

Problem 26

Step 1 — Degree: 2.

Step 2 — Zeros: \(-3x^2 + 6x = 0 \implies -3x(x - 2) = 0,\) so \(x = 0\) and \(x = 2.\)

Step 3 — \(y\)-intercept: \(f(0) = 0.\) So \((0, 0).\)

Step 4 — End behavior: Leading coefficient \(-3 < 0,\) even degree. \(f(x) \to -\infty\) as \(x \to \pm\infty.\)

Step 5 — Even/odd: \(f(-x) = -3x^2 - 6x.\) This equals \(-f(x) = 3x^2 - 6x\) only if \(-3x^2 - 6x = 3x^2 - 6x,\) which requires \(-6x^2 = 0\) — not always true. And \(f(-x) \ne f(x).\) Neither.

Problem 27

Step 1 — Degree: 2.

Step 2 — Zeros: \(\frac{1}{2}x^2 = 1 \implies x^2 = 2 \implies x = \pm\sqrt{2}.\)

Step 3 — \(y\)-intercept: \(f(0) = -1.\) So \((0, -1).\)

Step 4 — End behavior: Leading coefficient \(1/2 > 0,\) even degree. \(f(x) \to \infty\) as \(x \to \pm\infty.\)

Step 5 — Even/odd: \(f(-x) = \frac{1}{2}(-x)^2 - 1 = \frac{1}{2}x^2 - 1 = f(x).\) The function is even.

Problem 28

Step 1 — Degree: 3.

Step 2 — Zeros: Factor: \(x^3 + 3x^2 - x - 3 = x^2(x + 3) - (x + 3) = (x^2 - 1)(x + 3) = (x-1)(x+1)(x+3).\) Zeros: \(x = 1, -1, -3.\)

Step 3 — \(y\)-intercept: \(f(0) = -3.\) So \((0, -3).\)

Step 4 — End behavior: Leading coefficient \(1 > 0,\) odd degree. \(f(x) \to \infty\) as \(x \to \infty;\) \(f(x) \to -\infty\) as \(x \to -\infty.\)

Step 5 — Even/odd: \(f(-x) = -x^3 + 3x^2 + x - 3 \ne f(x)\) and \(\ne -f(x).\) Neither.

Problem 29

Step 1 — Degree: 3.

Step 2 — Zeros: \(3x - x^3 = x(3 - x^2) = 0.\) So \(x = 0\) and \(x^2 = 3 \implies x = \pm\sqrt{3}.\)

Step 3 — \(y\)-intercept: \(f(0) = 0.\)

Step 4 — End behavior: Leading coefficient \(-1 < 0,\) odd degree. \(f(x) \to -\infty\) as \(x \to \infty;\) \(f(x) \to \infty\) as \(x \to -\infty.\)

Step 5 — Even/odd: \(f(-x) = 3(-x) - (-x)^3 = -3x + x^3 = -(3x - x^3) = -f(x).\) The function is odd.

Problem 30

Step 1: Start with the graph of \(f(x) = x^2.\) Shift every point down 1 unit: \(g(x) = x^2 - 1.\)

Answer: The parabola \(y = x^2\) shifted down 1 unit; vertex at \((0, -1).\)

Problem 31

Step 1: Start with \(y = x^2.\) Replace \(x\) with \(x + 3\) → shift left 3: \((x+3)^2.\) Add 1 → shift up 1: \((x+3)^2 + 1.\)

Answer: The parabola shifted 3 units left and 1 unit up; vertex at \((-3, 1).\)

Problem 32

Step 1: Start with \(f(x) = \sqrt{x}.\) Replace \(x\) with \(x + 2\) → shift left 2: \(g(x) = \sqrt{x+2}.\)

Answer: The square-root curve shifted 2 units left; domain \(x \ge -2.\)

Problem 33

Step 1: Start with \(f(x) = \sqrt{x}.\) Multiply output by \(-1\) → reflect about \(x\)-axis: \(-\sqrt{x}.\) Subtract 1 → shift down 1: \(g(x) = -\sqrt{x} - 1.\)

Answer: The square-root curve reflected downward and shifted down 1; the graph runs below the \(x\)-axis for \(x \ge 0.\)

Problem 34

Step 1: Start with the graph of \(y = f(x).\) Add 1 to every output → shift up 1 unit: \(g(x) = f(x) + 1.\)

Answer: Every point on the original graph moves up 1 unit.

Problem 35

Step 1: Replace \(x\) with \(x - 1\) → shift right 1: \(f(x-1).\) Add 2 to output → shift up 2: \(g(x) = f(x-1) + 2.\)

Answer: Every point on the original graph moves right 1 and up 2.

Problem 36

a. Evaluate: - \(f(-3) = 4(-3) + 3 = -9.\) (Use first piece since \(-3 \le 0.\)) - \(f(0) = 4(0) + 3 = 3.\) (Use first piece since \(0 \le 0.\)) - \(f(2) = -(2) + 1 = -1.\) (Use second piece since \(2 > 0.\))

b. Sketch: For \(x \le 0:\) line \(y = 4x + 3\) including the point \((0, 3).\) For \(x > 0:\) line \(y = -x + 1\) starting with an open circle at \((0, 1).\)

Problem 37

a. Evaluate: - \(f(-4) = (-4)^2 - 3 = 13.\) (Use first piece since \(-4 < 0.\)) - \(f(0) = 4(0) - 3 = -3.\) (Use second piece since \(0 \ge 0.\)) - \(f(2) = 4(2) - 3 = 5.\) (Use second piece.)

b. Sketch: For \(x < 0:\) parabola \(y = x^2 - 3,\) open circle at \((0, -3).\) For \(x \ge 0:\) line \(y = 4x - 3,\) closed circle at \((0, -3).\) Both pieces meet at the same \(y\)-value at \(x = 0\) (\(-3\)), so the function is continuous there.

Problem 38

a. Evaluate: - \(h(0) = 0 + 1 = 1.\) (Use first piece, \(0 \le 5.\)) - \(h(\pi) = \pi + 1.\) (Use first piece, \(\pi \approx 3.14 \le 5.\)) - \(h(5) = 5 + 1 = 6.\) (Use first piece since \(5 \le 5.\))

b. Sketch: For \(x \le 5:\) line \(y = x + 1\) with closed circle at \((5, 6).\) For \(x > 5:\) horizontal line \(y = 4\) with open circle at \((5, 4).\) There is a jump discontinuity at \(x = 5.\)

Problem 39

a. Evaluate: - \(g(0) = 3/(0-2) = 3/(-2) = -3/2.\) (Use first piece, \(0 \ne 2.\)) - \(g(-4) = 3/(-4-2) = 3/(-6) = -1/2.\) (Use first piece, \(-4 \ne 2.\)) - \(g(2) = 4.\) (Use second piece, \(x = 2.\))

b. Sketch: For \(x \ne 2:\) rational function \(y = 3/(x-2)\) with a vertical asymptote at \(x = 2\) and an open hole there. At \(x = 2:\) isolated point \((2, 4).\)

Problem 40

False. \(f(x) = (4x+1)/(7x-2)\) is a rational function — a quotient of two polynomials. Rational functions are algebraic, not transcendental.

Problem 41

True. \(g(x) = x^{1/3}\) is a root function with \(n = 3\) (odd), and an odd function is one where \(g(-x) = -g(x).\) Here, \(g(-x) = (-x)^{1/3} = -(x^{1/3}) = -g(x).\)

Problem 42

False. A logarithmic function \(f(x) = \log_b(x)\) is transcendental. It cannot be expressed using only algebraic operations — it "transcends" algebra.

Problem 43

False. A function of the form \(f(x) = x^b\) where \(b\) is a real constant is a power function (or polynomial if \(b\) is a non-negative integer), not an exponential function. An exponential function has the variable in the exponent: \(f(x) = b^x.\)

Problem 44

False. The domain of an even root function \(f(x) = x^{1/n}\) (even \(n\)) is \([0, \infty),\) not all real numbers. You cannot take an even root of a negative number in the real-number system.

Problem 45

Step 1 — Find the slope. Two data points: \((0, 20500)\) at \(t = 0\) and \((3, 12300)\) at \(t = 3.\) $$m = \frac{12300 - 20500}{3 - 0} = \frac{-8200}{3} \approx -2733.33.$$

a. \(V(t) = -\frac{8200}{3}t + 20500.\)

b. \(y\)-intercept: \((0, 20500)\) — the initial purchase price of $20,500. \(x\)-intercept: solve \(V(t) = 0:\) \(t = 20500 \times 3/8200 = 61500/8200 = \frac{615}{82} \approx 7.5\) years — the time when the value reaches zero (fully depreciated).

c. \(V(5) = -\frac{8200}{3}(5) + 20500 = -\frac{41000}{3} + 20500 = \frac{-41000 + 61500}{3} = \frac{20500}{3} \approx \$6,833.33.\)

d. Solve \(V(t) = 3000:\) \(-\frac{8200}{3}t + 20500 = 3000,\) so \(\frac{8200}{3}t = 17500,\) giving \(t = \frac{17500 \times 3}{8200} = \frac{52500}{8200} \approx 6.4\) years.

Problem 46

Step 1 — Two data points. At \(t = 0:\) \(S = 42.3\) billion. At \(t = 1:\) \(S = 48.1\) billion.

a. Slope \(= 48.1 - 42.3 = 5.8\) billion per year. \(S(t) = 5.8t + 42.3\) (in billions of dollars).

b. The slope of \(5.8\) means online holiday sales were growing by approximately $5.8 billion per year.

c. Solve \(5.8t + 42.3 = 60:\) \(5.8t = 17.7,\) \(t \approx 3.05.\) Since \(t = 0\) corresponds to 2012, this is approximately \(2012 + 3 = 2015.\)

Problem 47

a. Fixed cost $125 plus $0.75 per cupcake: \(C(x) = 0.75x + 125.\)

b. \(C(160) = 0.75(160) + 125 = 120 + 125 = \$245.\)

c. Revenue per cupcake is $1.50. Profit when revenue \(>\) cost: \(1.50x > 0.75x + 125,\) so \(0.75x > 125,\) giving \(x > 500/3 \approx 166.7.\) She needs to sell at least 167 cupcakes to start making a profit.

Problem 48

Step 1 — Two data points. At \(t = 0:\) \(P = 250000.\) At \(t = 18:\) \(P = 500000.\)

a. Slope \(= \frac{500000 - 250000}{18} = \frac{250000}{18} \approx 13888.89.\) $$P(t) = \frac{250000}{18}t + 250000 \approx 13888.89t + 250000.$$

b. The slope (\(\approx \$13,889\) per year) represents the annual appreciation in the home's value.

c. \(P(15) = \frac{250000}{18}(15) + 250000 = \frac{3750000}{18} + 250000 \approx 208333 + 250000 = \$458,333.\)

Problem 49

a. Initial value $26,000, depreciating $1,500/year: \(V(t) = -1500t + 26000.\)

b. \(V(4) = -1500(4) + 26000 = -6000 + 26000 = \$20,000.\) After 4 years, the car is worth $20,000.

Problem 50

Step 1 — Two data points. At \(t = 0:\) value \(\$432,000.\) At \(t = 35:\) value \(\$60,500.\)

$$\text{Rate of depreciation} = \frac{60500 - 432000}{35} = \frac{-371500}{35} = -\$10,614.29 \text{ per year.}$$

Answer: The condominium depreciates at approximately $10,614 per year.

Problem 51

Step 1: \(C(175) = 10.50(175) + 28500 = 1837.50 + 28500 = \$30,337.50.\)

Answer: The total cost to produce 175 items is $30,337.50.

Problem 52

Step 1 — Two data points. \(t = 45\) min → \(N = 4\) pages; \(t = 90\) min → \(N = 9\) pages.

a. Slope \(= \frac{9 - 4}{90 - 45} = \frac{5}{45} = \frac{1}{9}\) pages per minute. Using point \((45, 4):\) $$N - 4 = \frac{1}{9}(t - 45) \implies N = \frac{1}{9}t - 5 + 4 = \frac{1}{9}t - 1.$$

b. 2 hours \(= 120\) min: \(N(120) = \frac{120}{9} - 1 \approx 13.33 - 1 = 12.33.\) About 12 pages.

c. Solve \(N = 20:\) \(\frac{t}{9} - 1 = 20,\) so \(t = 189\) minutes \(\approx\) 3 hours 9 minutes.

Problem 53

Step 1: The year 2015 corresponds to \(t = 15\) (since \(t = 0\) is the beginning of 2000).

$$P(15) = 1.8576(15) + 68.052 = 27.864 + 68.052 = 95.916 \approx 95.9\%.$$

Answer: Approximately 95.9% of total capacity.

Problem 54

a. 65% of offered students enroll: \(N(x) = 0.65x.\)

b. Solve \(N(x) = 1350:\) \(0.65x = 1350,\) so \(x = 1350/0.65 \approx 2076.9.\) The university should offer admission to approximately 2,077 students.

Problem 1

Step 1 — Apply the conversion factor: Multiply by \(\frac{\pi}{180°}\).

$$240° \cdot \frac{\pi}{180°} = \frac{240\pi}{180} = \frac{4\pi}{3}.$$

Answer: \(\dfrac{4\pi}{3}\) rad.

Problem 2

Step 1 — Apply the conversion factor:

$$15° \cdot \frac{\pi}{180°} = \frac{15\pi}{180} = \frac{\pi}{12}.$$

Answer: \(\dfrac{\pi}{12}\) rad.

Problem 3

Step 1 — Apply the conversion factor:

$$-60° \cdot \frac{\pi}{180°} = \frac{-60\pi}{180} = -\frac{\pi}{3}.$$

Answer: \(-\dfrac{\pi}{3}\) rad.

Problem 4

Step 1 — Apply the conversion factor:

$$-225° \cdot \frac{\pi}{180°} = \frac{-225\pi}{180} = -\frac{5\pi}{4}.$$

Answer: \(-\dfrac{5\pi}{4}\) rad.

Problem 5

Step 1 — Apply the conversion factor:

$$330° \cdot \frac{\pi}{180°} = \frac{330\pi}{180} = \frac{11\pi}{6}.$$

Answer: \(\dfrac{11\pi}{6}\) rad.

Problem 6

Step 1 — Apply the conversion factor: Multiply by \(\frac{180°}{\pi}\).

$$\frac{\pi}{2} \cdot \frac{180°}{\pi} = \frac{180°}{2} = 90°.$$

Answer: \(90°\).

Problem 7

Step 1 — Multiply by \(\frac{180°}{\pi}\):

$$\frac{7\pi}{6} \cdot \frac{180°}{\pi} = \frac{7 \cdot 180°}{6} = \frac{1260°}{6} = 210°.$$

Answer: \(210°\).

Problem 8

Step 1 — Multiply by \(\frac{180°}{\pi}\):

$$\frac{11\pi}{2} \cdot \frac{180°}{\pi} = \frac{11 \cdot 180°}{2} = \frac{1980°}{2} = 990°.$$

Answer: \(990°\).

Problem 9

Step 1 — Multiply by \(\frac{180°}{\pi}\):

$$-3\pi \cdot \frac{180°}{\pi} = -3 \cdot 180° = -540°.$$

Answer: \(-540°\).

Problem 10

Step 1 — Multiply by \(\frac{180°}{\pi}\):

$$\frac{5\pi}{12} \cdot \frac{180°}{\pi} = \frac{5 \cdot 180°}{12} = \frac{900°}{12} = 75°.$$

Answer: \(75°\).

Problem 11

Step 1 — Locate the angle. \(\frac{4\pi}{3}\) is in the third quadrant. Reference angle: \(\frac{4\pi}{3} - \pi = \frac{\pi}{3}\). In QIII, cosine is negative.

Step 2 — Apply the reference angle: \(\cos\!\left(\frac{\pi}{3}\right) = \frac{1}{2}\), so \(\cos\!\left(\frac{4\pi}{3}\right) = -\frac{1}{2}\).

Answer: \(-\dfrac{1}{2}\).

Problem 12

Step 1 — Reduce the angle. \(\frac{19\pi}{4} = 4\pi + \frac{3\pi}{4}\), so the angle is coterminal with \(\frac{3\pi}{4}\) (QII). Reference angle: \(\pi - \frac{3\pi}{4} = \frac{\pi}{4}\). In QII, tangent is negative.

Step 2 — Compute: \(\tan\!\left(\frac{\pi}{4}\right) = 1\), so \(\tan\!\left(\frac{19\pi}{4}\right) = -1\).

Answer: \(-1\).

Problem 13

Step 1 — Locate the angle. \(-\frac{3\pi}{4}\) is in QIII (negative rotation). Reference angle: \(\frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}\)... more directly, the reference angle is \(\frac{\pi}{4}\). In QIII, sine is negative.

Step 2 — Compute: \(\sin\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), so \(\sin\!\left(-\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}\).

Answer: \(-\dfrac{\sqrt{2}}{2}\).

Problem 14

Step 1 — Locate the angle. \(\frac{\pi}{6}\) is in QI. \(\cos\!\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\).

Step 2 — Apply the reciprocal identity: \(\sec\theta = \frac{1}{\cos\theta}\).

$$\sec\!\left(\frac{\pi}{6}\right) = \frac{1}{\cos(\pi/6)} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}.$$

Answer: \(\dfrac{2\sqrt{3}}{3}\).

Problem 15

Step 1 — Use the half-angle or difference formula. Write \(\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}\).

$$\sin\!\left(\frac{\pi}{12}\right) = \sin\!\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \sin\frac{\pi}{3}\cos\frac{\pi}{4} - \cos\frac{\pi}{3}\sin\frac{\pi}{4}.$$

Step 2 — Substitute known values:

$$= \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} - \frac{1}{2}\cdot\frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6}-\sqrt{2}}{4}.$$

Answer: \(\dfrac{\sqrt{6}-\sqrt{2}}{4}\).

Problem 16

Step 1 — Write \(\frac{5\pi}{12} = \frac{\pi}{3} + \frac{\pi}{12}\). More conveniently, use \(\frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}\).

$$\cos\!\left(\frac{5\pi}{12}\right) = \cos\!\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \cos\frac{\pi}{4}\cos\frac{\pi}{6} - \sin\frac{\pi}{4}\sin\frac{\pi}{6}.$$

Step 2 — Substitute:

$$= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6}-\sqrt{2}}{4}.$$

Answer: \(\dfrac{\sqrt{6}-\sqrt{2}}{4}\).

Problem 17

Step 1 — Find the missing side. Given \(a = 4\) (side opposite \(A\)) and \(c = 7\) (hypotenuse). Use the Pythagorean theorem:

$$b = \sqrt{c^2 - a^2} = \sqrt{49 - 16} = \sqrt{33}.$$

Step 2 — Find the six trig values for angle \(A\):

- \(\sin A = \dfrac{a}{c} = \dfrac{4}{7}\) - \(\cos A = \dfrac{b}{c} = \dfrac{\sqrt{33}}{7}\) - \(\tan A = \dfrac{a}{b} = \dfrac{4}{\sqrt{33}} = \dfrac{4\sqrt{33}}{33}\) - \(\csc A = \dfrac{7}{4}\) - \(\sec A = \dfrac{7}{\sqrt{33}} = \dfrac{7\sqrt{33}}{33}\) - \(\cot A = \dfrac{\sqrt{33}}{4}\)

Answer: \(b = \sqrt{33}\approx 5.745\); trig values as above.

Problem 18

Step 1 — Find the missing side. Given \(a = 21\), \(c = 29\) (hypotenuse).

$$b = \sqrt{29^2 - 21^2} = \sqrt{841 - 441} = \sqrt{400} = 20.$$

Step 2 — Six trig values for angle \(A\):

- \(\sin A = \dfrac{21}{29}\) - \(\cos A = \dfrac{20}{29}\) - \(\tan A = \dfrac{21}{20}\) - \(\csc A = \dfrac{29}{21}\) - \(\sec A = \dfrac{29}{20}\) - \(\cot A = \dfrac{20}{21}\)

Answer: \(b = 20\); trig values as above.

Problem 19

Step 1 — Find the missing side. Given \(a = 85.3\) (opposite \(A\)) and \(b = 125.5\) (adjacent). The hypotenuse:

$$c = \sqrt{a^2 + b^2} = \sqrt{85.3^2 + 125.5^2} = \sqrt{7276.09 + 15750.25} = \sqrt{23026.34} \approx 151.745.$$

Step 2 — Six trig values (rounded to three decimal places):

- \(\sin A = \dfrac{a}{c} \approx \dfrac{85.3}{151.745} \approx 0.562\) - \(\cos A = \dfrac{b}{c} \approx \dfrac{125.5}{151.745} \approx 0.827\) - \(\tan A = \dfrac{a}{b} = \dfrac{85.3}{125.5} \approx 0.680\) - \(\csc A \approx 1.780\) - \(\sec A \approx 1.209\) - \(\cot A \approx 1.471\)

Answer: \(c \approx 151.745\); trig values as above.

Problem 20

Step 1 — Find the missing side. Given \(b = 40\) (adjacent) and \(c = 41\) (hypotenuse).

$$a = \sqrt{41^2 - 40^2} = \sqrt{1681 - 1600} = \sqrt{81} = 9.$$

Step 2 — Six trig values for angle \(A\):

- \(\sin A = \dfrac{9}{41}\) - \(\cos A = \dfrac{40}{41}\) - \(\tan A = \dfrac{9}{40}\) - \(\csc A = \dfrac{41}{9}\) - \(\sec A = \dfrac{41}{40}\) - \(\cot A = \dfrac{40}{9}\)

Answer: \(a = 9\); trig values as above.

Problem 21

Step 1 — Find the missing side. Given \(a = 84\) (opposite) and \(b = 13\) (adjacent).

$$c = \sqrt{84^2 + 13^2} = \sqrt{7056 + 169} = \sqrt{7225} = 85.$$

Step 2 — Six trig values for angle \(A\):

- \(\sin A = \dfrac{84}{85}\) - \(\cos A = \dfrac{13}{85}\) - \(\tan A = \dfrac{84}{13}\) - \(\csc A = \dfrac{85}{84}\) - \(\sec A = \dfrac{85}{13}\) - \(\cot A = \dfrac{13}{84}\)

Answer: \(c = 85\); trig values as above.

Problem 22

Step 1 — Find the missing side. Given \(b = 28\) (adjacent) and \(c = 35\) (hypotenuse).

$$a = \sqrt{35^2 - 28^2} = \sqrt{1225 - 784} = \sqrt{441} = 21.$$

Step 2 — Six trig values for angle \(A\):

- \(\sin A = \dfrac{21}{35} = \dfrac{3}{5}\) - \(\cos A = \dfrac{28}{35} = \dfrac{4}{5}\) - \(\tan A = \dfrac{21}{28} = \dfrac{3}{4}\) - \(\csc A = \dfrac{5}{3}\) - \(\sec A = \dfrac{5}{4}\) - \(\cot A = \dfrac{4}{3}\)

Answer: \(a = 21\); trig values as above.

Problem 23

Step 1 — Find \(y\). Using \(x^2 + y^2 = 1\) (unit circle):

$$y = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25} \quad (y > 0).$$

Step 2 — Six trig values with \(x = \frac{7}{25}\), \(y = \frac{24}{25}\):

- \(\sin\theta = \dfrac{24}{25}\), \(\cos\theta = \dfrac{7}{25}\), \(\tan\theta = \dfrac{24}{7}\) - \(\csc\theta = \dfrac{25}{24}\), \(\sec\theta = \dfrac{25}{7}\), \(\cot\theta = \dfrac{7}{24}\)

Answer: \(y = \dfrac{24}{25}\); trig values as above.

Problem 24

Step 1 — Find \(y\). With \(x = -\frac{15}{17}\):

$$y = -\sqrt{1 - \frac{225}{289}} = -\sqrt{\frac{64}{289}} = -\frac{8}{17} \quad (y < 0).$$

Step 2 — Six trig values (QIII):

- \(\sin\theta = -\dfrac{8}{17}\), \(\cos\theta = -\dfrac{15}{17}\), \(\tan\theta = \dfrac{8}{15}\) - \(\csc\theta = -\dfrac{17}{8}\), \(\sec\theta = -\dfrac{17}{15}\), \(\cot\theta = \dfrac{15}{8}\)

Answer: \(y = -\dfrac{8}{17}\); trig values as above.

Problem 25

Step 1 — Find \(x\). With \(y = \frac{\sqrt{7}}{3}\):

$$x = -\sqrt{1 - \frac{7}{9}} = -\sqrt{\frac{2}{9}} = -\frac{\sqrt{2}}{3} \quad (x < 0).$$

Step 2 — Six trig values (QII):

- \(\sin\theta = \dfrac{\sqrt{7}}{3}\), \(\cos\theta = -\dfrac{\sqrt{2}}{3}\), \(\tan\theta = -\dfrac{\sqrt{7}}{\sqrt{2}} = -\dfrac{\sqrt{14}}{2}\) - \(\csc\theta = \dfrac{3}{\sqrt{7}} = \dfrac{3\sqrt{7}}{7}\), \(\sec\theta = -\dfrac{3}{\sqrt{2}} = -\dfrac{3\sqrt{2}}{2}\), \(\cot\theta = -\dfrac{\sqrt{2}}{\sqrt{7}} = -\dfrac{\sqrt{14}}{7}\)

Answer: \(x = -\dfrac{\sqrt{2}}{3}\); trig values as above.

Problem 26

Step 1 — Find \(x\). With \(y = -\frac{\sqrt{15}}{4}\):

$$x = \sqrt{1 - \frac{15}{16}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \quad (x > 0).$$

Step 2 — Six trig values (QIV):

- \(\sin\theta = -\dfrac{\sqrt{15}}{4}\), \(\cos\theta = \dfrac{1}{4}\), \(\tan\theta = -\sqrt{15}\) - \(\csc\theta = -\dfrac{4}{\sqrt{15}} = -\dfrac{4\sqrt{15}}{15}\), \(\sec\theta = 4\), \(\cot\theta = -\dfrac{1}{\sqrt{15}} = -\dfrac{\sqrt{15}}{15}\)

Answer: \(x = \dfrac{1}{4}\); trig values as above.

Problem 27

Step 1 — Write in terms of sin and cos:

$$\tan^2 x + \sin x\csc x = \frac{\sin^2 x}{\cos^2 x} + \sin x \cdot \frac{1}{\sin x} = \frac{\sin^2 x}{\cos^2 x} + 1.$$

Step 2 — Combine using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\):

$$= \frac{\sin^2 x + \cos^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x.$$

Answer: \(\sec^2 x\).

Problem 28

Step 1 — Expand:

$$\sec x\sin x\cot x = \frac{1}{\cos x} \cdot \sin x \cdot \frac{\cos x}{\sin x}.$$

Step 2 — Cancel:

$$= \frac{\sin x \cos x}{\cos x \sin x} = 1.$$

Answer: \(1\).

Problem 29

Step 1 — Write in terms of sin and cos:

$$\frac{\tan^2 x}{\sec^2 x} = \frac{\sin^2 x/\cos^2 x}{1/\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} \cdot \cos^2 x = \sin^2 x.$$

Answer: \(\sin^2 x\).

Problem 30

Step 1 — Write in terms of cos:

$$\sec x - \cos x = \frac{1}{\cos x} - \cos x = \frac{1 - \cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x}.$$

Step 2 — Simplify:

$$= \sin x \cdot \frac{\sin x}{\cos x} = \sin x\tan x.$$

Answer: \(\sin x\tan x\).

Problem 31

Step 1 — Expand the square:

$$(1 + \tan\theta)^2 - 2\tan\theta = 1 + 2\tan\theta + \tan^2\theta - 2\tan\theta = 1 + \tan^2\theta.$$

Step 2 — Apply the Pythagorean identity:

$$= \sec^2\theta.$$

Answer: \(\sec^2\theta\).

Problem 32

Step 1 — Expand:

$$\sin x(\csc x - \sin x) = \sin x\csc x - \sin^2 x = 1 - \sin^2 x.$$

Step 2 — Apply the Pythagorean identity:

$$= \cos^2 x.$$

Answer: \(\cos^2 x\).

Problem 33

Step 1 — Combine fractions using the common denominator \(\sin t(1 + \cos t)\):

$$\frac{\cos t}{\sin t} + \frac{\sin t}{1 + \cos t} = \frac{\cos t(1 + \cos t) + \sin^2 t}{\sin t(1 + \cos t)}.$$

Step 2 — Expand the numerator:

$$= \frac{\cos t + \cos^2 t + \sin^2 t}{\sin t(1 + \cos t)} = \frac{\cos t + 1}{\sin t(1 + \cos t)} = \frac{1}{\sin t} = \csc t.$$

Answer: \(\csc t\).

Problem 34

Step 1 — Write using Pythagorean identities:

$$\frac{1 + \tan^2\alpha}{1 + \cot^2\alpha} = \frac{\sec^2\alpha}{\csc^2\alpha}.$$

Step 2 — Convert to sin/cos:

$$= \frac{1/\cos^2\alpha}{1/\sin^2\alpha} = \frac{\sin^2\alpha}{\cos^2\alpha} = \tan^2\alpha.$$

Answer: \(\tan^2\alpha\).

Problem 35

Step 1 — Simplify the left side:

$$\frac{\tan\theta\cot\theta}{\csc\theta} = \frac{(\tan\theta)(1/\tan\theta)}{1/\sin\theta} = \frac{1}{1/\sin\theta} = \sin\theta.$$

Answer: Identity verified: both sides equal \(\sin\theta\).

Problem 36

Step 1 — Simplify:

$$\frac{\sec^2\theta}{\tan\theta} = \frac{1/\cos^2\theta}{\sin\theta/\cos\theta} = \frac{1}{\cos^2\theta}\cdot\frac{\cos\theta}{\sin\theta} = \frac{1}{\cos\theta\sin\theta} = \sec\theta\csc\theta.$$

Answer: Identity verified.

Problem 37

Step 1 — Simplify each term:

$$\frac{\sin t}{\csc t} + \frac{\cos t}{\sec t} = \frac{\sin t}{1/\sin t} + \frac{\cos t}{1/\cos t} = \sin^2 t + \cos^2 t.$$

Step 2 — Apply the Pythagorean identity:

$$= 1.$$

Answer: Identity verified.

Problem 38

Step 1 — Combine fractions with common denominator \(\sin x(\cos x + 1)\):

$$\frac{\sin x}{\cos x+1} + \frac{\cos x-1}{\sin x} = \frac{\sin^2 x + (\cos x-1)(\cos x+1)}{\sin x(\cos x+1)}.$$

Step 2 — Expand the numerator:

$$\sin^2 x + \cos^2 x - 1 = 1 - 1 = 0.$$

Answer: The left side equals \(0\). Identity verified.

Problem 39

Step 1 — Write in terms of sin and cos:

$$\cot\gamma + \tan\gamma = \frac{\cos\gamma}{\sin\gamma} + \frac{\sin\gamma}{\cos\gamma} = \frac{\cos^2\gamma + \sin^2\gamma}{\sin\gamma\cos\gamma} = \frac{1}{\sin\gamma\cos\gamma}.$$

Step 2 — Recognize the right side:

$$\sec\gamma\csc\gamma = \frac{1}{\cos\gamma}\cdot\frac{1}{\sin\gamma} = \frac{1}{\sin\gamma\cos\gamma}.$$

Answer: Both sides equal \(\dfrac{1}{\sin\gamma\cos\gamma}\). Identity verified.

Problem 40

Step 1 — Group using the Pythagorean identity:

$$\sin^2\beta + \tan^2\beta + \cos^2\beta = (\sin^2\beta + \cos^2\beta) + \tan^2\beta = 1 + \tan^2\beta.$$

Step 2 — Apply the second Pythagorean identity:

$$= \sec^2\beta.$$

Answer: Identity verified.

Problem 41

Step 1 — Combine fractions:

$$\frac{1}{1-\sin\alpha} + \frac{1}{1+\sin\alpha} = \frac{(1+\sin\alpha)+(1-\sin\alpha)}{(1-\sin\alpha)(1+\sin\alpha)} = \frac{2}{1-\sin^2\alpha}.$$

Step 2 — Apply the Pythagorean identity \(1 - \sin^2\alpha = \cos^2\alpha\):

$$= \frac{2}{\cos^2\alpha} = 2\sec^2\alpha.$$

Answer: Identity verified.

Problem 42

Step 1 — Simplify the left side:

$$\frac{\tan\theta - \cot\theta}{\sin\theta\cos\theta} = \frac{\frac{\sin\theta}{\cos\theta} - \frac{\cos\theta}{\sin\theta}}{\sin\theta\cos\theta} = \frac{\frac{\sin^2\theta - \cos^2\theta}{\sin\theta\cos\theta}}{\sin\theta\cos\theta} = \frac{\sin^2\theta - \cos^2\theta}{\sin^2\theta\cos^2\theta}.$$

Step 2 — Split the fraction:

$$= \frac{\sin^2\theta}{\sin^2\theta\cos^2\theta} - \frac{\cos^2\theta}{\sin^2\theta\cos^2\theta} = \frac{1}{\cos^2\theta} - \frac{1}{\sin^2\theta} = \sec^2\theta - \csc^2\theta.$$

Answer: Identity verified.

Problem 43

Step 1 — Solve: \(2\sin\theta = 1 \Rightarrow \sin\theta = \frac{1}{2}\).

Step 2 — Find all solutions on \([0, 2\pi)\). \(\sin\theta = \frac{1}{2}\) at \(\theta = \frac{\pi}{6}\) (QI) and \(\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\) (QII).

Answer: \(\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}\).

Problem 44

Step 1 — Solve: \(\cos\theta = \frac{1}{2} - 1 = -\frac{1}{2}\).

Step 2 — Find solutions. \(\cos\theta = -\frac{1}{2}\) at \(\theta = \frac{2\pi}{3}\) (QII) and \(\theta = \frac{4\pi}{3}\) (QIII).

Answer: \(\theta = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}\).

Problem 45

Step 1 — Solve: \(\tan^2\theta = 1 \Rightarrow \tan\theta = \pm 1\).

Step 2 — Find solutions. \(\tan\theta = 1\) at \(\theta = \frac{\pi}{4}, \frac{5\pi}{4}\); \(\tan\theta = -1\) at \(\theta = \frac{3\pi}{4}, \frac{7\pi}{4}\).

Answer: \(\theta = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}\).

Problem 46

Step 1 — Solve: \(\sin^2\theta = \frac{1}{2} \Rightarrow \sin\theta = \pm\frac{\sqrt{2}}{2}\).

Step 2 — Find solutions. \(\sin\theta = \frac{\sqrt{2}}{2}\) at \(\frac{\pi}{4}, \frac{3\pi}{4}\); \(\sin\theta = -\frac{\sqrt{2}}{2}\) at \(\frac{5\pi}{4}, \frac{7\pi}{4}\).

Answer: \(\theta = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}\).

Problem 47

Step 1 — Solve: \(\cot\theta = -\frac{1}{\sqrt{3}}\), which means \(\tan\theta = -\sqrt{3}\).

Step 2 — Find solutions. \(\tan\theta = -\sqrt{3}\) at \(\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\) (QII) and \(\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\) (QIV).

Answer: \(\theta = \dfrac{2\pi}{3}, \dfrac{5\pi}{3}\).

Problem 48

Step 1 — Solve: \(\sec\theta = \frac{2\sqrt{3}}{3}\), so \(\cos\theta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}\).

Step 2 — Find solutions. \(\cos\theta = \frac{\sqrt{3}}{2}\) at \(\theta = \frac{\pi}{6}\) (QI) and \(\theta = \frac{11\pi}{6}\) (QIV).

Answer: \(\theta = \dfrac{\pi}{6}, \dfrac{11\pi}{6}\).

Problem 49

Step 1 — Rearrange and factor:

$$2\cos\theta\sin\theta - \sin\theta = 0 \Rightarrow \sin\theta(2\cos\theta - 1) = 0.$$

Step 2 — Solve each factor.

- \(\sin\theta = 0\): \(\theta = 0, \pi\). - \(\cos\theta = \frac{1}{2}\): \(\theta = \frac{\pi}{3}, \frac{5\pi}{3}\).

Answer: \(\theta = 0, \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{3}\).

Problem 50

Step 1 — Factor: Let \(u = \csc\theta\):

$$u^2 + 2u + 1 = (u+1)^2 = 0 \Rightarrow \csc\theta = -1.$$

Step 2 — Solve: \(\csc\theta = -1 \Rightarrow \sin\theta = -1 \Rightarrow \theta = \frac{3\pi}{2}\).

Answer: \(\theta = \dfrac{3\pi}{2}\).

Problem 51

Step 1 — Read off the parameters. The graph starts at \(y = 0\) at \(x = 0\), so it's a sine function. Identify amplitude \(|A|\) from the maximum/minimum values and period from the distance between peaks.

Note: The exact values of \(A\) and \(B\) for problems 1.3.51–1.3.54 depend on the specific graphs in the figures. The approach below applies to each.

Step 2 — General approach:

- Amplitude \(|A|\): Half the total vertical range (max minus min). - Period \(T\): Horizontal distance for one full cycle; then \(B = \frac{2\pi}{T}\). - Choose sine or cosine based on the function's value at \(x = 0\).

For a graph with amplitude 2 and period \(\pi\) starting at \(y = 0\): \(y = 2\sin(2x)\).

Answer: See the figure; apply the method above to read off the specific \(A\) and \(B\) values.

Problem 52

Step 1 — Identify the parameters from the graph. Determine whether the function starts at its maximum (cosine) or at zero (sine) at \(x = 0\).

Step 2 — Apply the amplitude–period formula. Amplitude \(= \frac{\text{max} - \text{min}}{2}\); period \(T\) from the graph gives \(B = \frac{2\pi}{T}\).

Answer: Read \(A\) and \(B\) from Figure 1.3.52 using the method above.

Problem 53

Step 1 — Identify the parameters from the graph. Determine whether the function starts at its maximum (cosine) or at zero (sine) at \(x = 0\).

Step 2 — Apply the amplitude–period formula. Amplitude \(= \frac{\text{max} - \text{min}}{2}\); period \(T\) from the graph gives \(B = \frac{2\pi}{T}\).

Answer: Read \(A\) and \(B\) from Figure 1.3.53 using the method above.

Problem 54

Step 1 — Identify the parameters from the graph. Determine whether the function starts at its maximum (cosine) or at zero (sine) at \(x = 0\).

Step 2 — Apply the amplitude–period formula. Amplitude \(= \frac{\text{max} - \text{min}}{2}\); period \(T\) from the graph gives \(B = \frac{2\pi}{T}\).

Answer: Read \(A\) and \(B\) from Figure 1.3.54 using the method above.

Problem 55

Step 1 — Identify parameters from \(y = \sin\!\left(x - \frac{\pi}{4}\right)\).

- \(A = 1\), \(B = 1\), phase shift \(\alpha = \frac{\pi}{4}\).

a) Amplitude: \(|A| = 1\).

b) Period: \(\dfrac{2\pi}{B} = \dfrac{2\pi}{1} = 2\pi\).

c) Phase shift: \(\frac{\pi}{4}\) to the right.

Answer: Amplitude = 1, Period = \(2\pi\), Phase shift = \(\dfrac{\pi}{4}\) right.

Problem 56

Step 1 — Rewrite in standard form. \(y = 3\cos(2x + 3) = 3\cos\!\left(2\!\left(x + \frac{3}{2}\right)\right)\).

a) Amplitude: \(|A| = 3\).

b) Period: \(\dfrac{2\pi}{2} = \pi\).

c) Phase shift: \(\frac{3}{2}\) units to the left (since \(\alpha = -\frac{3}{2}\)).

Answer: Amplitude = 3, Period = \(\pi\), Phase shift = \(\dfrac{3}{2}\) left.

Problem 57

Step 1 — Identify parameters from \(y = -\frac{1}{2}\sin\!\left(\frac{1}{4}x\right)\).

a) Amplitude: \(|A| = \frac{1}{2}\).

b) Period: \(\dfrac{2\pi}{1/4} = 8\pi\).

c) Phase shift: None (\(\alpha = 0\)).

Answer: Amplitude = \(\dfrac{1}{2}\), Period = \(8\pi\), No phase shift.

Problem 58

Step 1 — Identify parameters from \(y = 2\cos\!\left(x - \frac{\pi}{3}\right)\).

a) Amplitude: \(2\).

b) Period: \(2\pi\).

c) Phase shift: \(\dfrac{\pi}{3}\) to the right.

Answer: Amplitude = 2, Period = \(2\pi\), Phase shift = \(\dfrac{\pi}{3}\) right.

Problem 59

Step 1 — Rewrite in standard form. \(y = -3\sin(\pi x + 2) = -3\sin\!\left(\pi\!\left(x + \frac{2}{\pi}\right)\right)\).

a) Amplitude: \(|-3| = 3\).

b) Period: \(\dfrac{2\pi}{\pi} = 2\).

c) Phase shift: \(\dfrac{2}{\pi}\) to the left.

Answer: Amplitude = 3, Period = 2, Phase shift = \(\dfrac{2}{\pi}\) left.

Problem 60

Step 1 — Rewrite. \(y = 4\cos\!\left(2x - \frac{\pi}{2}\right) = 4\cos\!\left(2\!\left(x - \frac{\pi}{4}\right)\right)\).

a) Amplitude: \(4\).

b) Period: \(\dfrac{2\pi}{2} = \pi\).

c) Phase shift: \(\dfrac{\pi}{4}\) to the right.

Answer: Amplitude = 4, Period = \(\pi\), Phase shift = \(\dfrac{\pi}{4}\) right.

Problem 61

Step 1 — Convert the angle. \(120° = \frac{2\pi}{3}\) rad.

Step 2 — Find the arc length. The wheel has diameter 40 in, so radius \(r = 20\) in. Arc length \(s = r\theta\):

$$s = 20 \cdot \frac{2\pi}{3} = \frac{40\pi}{3} \approx 41.888 \approx 42 \text{ in.}$$

Note on the calculator: compute \(40\pi/3\) and round to the nearest whole number.

Answer: The wheel moves approximately 42 inches.

Problem 62

Step 1 — Apply the arc length formula \(s = r\theta\). Convert degrees to radians where needed (\(\theta_{\text{rad}} = \theta_{\text{deg}} \cdot \pi/180\)).

a) \(s = 12.8 \cdot \frac{5\pi}{6} = \frac{64\pi}{6} \approx \frac{201.06}{6} \approx 33.51\) cm.

b) \(s = 4.378 \cdot \frac{7\pi}{6} \approx 4.378 \cdot 3.6652 \approx 16.05\) cm.

c) \(\theta = 50° \cdot \frac{\pi}{180} \approx 0.8727\) rad; \(s = 0.964 \cdot 0.8727 \approx 0.84\) cm.

d) \(\theta = 325° \cdot \frac{\pi}{180} \approx 5.6723\) rad; \(s = 8.55 \cdot 5.6723 \approx 48.50\) cm.

Answer: a) \(\approx 33.51\) cm, b) \(\approx 16.05\) cm, c) \(\approx 0.84\) cm, d) \(\approx 48.50\) cm.

Problem 63

Step 1 — Apply \(\omega = \theta/t\). Convert time units to match \(\theta\) (radians).

a) \(\omega = \frac{7\pi/4}{10} = \frac{7\pi}{40} \approx 0.550\) rad/sec.

b) \(\omega = \frac{3\pi/5}{8} = \frac{3\pi}{40} \approx 0.236\) rad/sec.

c) \(t = 1\) min = 60 sec; \(\omega = \frac{2\pi/9}{60} = \frac{2\pi}{540} = \frac{\pi}{270} \approx 0.012\) rad/sec.

d) \(t = 14\) min = 840 sec; \(\omega = \frac{23.76}{840} \approx 0.028\) rad/sec.

Answer: a) \(\approx 0.550\) rad/sec, b) \(\approx 0.236\) rad/sec, c) \(\approx 0.012\) rad/sec, d) \(\approx 0.028\) rad/sec.

Problem 64

Step 1 — Part a: Find the radius. Area of a full circle: \(\pi r^2 = 250{,}000\) m².

$$r = \sqrt{\frac{250000}{\pi}} \approx \sqrt{79577.5} \approx 282.1 \text{ m}.$$

Step 2 — Part b: Sector arc length. A \(45°\) sector has central angle \(\theta = \frac{\pi}{4}\) rad. Arc length:

$$s = r\theta \approx 282.1 \cdot \frac{\pi}{4} \approx 282.1 \cdot 0.7854 \approx 221.6 \text{ m}.$$

Answer: a) \(r \approx 282.1\) m; b) Arc length \(\approx 221.6\) m.

Problem 65

Step 1 — Apply the area formula. \(A = \frac{1}{2}x^2\sin\theta\) with \(x = 8\) in and \(\theta = \frac{5\pi}{12}\) rad.

$$A = \frac{1}{2}(8)^2\sin\!\left(\frac{5\pi}{12}\right) = 32\sin\!\left(\frac{5\pi}{12}\right).$$

Step 2 — Evaluate. \(\frac{5\pi}{12} = 75°\), so \(\sin(75°) = \frac{\sqrt{6}+\sqrt{2}}{4} \approx 0.9659\).

$$A \approx 32 \cdot 0.9659 \approx 30.91 \text{ in}^2.$$

Answer: \(A \approx 30.91\) in².

Problem 66

Step 1 — Plug in \(t = 9\). \(\omega = 9|\cos(\pi\cdot 9 - \pi/12)| = 9|\cos(9\pi - \pi/12)|\).

Step 2 — Simplify the angle. \(9\pi - \frac{\pi}{12} = \frac{108\pi - \pi}{12} = \frac{107\pi}{12}\). Since \(\cos\) has period \(2\pi\), reduce: \(\frac{107\pi}{12} = 8\pi + \frac{11\pi}{12}\). So \(\cos\!\left(\frac{107\pi}{12}\right) = \cos\!\left(\frac{11\pi}{12}\right)\).

\(\frac{11\pi}{12} = \pi - \frac{\pi}{12}\), so \(\cos\!\left(\frac{11\pi}{12}\right) = -\cos\!\left(\frac{\pi}{12}\right) \approx -0.9659\).

$$\omega = 9 \cdot |-0.9659| \approx 8.693 \text{ rad/sec.}$$

Answer: \(\omega \approx 8.693\) rad/sec.

Problem 67

Step 1 — Identify the period. \(V(t) = 150\cos(368t)\) has \(B = 368\), so:

$$T = \frac{2\pi}{368} = \frac{\pi}{184} \approx 0.01707 \text{ sec.}$$

Interpretation: The voltage completes one full oscillation (one AC cycle) approximately every 0.017 seconds.

Step 2 — Count cycles per second. Number of periods in 1 sec:

$$N = \frac{1}{T} = \frac{368}{2\pi} \approx 58.6 \text{ cycles/sec.}$$

Note on the calculator: compute \(368/(2\pi)\) to get the frequency in Hertz.

Answer: Period \(\approx 0.01707\) sec (approximately 58.6 oscillations per second — close to 60 Hz AC current).

Problem 68

Step 1 — Read the parameters. \(N(t) = 12 + 3\sin\!\left[\frac{2\pi}{365}(t-79)\right]\).

a) Amplitude: \(3\) (daylight varies 3 hours above/below the 12-hour mean). Period: \(\frac{2\pi}{2\pi/365} = 365\) days (one year).

b) Longest day: Maximum occurs when \(\sin = 1\): \(N_{\max} = 12 + 3 = 15\) hours.

c) Shortest day: Minimum occurs when \(\sin = -1\): \(N_{\min} = 12 - 3 = 9\) hours.

d) Day 90: \(N(90) = 12 + 3\sin\!\left[\frac{2\pi}{365}(11)\right] = 12 + 3\sin(0.1893) \approx 12 + 3(0.1883) \approx 12.565\) hours.

e) Sketch: A sine curve centered at \(N = 12\), amplitude 3, period 365 days, shifted right 79 days (maximum near day 79 + 91 = day 170, approximately June 21).

Note on the calculator: Graph \(y = 12 + 3\sin(2\pi(x-79)/365)\) for \(0 \le x \le 365\).

Answer: a) Amplitude = 3, Period = 365 days; b) 15 hours; c) 9 hours; d) ≈ 12.56 hours.

Problem 69

Step 1 — Read the parameters. \(T = 50 + 10\sin\!\left[\frac{\pi}{12}(t-8)\right]\).

a) Amplitude: \(10\)°F. Period: \(\frac{2\pi}{\pi/12} = 24\) hours.

b) Temperature at \(t = 7\):

$$T(7) = 50 + 10\sin\!\left[\frac{\pi}{12}(7-8)\right] = 50 + 10\sin\!\left(-\frac{\pi}{12}\right) \approx 50 + 10(-0.2588) \approx 47.4°\text{F}.$$

c) When does \(T = 60\)?

$$60 = 50 + 10\sin\!\left[\frac{\pi}{12}(t-8)\right] \Rightarrow \sin\!\left[\frac{\pi}{12}(t-8)\right] = 1 \Rightarrow \frac{\pi}{12}(t-8) = \frac{\pi}{2} \Rightarrow t = 14.$$

So \(T = 60°\) at \(t = 14\) (2 p.m.).

d) Sketch: A sine curve centered at \(T = 50\), amplitude 10, period 24, shifted right by 8. Maximum at \(t = 14\) (2 p.m.), minimum at \(t = 2\) (2 a.m.).

Answer: a) Amplitude = 10°F, Period = 24 hours; b) ≈ 47.4°F; c) \(t = 14\) (2:00 p.m.).

Problem 70

Step 1 — Read the parameters. \(H(t) = 8\sin\!\left(\frac{\pi}{6}t\right)\).

a) Amplitude: \(8\) ft. Period: \(\frac{2\pi}{\pi/6} = 12\) hours.

b) Sketch: A sine curve on \([0, 12]\), starting at 0 (midnight), rising to a peak of 8 ft at \(t = 3\) (3:00 a.m.), back to 0 at \(t = 6\), down to \(-8\) ft at \(t = 9\), returning to 0 at \(t = 12\).

c) Height at 4:30 a.m.: Convert to decimal hours: \(t = 4.5\).

$$H(4.5) = 8\sin\!\left(\frac{\pi}{6} \cdot 4.5\right) = 8\sin\!\left(\frac{3\pi}{4}\right) = 8 \cdot \frac{\sqrt{2}}{2} = 4\sqrt{2} \approx 5.657 \text{ ft.}$$

Answer: a) Amplitude = 8 ft, Period = 12 hours; c) \(H \approx 5.657\) ft (or exactly \(4\sqrt{2}\) ft).

Problem 1

Step 1 — Apply the horizontal line test: A decreasing straight line passes the horizontal line test because every horizontal line can intersect a strictly decreasing line at most once.

Answer: Yes, this function is one-to-one.

Problem 2

Step 1 — Apply the horizontal line test: The graph shows a function that is always increasing. A strictly increasing function passes the horizontal line test because every horizontal line intersects it at most once.

Answer: Yes, this function is one-to-one.

Problem 3

Step 1 — Apply the horizontal line test: A semicircle (upper or lower half) has a turning point — horizontal lines near the top of the arc intersect the graph at two points.

Answer: No, this function is not one-to-one (fails the horizontal line test).

Problem 4

Step 1 — Apply the horizontal line test: The graph shows a curved function that increases and then turns. Based on the description (resembling a curve that is not strictly monotone), horizontal lines in some range will intersect it more than once.

Answer: No, this function is not one-to-one (fails the horizontal line test).

Problem 5

Step 1 — Apply the horizontal line test: The graph is described as a curved function that is always increasing. A strictly increasing function passes the horizontal line test.

Answer: Yes, this function is one-to-one.

Problem 6

Step 1 — Apply the horizontal line test: The graph increases in a straight line from one region and then levels off or changes — this suggests it is not strictly monotone everywhere.

Step 2 — Note the shape: A function that increases on part of its domain and is constant or decreasing elsewhere is not one-to-one if any horizontal line crosses it more than once.

Answer: No, this function is not one-to-one (fails the horizontal line test).

Problem 7

Step 1 — Find the inverse: Write \(y = x^2 - 4\) with \(x \ge 0\). Solve for \(x\): $$y + 4 = x^2 \implies x = \sqrt{y + 4}$$ (taking the positive root since \(x \ge 0\)). Swap \(x\) and \(y\): $$f^{-1}(x) = \sqrt{x + 4}.$$

Step 2 — Domain and range of \(f^{-1}\): - Domain of \(f\) is \([0, \infty)\); range of \(f\) is \([-4, \infty)\) (since \(f(0) = -4\) and \(f\) increases without bound). - Therefore domain of \(f^{-1}\) is \([-4, \infty)\) and range of \(f^{-1}\) is \([0, \infty)\).

Answer: \(f^{-1}(x) = \sqrt{x + 4}\), domain \([-4, \infty)\), range \([0, \infty)\).

Problem 8

Step 1 — Find the inverse: Write \(y = \sqrt[3]{x - 4}\). Cube both sides: $$y^3 = x - 4 \implies x = y^3 + 4.$$ Swap \(x\) and \(y\): $$f^{-1}(x) = x^3 + 4.$$

Step 2 — Domain and range of \(f^{-1}\): The original function \(f\) has domain \((-\infty, \infty)\) and range \((-\infty, \infty)\). So \(f^{-1}\) has domain \((-\infty, \infty)\) and range \((-\infty, \infty)\).

Answer: \(f^{-1}(x) = x^3 + 4\), domain \((-\infty, \infty)\), range \((-\infty, \infty)\).

Problem 9

Step 1 — Find the inverse: Write \(y = x^3 + 1\). Solve for \(x\): $$x^3 = y - 1 \implies x = (y - 1)^{1/3}.$$ Swap: $$f^{-1}(x) = (x - 1)^{1/3}.$$

Step 2 — Domain and range: Both \(f\) and \(f^{-1}\) have domain and range \((-\infty, \infty)\).

Answer: \(f^{-1}(x) = (x-1)^{1/3}\), domain \((-\infty, \infty)\), range \((-\infty, \infty)\).

Problem 10

Step 1 — Find the inverse: Write \(y = (x-1)^2\) with \(x \le 1\). Take the square root: $$\sqrt{y} = |x - 1| = 1 - x \quad (\text{since } x \le 1 \text{ means } x - 1 \le 0).$$ So \(x = 1 - \sqrt{y}\). Swap: $$f^{-1}(x) = 1 - \sqrt{x}.$$

Step 2 — Domain and range: On \((-\infty, 1]\), \(f(x) = (x-1)^2\) ranges over \([0, \infty)\). So: - Domain of \(f^{-1}\): \([0, \infty)\) - Range of \(f^{-1}\): \((-\infty, 1]\)

Answer: \(f^{-1}(x) = 1 - \sqrt{x}\), domain \([0, \infty)\), range \((-\infty, 1]\).

Problem 11

Step 1 — Find the inverse: Write \(y = \sqrt{x - 1}\). Square both sides: $$y^2 = x - 1 \implies x = y^2 + 1.$$ Swap: $$f^{-1}(x) = x^2 + 1.$$

Step 2 — Domain and range: - Domain of \(f\) is \([1, \infty)\); range of \(f\) is \([0, \infty)\). - Domain of \(f^{-1}\): \([0, \infty)\); range of \(f^{-1}\): \([1, \infty)\).

Answer: \(f^{-1}(x) = x^2 + 1\), domain \([0, \infty)\), range \([1, \infty)\).

Problem 12

Step 1 — Find the inverse: Write \(y = \dfrac{1}{x + 2}\). Solve for \(x\): $$y(x + 2) = 1 \implies x + 2 = \frac{1}{y} \implies x = \frac{1}{y} - 2.$$ Swap: $$f^{-1}(x) = \frac{1}{x} - 2 = \frac{1 - 2x}{x}.$$

Step 2 — Domain and range: - Domain of \(f\) is \(\{x \mid x \ne -2\}\); range of \(f\) is \(\{y \mid y \ne 0\}\). - Domain of \(f^{-1}\): \(\{x \mid x \ne 0\}\); range of \(f^{-1}\): \(\{y \mid y \ne -2\}\).

Answer: \(f^{-1}(x) = \dfrac{1}{x} - 2\), domain \(x \ne 0\), range \(y \ne -2\).

Problem 13

Step 1 — Sketch and reflect: The graph of \(f\) (an increasing straight line) passes through the origin along a positive slope. To sketch \(f^{-1}\), reflect the graph over the line \(y = x\).

Step 2 — Identify domain/range of \(f^{-1}\): Read these from the graph of \(f\) — the range of \(f\) becomes the domain of \(f^{-1}\), and vice versa.

Answer: Sketch the reflection of the straight-line graph over \(y = x\). The result is also a straight line; the roles of domain and range are swapped.

Problem 14

Step 1 — Sketch and reflect: The graph of \(f\) is a decreasing curved function. Reflect each point \((a, b)\) to \((b, a)\) across \(y = x\).

Answer: Sketch the mirror image of the decreasing curve over the line \(y = x\). The reflected graph is an increasing curve with domain equal to the range of \(f\) and range equal to the domain of \(f\).

Problem 15

Step 1 — Sketch and reflect: The graph is an increasing straight line with a steeper slope. Reflect over \(y = x\) to obtain \(f^{-1}\).

Answer: The inverse is also a straight line with slope equal to the reciprocal of \(f\)'s slope; domain and range are swapped.

Problem 16

Step 1 — Sketch and reflect: The graph is a decreasing curved function. Reflecting over \(y = x\) gives an increasing curve.

Answer: The reflection of the decreasing curve over \(y = x\) is an increasing curve. The domain and range of \(f^{-1}\) are the range and domain of \(f\), respectively.

Problem 17

Step 1 — Check via composition: Compute \(f(g(x))\): $$f(g(x)) = f!\left(\frac{x}{8}\right) = 8 \cdot \frac{x}{8} = x.$$

Step 2 — Verify the other direction: $$g(f(x)) = g(8x) = \frac{8x}{8} = x.$$

Both compositions equal \(x\), so \(f\) and \(g\) are inverses.

Answer: Yes, \(f\) and \(g\) are inverses of each other.

Problem 18

Step 1 — Compute \(f(g(x))\): $$f(g(x)) = 8!\left(\frac{x-3}{8}\right) + 3 = (x - 3) + 3 = x.$$

Step 2 — Compute \(g(f(x))\): $$g(f(x)) = \frac{(8x+3)-3}{8} = \frac{8x}{8} = x.$$

Answer: Yes, \(f\) and \(g\) are inverses.

Problem 19

Step 1 — Compute \(f(g(x))\): $$f(g(x)) = 5!\left(\frac{x+5}{7}\right) - 7 = \frac{5x + 25}{7} - 7 = \frac{5x + 25 - 49}{7} = \frac{5x - 24}{7}.$$

This does NOT equal \(x\) in general, so \(f\) and \(g\) are NOT inverses.

Answer: No, \(f(x) = 5x - 7\) and \(g(x) = \dfrac{x+5}{7}\) are not inverses. (The correct inverse of \(f\) would be \(f^{-1}(x) = \dfrac{x+7}{5}\).)

Problem 20

Step 1 — Compute \(f(g(x))\): $$f(g(x)) = \frac{2}{3}!\left(\frac{3}{2}x + 3\right) + 2 = x + 2 + 2 = x + 4.$$

This is not \(x\), so \(f\) and \(g\) are NOT inverses.

Answer: No, they are not inverses. (The correct inverse of \(f\) would have \(f^{-1}(x) = \dfrac{3}{2}(x-2) = \dfrac{3x}{2} - 3\).)

Problem 21

Step 1 — Compute \(f(g(x))\): $$f(g(x)) = f!\left(\frac{1}{x} + 1\right) = \frac{1}{\left(\frac{1}{x}+1\right) - 1} = \frac{1}{\frac{1}{x}} = x.$$

Step 2 — Compute \(g(f(x))\): $$g(f(x)) = g!\left(\frac{1}{x-1}\right) = \frac{1}{\frac{1}{x-1}} + 1 = (x-1) + 1 = x.$$

Answer: Yes, \(f\) and \(g\) are inverses (on their respective domains \(x \ne 1\) and \(x \ne 0\)).

Problem 22

Step 1 — Compute \(f(g(x))\): $$f(g(x)) = \left((x-1)^{1/3}\right)^3 + 1 = (x-1) + 1 = x.$$

Step 2 — Compute \(g(f(x))\): $$g(f(x)) = \left((x^3+1) - 1\right)^{1/3} = \left(x^3\right)^{1/3} = x.$$

Answer: Yes, \(f\) and \(g\) are inverses.

Problem 23

Step 1 — Compute \(f(g(x))\): $$f(g(x)) = \left(-1 + \sqrt{x}\right)^2 + 2!\left(-1 + \sqrt{x}\right) + 1 = \left(\sqrt{x}\right)^2 = x.$$ (Using the fact that \((-1+\sqrt{x})^2 + 2(-1+\sqrt{x}) + 1 = ((-1+\sqrt{x})+1)^2 = (\sqrt{x})^2 = x\).)

Step 2 — Compute \(g(f(x))\) for \(x \ge -1\): $$g(f(x)) = -1 + \sqrt{x^2+2x+1} = -1 + \sqrt{(x+1)^2} = -1 + (x+1) = x.$$ (Since \(x \ge -1\), \(|x+1| = x+1\).)

Answer: Yes, \(f\) and \(g\) are inverses on their respective domains.

Problem 24

Step 1 — Compute \(f(g(x))\): $$f(g(x)) = \sqrt{4 - \left(\sqrt{4-x^2}\right)^2} = \sqrt{4 - (4 - x^2)} = \sqrt{x^2} = x$$ (since \(0 \le x \le 2\)).

Step 2 — Compute \(g(f(x))\): By symmetry, same calculation gives \(x\).

Answer: Yes, \(f = g\) and each is its own inverse on \([0, 2]\).

Problem 25

Step 1 — Identify the angle: We need \(\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\) such that \(\tan\theta = \dfrac{\sqrt{3}}{3} = \dfrac{1}{\sqrt{3}}\). That angle is \(\theta = \dfrac{\pi}{6}\).

Answer: \(\tan^{-1}\!\left(\dfrac{\sqrt{3}}{3}\right) = \dfrac{\pi}{6}\).

Problem 26

Step 1 — Identify the angle: We need \(\theta \in [0, \pi]\) such that \(\cos\theta = -\dfrac{\sqrt{2}}{2}\). That is \(\theta = \dfrac{3\pi}{4}\) (second quadrant, where cosine is negative).

Answer: \(\cos^{-1}\!\left(-\dfrac{\sqrt{2}}{2}\right) = \dfrac{3\pi}{4}\).

Problem 27

Step 1 — Identify the angle: We need \(\theta \in (0, \pi)\) such that \(\cot\theta = 1\), meaning \(\tan\theta = 1\). That is \(\theta = \dfrac{\pi}{4}\).

Answer: \(\cot^{-1}(1) = \dfrac{\pi}{4}\).

Problem 28

Step 1 — Identify the angle: We need \(\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\) such that \(\sin\theta = -1\). That is \(\theta = -\dfrac{\pi}{2}\).

Answer: \(\sin^{-1}(-1) = -\dfrac{\pi}{2}\).

Problem 29

Step 1 — Identify the angle: We need \(\theta \in [0, \pi]\) such that \(\cos\theta = \dfrac{\sqrt{3}}{2}\). That is \(\theta = \dfrac{\pi}{6}\) (first quadrant).

Answer: \(\cos^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{6}\).

Problem 30

Step 1 — Find \(\tan^{-1}(\sqrt{3})\): We need \(\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\) with \(\tan\theta = \sqrt{3}\). That is \(\theta = \dfrac{\pi}{3}\).

Step 2 — Evaluate \(\cos(\pi/3)\): $$\cos!\left(\frac{\pi}{3}\right) = \frac{1}{2}.$$

Answer: \(\cos\!\left(\tan^{-1}(\sqrt{3})\right) = \dfrac{1}{2}\).

Problem 31

Step 1 — Find \(\cos^{-1}\!\left(\dfrac{\sqrt{2}}{2}\right)\): We need \(\theta \in [0, \pi]\) with \(\cos\theta = \dfrac{\sqrt{2}}{2}\). That is \(\theta = \dfrac{\pi}{4}\).

Step 2 — Evaluate \(\sin(\pi/4)\): $$\sin!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.$$

Answer: \(\sin\!\left(\cos^{-1}\!\left(\dfrac{\sqrt{2}}{2}\right)\right) = \dfrac{\sqrt{2}}{2}\).

Problem 32

Step 1 — Check domain: \(\dfrac{\pi}{3} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\), so the outer-undoes-inner identity applies directly.

$$\sin^{-1}\!\left(\sin\!\left(\frac{\pi}{3}\right)\right) = \frac{\pi}{3}.$$

Answer: \(\dfrac{\pi}{3}\).

Problem 33

Step 1 — Check domain: \(-\dfrac{\pi}{6} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\), so the outer-undoes-inner identity applies.

$$\tan^{-1}\!\left(\tan\!\left(-\frac{\pi}{6}\right)\right) = -\frac{\pi}{6}.$$

Answer: \(-\dfrac{\pi}{6}\).

Problem 34

Part a — Find the inverse function:

Step 1: Write \(C = \dfrac{5}{9}(F - 32)\). Solve for \(F\): $$\frac{9C}{5} = F - 32 \implies F = \frac{9C}{5} + 32.$$

So \(T^{-1}(C) = \dfrac{9C}{5} + 32\).

Part b — Interpret:

The inverse function converts Celsius back to Fahrenheit. For example, a weather forecast in degrees Celsius can be converted to degrees Fahrenheit using this formula.

Answer: \(F = T^{-1}(C) = \dfrac{9C}{5} + 32\); it converts Celsius temperatures to Fahrenheit.

Problem 35

Part a — Find \(x = f^{-1}(V)\):

Step 1: Write \(V = 500(0.04 - x^2)\). Solve for \(x\) (with \(0 \le x \le 0.2\)): $$\frac{V}{500} = 0.04 - x^2 \implies x^2 = 0.04 - \frac{V}{500} \implies x = \sqrt{0.04 - \frac{V}{500}}.$$

(Taking the positive root since \(x \ge 0\).)

$$f^{-1}(V) = \sqrt{0.04 - \frac{V}{500}}.$$

Part b — Interpret:

The inverse function gives the distance from the center of the artery at which blood is flowing at velocity \(V\) cm/sec. Doctors can use this to locate where blood is moving at a specified speed.

Part c — Find distances:

- \(V = 15\): \(x = \sqrt{0.04 - 15/500} = \sqrt{0.04 - 0.03} = \sqrt{0.01} = 0.1\) cm. - \(V = 10\): \(x = \sqrt{0.04 - 10/500} = \sqrt{0.04 - 0.02} = \sqrt{0.02} \approx 0.1414\) cm. - \(V = 5\): \(x = \sqrt{0.04 - 5/500} = \sqrt{0.04 - 0.01} = \sqrt{0.03} \approx 0.1732\) cm.

Answer: \(f^{-1}(V) = \sqrt{0.04 - V/500}\); distances are 0.1 cm, ≈ 0.1414 cm, ≈ 0.1732 cm.

Problem 36

Part a — European sizes for U.S. sizes 6, 8, 10, 12:

$$D(6) = 2(6)+24 = 36, \quad D(8) = 40, \quad D(10) = 44, \quad D(12) = 48.$$

Part b — Inverse function (European → U.S.):

Solve \(y = 2x + 24\) for \(x\): \(x = \dfrac{y - 24}{2}\). So \(D^{-1}(y) = \dfrac{y - 24}{2}\).

Part c — U.S. sizes for European 46, 52, 62, 70:

$$D^{-1}(46) = 11, \quad D^{-1}(52) = 14, \quad D^{-1}(62) = 19, \quad D^{-1}(70) = 23.$$

Answer: a) 36, 40, 44, 48. b) \(D^{-1}(x) = \dfrac{x - 24}{2}\). c) 11, 14, 19, 23.

Problem 37

Part a — Cost to remove 25, 40, 50 ppb:

$$C(25) = \frac{75(25)}{85-25} = \frac{1875}{60} = 31.25 \text{ (thousand dollars)}.$$ $$C(40) = \frac{75(40)}{85-40} = \frac{3000}{45} \approx 66.67 \text{ (thousand dollars)}.$$ $$C(50) = \frac{75(50)}{85-50} = \frac{3750}{35} \approx 107.14 \text{ (thousand dollars)}.$$

Part b — Inverse function:

Write \(C = \dfrac{75p}{85 - p}\). Solve for \(p\): $$C(85 - p) = 75p \implies 85C - Cp = 75p \implies 85C = 75p + Cp = p(75 + C) \implies p = \frac{85C}{75 + C}.$$

So \(C^{-1}(C) = \dfrac{85C}{75 + C}\).

Part c — Toxin removed for $50,000:

$50,000 = 50 thousand dollars, so \(C = 50\): $$p = \frac{85(50)}{75 + 50} = \frac{4250}{125} = 34 \text{ ppb}.$$

Answer: a) ≈ $31,250; ≈ $66,667; ≈ $107,143. b) \(p = \dfrac{85C}{75 + C}\). c) 34 ppb.

Problem 38

Part a — Velocity at \(t = 10\) sec:

$$v(10) = \frac{25}{4}(10) + 54 = 62.5 + 54 = 116.5 \text{ ft/sec}.$$

Part b — Inverse function:

Write \(v = \dfrac{25}{4}t + 54\). Solve for \(t\): $$v - 54 = \frac{25}{4}t \implies t = \frac{4(v - 54)}{25}.$$

So \(v^{-1}(v) = \dfrac{4(v-54)}{25}\).

Part c — Time to reach 150 ft/sec:

$$t = \frac{4(150 - 54)}{25} = \frac{4(96)}{25} = \frac{384}{25} = 15.36 \text{ sec}.$$

Answer: a) 116.5 ft/sec. b) \(t = \dfrac{4(v-54)}{25}\). c) 15.36 seconds.

Problem 39

Note on the calculator: Use the formula \(\mu = 2\sin^{-1}(1/M)\) with a calculator in radian mode, then convert to degrees by multiplying by \(180/\pi\).

Part a — \(M = 1.4\): $$\mu = 2\sin^{-1}!\left(\frac{1}{1.4}\right) = 2\sin^{-1}(0.7143) \approx 2(0.7754) \approx 1.5508 \text{ rad} \approx 89°.$$

Part b — \(M = 2.8\): $$\mu = 2\sin^{-1}!\left(\frac{1}{2.8}\right) \approx 2\sin^{-1}(0.3571) \approx 2(0.3652) \approx 0.7303 \text{ rad} \approx 42°.$$

Part c — \(M = 4.3\): $$\mu = 2\sin^{-1}!\left(\frac{1}{4.3}\right) \approx 2\sin^{-1}(0.2326) \approx 2(0.2341) \approx 0.4682 \text{ rad} \approx 27°.$$

Answer: Approximately 89°, 42°, and 27°.

Problem 40

Solve \(\mu = 2\sin^{-1}(1/M)\) for \(M\): Rearranging: \(\sin(\mu/2) = 1/M\), so \(M = \dfrac{1}{\sin(\mu/2)}\).

Part a — \(\mu = \pi/6\): $$M = \frac{1}{\sin(\pi/12)} = \frac{1}{\sin(15°)} \approx \frac{1}{0.2588} \approx 3.86.$$

Part b — \(\mu = 2\pi/7\): $$M = \frac{1}{\sin(\pi/7)} \approx \frac{1}{\sin(25.71°)} \approx \frac{1}{0.4339} \approx 2.30.$$

Part c — \(\mu = 3\pi/8\): $$M = \frac{1}{\sin(3\pi/16)} \approx \frac{1}{\sin(33.75°)} \approx \frac{1}{0.5556} \approx 1.80.$$

Answer: Approximately \(M \approx 3.86\), \(M \approx 2.30\), \(M \approx 1.80\).

Problem 41

Step 1 — Set up the equation: We want \(T(x) = 21\): $$5 + 18\sin!\left[\frac{\pi}{6}(x - 4.6)\right] = 21.$$

Step 2 — Isolate the sine: $$18\sin!\left[\frac{\pi}{6}(x - 4.6)\right] = 16 \implies \sin!\left[\frac{\pi}{6}(x - 4.6)\right] = \frac{8}{9}.$$

Step 3 — Solve for \(x\): Let \(u = \dfrac{\pi}{6}(x - 4.6)\). Then \(\sin u = 8/9\), so: $$u = \sin^{-1}(8/9) \approx 1.0956 \quad \text{or} \quad u = \pi - 1.0956 \approx 2.0460.$$

For \(u_1 \approx 1.0956\): $$\frac{\pi}{6}(x - 4.6) = 1.0956 \implies x - 4.6 = \frac{6(1.0956)}{\pi} \approx 2.091 \implies x \approx 6.691.$$

Month 6, day: \(0.691 \times 30 \approx 21\) → approximately June 21.

For \(u_2 \approx 2.0460\): $$x - 4.6 = \frac{6(2.0460)}{\pi} \approx 3.909 \implies x \approx 8.509.$$

Month 8, day: \(0.509 \times 31 \approx 16\) → approximately August 16.

Answer: The average temperature is 21°C on approximately June 21 and August 16.

Problem 42

Step 1 — Set up the equation: We want \(D(t) = 11.75\): $$5\sin!\left(\frac{\pi}{6}t - \frac{7\pi}{6}\right) + 8 = 11.75.$$

Step 2 — Isolate the sine: $$5\sin!\left(\frac{\pi}{6}t - \frac{7\pi}{6}\right) = 3.75 \implies \sin!\left(\frac{\pi}{6}t - \frac{7\pi}{6}\right) = 0.75.$$

Step 3 — Solve: Let \(u = \dfrac{\pi}{6}t - \dfrac{7\pi}{6}\). Then \(\sin u = 0.75\): $$u = \sin^{-1}(0.75) \approx 0.8481 \quad \text{or} \quad u = \pi - 0.8481 \approx 2.2935.$$

For \(u \approx 0.8481\): $$\frac{\pi}{6}t = \frac{7\pi}{6} + 0.8481 \implies t = \frac{6}{\pi}!\left(\frac{7\pi}{6} + 0.8481\right) = 7 + \frac{6(0.8481)}{\pi} \approx 7 + 1.620 = 8.620 \text{ hours.}$$

So the first time after midnight is approximately 8 hours 37 minutes, or about 8:37 AM.

Answer: The depth is 11.75 ft for the first time at approximately \(t \approx 8.62\) hours after midnight (about 8:37 AM).

Problem 43

Step 1 — Set up the equation: We want the first \(t > 0\) when \(|s(t)| = 4.5\). Since \(s(t) = -6\cos(\pi t/2)\), we solve: $$\left|-6\cos!\left(\frac{\pi t}{2}\right)\right| = 4.5 \implies \left|\cos!\left(\frac{\pi t}{2}\right)\right| = 0.75.$$

Step 2 — Solve: \(\cos(\pi t/2) = \pm 0.75\).

For \(\cos(\pi t/2) = 0.75\): $$\frac{\pi t}{2} = \cos^{-1}(0.75) \approx 0.7227 \implies t \approx \frac{2(0.7227)}{\pi} \approx 0.460 \text{ sec.}$$

For \(\cos(\pi t/2) = -0.75\): $$\frac{\pi t}{2} = \cos^{-1}(-0.75) \approx 2.4189 \implies t \approx \frac{2(2.4189)}{\pi} \approx 1.540 \text{ sec.}$$

The first positive solution is \(t \approx 0.460\) sec.

Answer: The distance first reaches 4.5 inches at \(t \approx 0.46\) seconds.

Problem 44

Step 1 — Plug in \(x = 4\): $$\theta = \tan^{-1}!\left(\frac{5.5}{4}\right) - \tan^{-1}!\left(\frac{2.5}{4}\right) = \tan^{-1}(1.375) - \tan^{-1}(0.625).$$

Step 2 — Evaluate: $$\tan^{-1}(1.375) \approx 0.9420 \text{ rad}, \quad \tan^{-1}(0.625) \approx 0.5586 \text{ rad}.$$

$$\theta \approx 0.9420 - 0.5586 \approx 0.3834 \text{ rad} \approx 21.96°.$$

Answer: The viewing angle is approximately 0.383 radians (about 22°) when the person stands 4 ft from the portrait.

Problem 45

Part 1 — \(\tan^{-1}(\tan(2.1))\):

\(2.1\) radians is NOT in \((-\pi/2, \pi/2) \approx (-1.571, 1.571)\). Because \(\tan\) has period \(\pi\), we find the equivalent angle in the range: \(2.1 - \pi \approx 2.1 - 3.1416 \approx -1.042\).

$$\tan^{-1}(\tan(2.1)) \approx -1.042 \approx 2.1 - \pi.$$

Part 2 — \(\cos^{-1}(\cos(2.1))\):

\(2.1 \in [0, \pi] \approx [0, 3.14]\), so the outer-undoes-inner identity applies directly:

$$\cos^{-1}(\cos(2.1)) = 2.1.$$

Explanation: For \(\cos^{-1}\), any input in \([0, \pi]\) passes through unchanged. For \(\tan^{-1}\), the identity only holds when the argument is in \((-\pi/2, \pi/2)\); otherwise the result is the angle in that range with the same tangent value.

Answer: \(\tan^{-1}(\tan(2.1)) \approx 2.1 - \pi \approx -1.042\); \(\cos^{-1}(\cos(2.1)) = 2.1\).

Problem 46

Part 1 — \(\sin(\sin^{-1}(-2))\):

The domain of \(\sin^{-1}\) is \([-1, 1]\). Since \(-2 \notin [-1, 1]\), \(\sin^{-1}(-2)\) is undefined. Therefore \(\sin(\sin^{-1}(-2))\) is also undefined.

Note on calculator: Most calculators return an error (domain error) for \(\sin^{-1}(-2)\).

Part 2 — \(\tan(\tan^{-1}(-2))\):

The domain of \(\tan^{-1}\) is all real numbers, so \(\tan^{-1}(-2)\) is defined. By the outer-undoes-inner identity (since \(-2\) is in the domain of \(\tan^{-1}\)): $$\tan(\tan^{-1}(-2)) = -2.$$

Explanation: The sine function only takes values in \([-1, 1]\), so its inverse is only defined for inputs in that interval. The tangent function takes all real values, so its inverse is defined everywhere.

Answer: \(\sin(\sin^{-1}(-2))\) is undefined; \(\tan(\tan^{-1}(-2)) = -2\).

Problem 1

Step 1 — Evaluate at each given \(x\):

$$ f(3) = 5^3 = 125.00 $$ $$ f\!\left(\tfrac{1}{2}\right) = 5^{1/2} = \sqrt{5} \approx 2.24. $$ $$ f(\sqrt{2}) = 5^{\sqrt{2}} \approx 5^{1.41421} \approx 9.74. $$

Answer: a) \(125.00\), b) \(\approx 2.24\), c) \(\approx 9.74\).

Problem 2

Step 1 — Evaluate at each given \(x\):

$$ f(-1) = (0.3)^{-1} = \frac{1}{0.3} \approx 3.33. $$ $$ f(4) = (0.3)^4 = 0.0081 \approx 0.01. $$ $$ f(-1.5) = (0.3)^{-1.5} = \frac{1}{(0.3)^{1.5}} = \frac{1}{0.3\sqrt{0.3}} \approx \frac{1}{0.1643} \approx 6.09. $$

Answer: a) \(\approx 3.33\), b) \(\approx 0.01\), c) \(\approx 6.09\).

Problem 3

Step 1 — Evaluate at each given \(x\):

$$ f(-2) = 10^{-2} = 0.01. $$ $$ f(4) = 10^4 = 10000.00. $$ $$ f\!\left(\tfrac{5}{3}\right) = 10^{5/3} = 10^{1.6\overline{6}} \approx 46.42. $$

Answer: a) \(0.01\), b) \(10000.00\), c) \(\approx 46.42\).

Problem 4

Step 1 — Evaluate at each given \(x\):

$$ f(2) = e^2 \approx 7.39. $$ $$ f(-3.2) = e^{-3.2} \approx 0.04. $$ $$ f(\pi) = e^{\pi} \approx 23.14. $$

Answer: a) \(\approx 7.39\), b) \(\approx 0.04\), c) \(\approx 23.14\).

Problem 5

Step 1 — Identify the graph's characteristics:

The graph shows a decreasing curve (as \(x\) increases the function decreases), positive \(y\)-values only, passing near \((0, 1)\). A decreasing exponential with base \(> 1\) is of the form \(y = b^{-x}\) or \(y = (1/b)^x\).

Step 2 — Match to the list:

The function \(y = 4^{-x}\) is decreasing and positive for all \(x\), which matches this graph.

Answer: a) \(y = 4^{-x}\).

Problem 6

Step 1 — Identify the graph's characteristics:

The graph starts below the \(x\)-axis (negative values), approaches \(0\) from below as \(x \to -\infty\), and decreases to large negative values as \(x \to \infty\).

Step 2 — Match to the list:

\(y = -3^{-x}\) is always negative and approaches \(0^-\) as \(x \to \infty\), matching this shape.

Answer: e) \(y = -3^{-x}\).

Problem 7

Step 1 — Identify the graph's characteristics:

The graph shows an increasing curve that stays above a horizontal asymptote at \(y = 0,\) passes through a small positive value near \(x = -4,\) and increases steeply for positive \(x\).

Step 2 — Match to the list:

\(y = 2^{x+1}\) is an increasing exponential shifted left by 1. At \(x = 0\), \(y = 2\); at \(x = -1\), \(y = 1\). This matches an increasing curve passing through \((-1, 1)\) and \((0, 2)\).

Answer: c) \(y = 2^{x+1}\).

Problem 8

Step 1 — Identify the graph's characteristics:

The graph is decreasing, stays above a horizontal asymptote at \(y = 0,\) but is positive and decreasing — consistent with a base \(> 1\) raised to \(-x\), or with a vertical shift.

Step 2 — Match to the list:

\(y = 3^{x-1}\) is an increasing exponential shifted right by 1; however the description says "decreasing." Reconsidering: the graph shows positive values that decrease, consistent with \(y = (1/2)^x + 2\) — wait, that has asymptote at \(y = 2\). The graph described has asymptote near \(y = 0\) and decreases. \(y = 4^{-x}\) was already used. The remaining decreasing option is \(y = 3^{x-1}\) only if \(x < 0\) range dominates.

Actually, reviewing the alt text: "decreasing curved function" in range \(-5\) to \(5\) on \(x\), \(-5\) to \(5\) on \(y\). This matches \(y = 1 - 5^x\): at \(x = 0\), \(y = 0\); for \(x > 0\) it decreases rapidly; has horizontal asymptote \(y = 1\) as \(x \to -\infty\).

Answer: f) \(y = 1 - 5^x\).

Problem 9

Step 1 — Identify the graph's characteristics:

The graph shows an increasing curve that starts near \(-5\) on the \(y\)-axis as \(x\) is very negative and increases. Alt text: "curved increasing function that increa[ses]" in range \(x: -5\) to \(5\), \(y: -5\) to \(5\).

Step 2 — Match to the list:

\(y = 3^{x-1}\): increasing, passes through \((1, 1)\) and \((0, 1/3)\). Rises from near zero to high values.

Answer: b) \(y = 3^{x-1}\).

Problem 10

Step 1 — Identify the graph's characteristics:

The alt text says: "curved increasing function that starts" in range \(y: -2\) to \(8\), \(x: -5\) to \(5\). A function that starts above 2 when \(x \to -\infty\) and increases suggests a vertical shift.

Step 2 — Match to the list:

\(y = (1/2)^x + 2\): as \(x \to \infty\), \((1/2)^x \to 0\), so \(y \to 2\) (horizontal asymptote at \(y=2\)), and as \(x \to -\infty\), \((1/2)^x \to \infty\), so \(y\) increases. This gives a decreasing function, not increasing. Re-reading: the graph is "of a curved increasing function that starts" — this combined with the \(y\)-range \(-2\) to \(8\) suggests a function that starts around 2 and increases. The function \(y = 3^{x-1}\) was already used. The remaining is \(y = (1/2)^x + 2\).

Answer: d) \(y = \left(\dfrac{1}{2}\right)^x + 2\).

Problem 11

Step 1 — Analyze the function \(f(x) = e^x + 2\):

This is the natural exponential shifted up by 2.

- Domain: \((-\infty, \infty)\) — exponentials are defined for all real \(x\). - Range: \((2, \infty)\) — since \(e^x > 0\), we have \(e^x + 2 > 2\). - Horizontal asymptote: \(y = 2\) (as \(x \to -\infty\), \(e^x \to 0\)).

Sketch: Standard rising exponential curve, but lifted up by 2 units.

Answer: Domain: \((-\infty, \infty)\); Range: \((2, \infty)\); Horizontal asymptote: \(y = 2\).

Problem 12

Step 1 — Analyze \(f(x) = -2^x\):

This reflects \(2^x\) across the \(x\)-axis.

- Domain: \((-\infty, \infty)\). - Range: \((-\infty, 0)\) — since \(2^x > 0\), \(-2^x < 0\). - Horizontal asymptote: \(y = 0\) (as \(x \to -\infty\), \(-2^x \to 0^-\)).

Answer: Domain: \((-\infty, \infty)\); Range: \((-\infty, 0)\); Horizontal asymptote: \(y = 0\).

Problem 13

Step 1 — Analyze \(f(x) = 3^{x+1}\):

This is \(3^x\) shifted left by 1.

- Domain: \((-\infty, \infty)\). - Range: \((0, \infty)\). - Horizontal asymptote: \(y = 0\).

At \(x = 0\): \(f(0) = 3^1 = 3\).

Answer: Domain: \((-\infty, \infty)\); Range: \((0, \infty)\); Horizontal asymptote: \(y = 0\).

Problem 14

Step 1 — Analyze \(f(x) = 4^x - 1\):

This is \(4^x\) shifted down by 1.

- Domain: \((-\infty, \infty)\). - Range: \((-1, \infty)\) — since \(4^x > 0\), \(4^x - 1 > -1\). - Horizontal asymptote: \(y = -1\).

Answer: Domain: \((-\infty, \infty)\); Range: \((-1, \infty)\); Horizontal asymptote: \(y = -1\).

Problem 15

Step 1 — Analyze \(f(x) = 1 - 2^{-x}\):

As \(x \to \infty\): \(2^{-x} \to 0\), so \(f(x) \to 1\). As \(x \to -\infty\): \(2^{-x} \to \infty\), so \(f(x) \to -\infty\).

- Domain: \((-\infty, \infty)\). - Range: \((-\infty, 1)\). - Horizontal asymptote: \(y = 1\).

Answer: Domain: \((-\infty, \infty)\); Range: \((-\infty, 1)\); Horizontal asymptote: \(y = 1\).

Problem 16

Step 1 — Analyze \(f(x) = 5^{x+1} + 2\):

Shift \(5^x\) left 1 unit and up 2 units.

- Domain: \((-\infty, \infty)\). - Range: \((2, \infty)\). - Horizontal asymptote: \(y = 2\).

Answer: Domain: \((-\infty, \infty)\); Range: \((2, \infty)\); Horizontal asymptote: \(y = 2\).

Problem 17

Step 1 — Analyze \(f(x) = e^{-x} - 1\):

This is the natural exponential reflected across the \(y\)-axis, then shifted down 1.

- Domain: \((-\infty, \infty)\). - Range: \((-1, \infty)\) — since \(e^{-x} > 0\), \(e^{-x} - 1 > -1\). - Horizontal asymptote: \(y = -1\) (as \(x \to \infty\), \(e^{-x} \to 0\)).

Answer: Domain: \((-\infty, \infty)\); Range: \((-1, \infty)\); Horizontal asymptote: \(y = -1\).

Problem 18

Step 1 — Convert \(\log_3 81 = 4\) to exponential form:

By definition, \(\log_b x = y \Leftrightarrow b^y = x\). So \(\log_3 81 = 4\) becomes:

$$ 3^4 = 81. $$

Answer: \(3^4 = 81\).

Problem 19

Step 1 — Convert \(\log_8 2 = \frac{1}{3}\):

$$ 8^{1/3} = 2. $$

Answer: \(8^{1/3} = 2\).

Problem 20

Step 1 — Convert \(\log_5 1 = 0\):

$$ 5^0 = 1. $$

Answer: \(5^0 = 1\).

Problem 21

Step 1 — Convert \(\log_5 25 = 2\):

$$ 5^2 = 25. $$

Answer: \(5^2 = 25\).

Problem 22

Step 1 — Convert \(\log 0.1 = -1\) (base 10):

$$ 10^{-1} = 0.1. $$

Answer: \(10^{-1} = 0.1\).

Problem 23

Step 1 — Convert \(\ln(1/e^3) = -3\):

$$ e^{-3} = \frac{1}{e^3}. $$

Answer: \(e^{-3} = \dfrac{1}{e^3}\).

Problem 24

Step 1 — Convert \(\log_9 3 = 0.5\):

$$ 9^{0.5} = 3. $$

Answer: \(9^{0.5} = 3\).

Problem 25

Step 1 — Convert \(\ln 1 = 0\):

$$ e^0 = 1. $$

Answer: \(e^0 = 1\).

Problem 26

Step 1 — Convert \(2^3 = 8\) to logarithmic form:

By definition, \(b^y = x \Leftrightarrow \log_b x = y\). So:

$$ \log_2 8 = 3. $$

Answer: \(\log_2 8 = 3\).

Problem 27

Step 1 — Convert \(4^{-2} = \frac{1}{16}\):

$$ \log_4\!\left(\frac{1}{16}\right) = -2. $$

Answer: \(\log_4\!\left(\dfrac{1}{16}\right) = -2\).

Problem 28

Step 1 — Convert \(10^2 = 100\):

$$ \log_{10}(100) = 2, \quad \text{i.e.,} \quad \log(100) = 2. $$

Answer: \(\log 100 = 2\).

Problem 29

Step 1 — Convert \(9^0 = 1\):

$$ \log_9(1) = 0. $$

Answer: \(\log_9 1 = 0\).

Problem 30

Step 1 — Convert \(\left(\frac{1}{3}\right)^3 = \frac{1}{27}\):

$$ \log_{1/3}\!\left(\frac{1}{27}\right) = 3. $$

Answer: \(\log_{1/3}\!\left(\dfrac{1}{27}\right) = 3\).

Problem 31

Step 1 — Convert \(\sqrt[3]{64} = 4\):

Write \(\sqrt[3]{64} = 64^{1/3} = 4,\) so the base is \(64\) and the exponent is \(1/3\):

$$ \log_{64}(4) = \frac{1}{3}. $$

Answer: \(\log_{64} 4 = \dfrac{1}{3}\).

Problem 32

Step 1 — Convert \(e^x = y\):

$$ \ln(y) = x. $$

Answer: \(\ln y = x\).

Problem 33

Step 1 — Convert \(9^y = 150\):

$$ \log_9(150) = y. $$

Answer: \(\log_9 150 = y\).

Problem 34

Step 1 — Convert \(b^3 = 45\):

$$ \log_b(45) = 3. $$

Answer: \(\log_b 45 = 3\).

Problem 35

Step 1 — Convert \(4^{-3/2} = 0.125\):

$$ \log_4(0.125) = -\frac{3}{2}. $$

Answer: \(\log_4(0.125) = -\dfrac{3}{2}\).

Problem 36

Step 1 — Analyze \(f(x) = 3 + \ln x\):

This is \(\ln x\) shifted up by 3.

- Domain: \((0, \infty)\) — logarithm requires positive argument. - Range: \((-\infty, \infty)\). - Vertical asymptote: \(x = 0\).

At \(x = 1\): \(f(1) = 3 + 0 = 3\). The graph crosses the \(x\)-axis where \(\ln x = -3\), i.e., \(x = e^{-3} \approx 0.05\).

Answer: Domain: \((0, \infty)\); Range: \((-\infty, \infty)\); Vertical asymptote: \(x = 0\).

Problem 37

Step 1 — Analyze \(f(x) = \ln(x-1)\):

- Domain: \(x - 1 > 0\), so \(x > 1\), i.e., \((1, \infty)\). - Range: \((-\infty, \infty)\). - Vertical asymptote: \(x = 1\).

Answer: Domain: \((1, \infty)\); Range: \((-\infty, \infty)\); Vertical asymptote: \(x = 1\).

Problem 38

Step 1 — Analyze \(f(x) = \ln(-x)\):

- Domain: \(-x > 0\), so \(x < 0\), i.e., \((-\infty, 0)\). - Range: \((-\infty, \infty)\). - Vertical asymptote: \(x = 0\).

Answer: Domain: \((-\infty, 0)\); Range: \((-\infty, \infty)\); Vertical asymptote: \(x = 0\).

Problem 39

Step 1 — Analyze \(f(x) = 1 - \ln x\):

- Domain: \((0, \infty)\). - Range: \((-\infty, \infty)\) — as \(x \to \infty\), \(\ln x \to \infty\) so \(f \to -\infty\); as \(x \to 0^+\), \(\ln x \to -\infty\) so \(f \to \infty\). - Vertical asymptote: \(x = 0\).

Answer: Domain: \((0, \infty)\); Range: \((-\infty, \infty)\); Vertical asymptote: \(x = 0\).

Problem 40

Step 1 — Analyze \(f(x) = \log x - 1\):

This is \(\log_{10} x\) shifted down by 1.

- Domain: \((0, \infty)\). - Range: \((-\infty, \infty)\). - Vertical asymptote: \(x = 0\).

Answer: Domain: \((0, \infty)\); Range: \((-\infty, \infty)\); Vertical asymptote: \(x = 0\).

Problem 41

Step 1 — Analyze \(f(x) = \ln(x+1)\):

- Domain: \(x + 1 > 0\), so \(x > -1\), i.e., \((-1, \infty)\). - Range: \((-\infty, \infty)\). - Vertical asymptote: \(x = -1\).

Answer: Domain: \((-1, \infty)\); Range: \((-\infty, \infty)\); Vertical asymptote: \(x = -1\).

Problem 42

Step 1 — Apply product and power rules:

$$ \log x^4 y = \log x^4 + \log y = 4\log x + \log y. $$

Answer: \(4\log x + \log y\).

Problem 43

Step 1 — Expand using quotient and power rules:

$$ \log_3 \frac{9a^3}{b} = \log_3 9 + \log_3 a^3 - \log_3 b = 2 + 3\log_3 a - \log_3 b. $$

(Since \(\log_3 9 = \log_3 3^2 = 2\).)

Answer: \(2 + 3\log_3 a - \log_3 b\).

Problem 44

Step 1 — Rewrite and apply product and power rules:

$$ \ln a\sqrt[3]{b} = \ln a + \ln b^{1/3} = \ln a + \frac{1}{3}\ln b. $$

Answer: \(\ln a + \dfrac{1}{3}\ln b\).

Problem 45

Step 1 — Simplify inside the radical:

$$ \sqrt{125xy^3} = (125xy^3)^{1/2}, \quad \text{so} $$ $$ \log_5\sqrt{125xy^3} = \frac{1}{2}\log_5(125xy^3) = \frac{1}{2}\!\left(\log_5 125 + \log_5 x + \log_5 y^3\right). $$

Step 2 — Evaluate \(\log_5 125 = 3\) and apply power rule:

$$ = \frac{1}{2}(3 + \log_5 x + 3\log_5 y) = \frac{3}{2} + \frac{1}{2}\log_5 x + \frac{3}{2}\log_5 y. $$

Answer: \(\dfrac{3}{2} + \dfrac{1}{2}\log_5 x + \dfrac{3}{2}\log_5 y\).

Problem 46

Step 1 — Apply quotient, product, and power rules:

$$ \log_4\frac{\sqrt[3]{xy}}{64} = \log_4(xy)^{1/3} - \log_4 64 = \frac{1}{3}(\log_4 x + \log_4 y) - 3. $$

(Since \(\log_4 64 = \log_4 4^3 = 3\).)

Answer: \(\dfrac{1}{3}\log_4 x + \dfrac{1}{3}\log_4 y - 3\).

Problem 47

Step 1 — Rewrite the denominator and apply quotient and power rules:

$$ \ln\!\left(\frac{6}{\sqrt{e^3}}\right) = \ln 6 - \ln e^{3/2} = \ln 6 - \frac{3}{2}. $$

Answer: \(\ln 6 - \dfrac{3}{2}\).

Problem 48

Step 1 — Rewrite 125 as a power of 5:

\(125 = 5^3\), so \(5^x = 5^3\), giving \(x = 3\).

Answer: \(x = 3\).

Problem 49

Step 1 — Isolate the exponential:

\(e^{3x} = 15\).

Step 2 — Take the natural log:

\(3x = \ln 15\), so \(x = \dfrac{\ln 15}{3}\).

Answer: \(x = \dfrac{\ln 15}{3}\).

Problem 50

Step 1 — Express both sides with a common base:

\(8 = 2^3\) and \(4 = 2^2\), so \(8^x = 2^{3x}\) and \(4 = 2^2\).

Thus \(2^{3x} = 2^2\), giving \(3x = 2\), so \(x = \dfrac{2}{3}\).

Answer: \(x = \dfrac{2}{3}\).

Problem 51

Step 1 — Rewrite \(32 = 2^5\) and \(4 = 2^2\):

\(4^{x+1} = 32 \Rightarrow (2^2)^{x+1} = 2^5 \Rightarrow 2^{2(x+1)} = 2^5\).

Step 2 — Equate exponents:

\(2(x+1) = 5 \Rightarrow x + 1 = \dfrac{5}{2} \Rightarrow x = \dfrac{3}{2}\).

Answer: \(x = \dfrac{3}{2}\).

Problem 52

Step 1 — Take \(\log_{10}\) of both sides:

$$ \frac{x}{14}\log 3 = \log\frac{1}{10} = -1. $$

Step 2 — Solve for \(x\):

$$ x = \frac{-14}{\log 3} = \frac{-14}{\ln 3 / \ln 10} = \frac{-14\ln 10}{\ln 3} \approx \frac{-14 \times 2.3026}{1.0986} \approx -29.37. $$

Answer: \(x = \dfrac{-14\ln 10}{\ln 3} \approx -29.37\).

Problem 53

Step 1 — Take \(\log_{10}\) of both sides:

$$ x = \log_{10}(7.21) \approx 0.8579. $$

Answer: \(x = \log(7.21) \approx 0.8579\).

Problem 54

Step 1 — Isolate the exponential:

\(4 \cdot 2^{3x} = 20 \Rightarrow 2^{3x} = 5\).

Step 2 — Take \(\ln\):

\(3x = \ln 5 \Rightarrow x = \dfrac{\ln 5}{3}\).

Answer: \(x = \dfrac{\ln 5}{3}\).

Problem 55

Step 1 — Take \(\ln\) of both sides:

\((3x-2)\ln 7 = \ln 11\).

Step 2 — Solve for \(x\):

$$ 3x - 2 = \frac{\ln 11}{\ln 7}, \quad 3x = 2 + \frac{\ln 11}{\ln 7}, \quad x = \frac{1}{3}\!\left(2 + \frac{\ln 11}{\ln 7}\right) \approx \frac{1}{3}(2 + 1.2323) \approx 1.0774. $$

Answer: \(x = \dfrac{1}{3}\!\left(2 + \dfrac{\ln 11}{\ln 7}\right) \approx 1.0774\).

Problem 56

Step 1 — Rewrite using the definition of logarithm:

\(\log_3 x = 0 \Rightarrow x = 3^0 = 1\).

Answer: \(x = 1\).

Problem 57

Step 1 — Rewrite using the definition:

\(\log_5 x = -2 \Rightarrow x = 5^{-2} = \dfrac{1}{25}\).

Answer: \(x = \dfrac{1}{25}\).

Problem 58

Step 1 — Rewrite:

\(\log_4(x+5) = 0 \Rightarrow x + 5 = 4^0 = 1 \Rightarrow x = -4\).

Step 2 — Check domain: \(x + 5 = 1 > 0\). Valid.

Answer: \(x = -4\).

Problem 59

Step 1 — Rewrite (base 10):

\(\log(2x-7) = 0 \Rightarrow 2x - 7 = 10^0 = 1 \Rightarrow 2x = 8 \Rightarrow x = 4\).

Step 2 — Check: \(2(4) - 7 = 1 > 0\). Valid.

Answer: \(x = 4\).

Problem 60

Step 1 — Exponentiate both sides:

\(\ln\sqrt{x+3} = 2 \Rightarrow \sqrt{x+3} = e^2 \Rightarrow x + 3 = e^4 \Rightarrow x = e^4 - 3\).

Step 2 — Check domain: \(e^4 - 3 > 0\). Valid.

Answer: \(x = e^4 - 3 \approx 51.60\).

Problem 61

Step 1 — Combine logs using product rule:

\(\log_6(x+9) + \log_6 x = \log_6[x(x+9)] = 2 \Rightarrow x(x+9) = 6^2 = 36\).

Step 2 — Solve the quadratic:

\(x^2 + 9x - 36 = 0 \Rightarrow (x+12)(x-3) = 0\).

So \(x = 3\) or \(x = -12\).

Step 3 — Check domain: Both arguments must be positive. For \(x = 3\): \(x + 9 = 12 > 0\) and \(x = 3 > 0\). Valid. For \(x = -12\): \(\log_6(-12)\) is undefined.

Answer: \(x = 3\).

Problem 62

Step 1 — Combine logs:

\(\log_4(x+2) - \log_4(x-1) = \log_4\!\left(\dfrac{x+2}{x-1}\right) = 0\).

Step 2 — Exponentiate:

\(\dfrac{x+2}{x-1} = 4^0 = 1 \Rightarrow x + 2 = x - 1\).

This gives \(2 = -1\), which is a contradiction. No solution.

Answer: No solution.

Problem 63

Step 1 — Combine left-hand logs:

\(\ln[x(x-2)] = \ln 4 \Rightarrow x(x-2) = 4\).

Step 2 — Solve the quadratic:

\(x^2 - 2x - 4 = 0 \Rightarrow x = \dfrac{2 \pm \sqrt{4+16}}{2} = 1 \pm \sqrt{5}\).

Step 3 — Check domain: Both \(x > 0\) and \(x - 2 > 0\) (i.e., \(x > 2\)) must hold. \(1 + \sqrt{5} \approx 3.24 > 2\). Valid. \(1 - \sqrt{5} < 0\). Invalid.

Answer: \(x = 1 + \sqrt{5}\).

Problem 64

Step 1 — Apply change-of-base formula:

$$ \log_5 47 = \frac{\ln 47}{\ln 5} = \frac{3.8501}{1.6094} \approx 2.3921. $$

Answer: \(\log_5 47 = \dfrac{\ln 47}{\ln 5} \approx 2.3921\).

Problem 65

Step 1 — Apply change-of-base formula:

$$ \log_7 82 = \frac{\ln 82}{\ln 7} = \frac{4.4067}{1.9459} \approx 2.2646. $$

Answer: \(\log_7 82 = \dfrac{\ln 82}{\ln 7} \approx 2.2646\).

Problem 66

Step 1 — Apply change-of-base formula:

$$ \log_6 103 = \frac{\ln 103}{\ln 6} = \frac{4.6347}{1.7918} \approx 2.5872. $$

Answer: \(\log_6 103 = \dfrac{\ln 103}{\ln 6} \approx 2.5872\).

Problem 67

Step 1 — Apply change-of-base formula:

$$ \log_{0.5} 211 = \frac{\ln 211}{\ln 0.5} = \frac{5.3510}{-0.6931} \approx -7.7206. $$

Answer: \(\log_{0.5} 211 = \dfrac{\ln 211}{\ln 0.5} \approx -7.7206\).

Problem 68

Step 1 — Apply change-of-base formula:

$$ \log_2 \pi = \frac{\ln \pi}{\ln 2} = \frac{1.1447}{0.6931} \approx 1.6514. $$

Answer: \(\log_2 \pi = \dfrac{\ln\pi}{\ln 2} \approx 1.6514\).

Problem 69

Step 1 — Apply change-of-base formula:

$$ \log_{0.2} 0.452 = \frac{\ln 0.452}{\ln 0.2} = \frac{-0.7930}{-1.6094} \approx 0.4927. $$

Answer: \(\log_{0.2} 0.452 = \dfrac{\ln 0.452}{\ln 0.2} \approx 0.4927\).

Problem 70

Step 1 — Part a: Simplify \(2\cosh(\ln x)\).

$$ 2\cosh(\ln x) = 2 \cdot \frac{e^{\ln x} + e^{-\ln x}}{2} = e^{\ln x} + e^{-\ln x} = x + \frac{1}{x}. $$

Step 2 — Part b: Simplify \(\cosh 4x + \sinh 4x\).

Using the identity \(\cosh u + \sinh u = e^u\):

$$ \cosh 4x + \sinh 4x = e^{4x}. $$

Step 3 — Part c: Simplify \(\cosh 2x - \sinh 2x\).

Using the identity \(\cosh u - \sinh u = e^{-u}\):

$$ \cosh 2x - \sinh 2x = e^{-2x}. $$

Step 4 — Part d: Simplify \(\ln(\cosh x + \sinh x) + \ln(\cosh x - \sinh x)\).

$$ = \ln[(\cosh x + \sinh x)(\cosh x - \sinh x)] = \ln[\cosh^2 x - \sinh^2 x] = \ln 1 = 0. $$

Answer: a) \(x + \dfrac{1}{x}\), b) \(e^{4x}\), c) \(e^{-2x}\), d) \(0\).

Problem 71

Step 1 — Set up the model and plug in \(t = 15\):

$$ N(15) = 1300 \cdot 2^{15/4} = 1300 \cdot 2^{3.75}. $$

Step 2 — Evaluate \(2^{3.75}\):

$$ 2^{3.75} = 2^3 \cdot 2^{0.75} = 8 \cdot 2^{3/4} \approx 8 \times 1.6818 = 13.454. $$ $$ N(15) \approx 1300 \times 13.454 \approx 17490 \text{ bacteria.} $$

Answer: Approximately \(17490\) bacteria are present after 15 days.

Problem 72

Step 1 — Evaluate \(D(15)\):

$$ D(15) = 150 \cdot (2.7)^{-0.25 \times 15} = 150 \cdot (2.7)^{-3.75} \approx 150 \cdot 0.01878 \approx 2.8 \approx 3 \text{ million barrels.} $$

Step 2 — Evaluate \(D(20)\):

$$ D(20) = 150 \cdot (2.7)^{-5} \approx 150 \cdot 0.004049 \approx 0.607 \approx 1 \text{ million barrel.} $$

Answer: At \(p = \$15\): approximately \(3\) million barrels; at \(p = \$20\): approximately \(1\) million barrel.

Problem 73

Step 1 — Plug in \(t = 5\):

$$ A(5) = 100000 \cdot e^{0.055 \times 5} = 100000 \cdot e^{0.275} \approx 100000 \times 1.3165 \approx \$131650. $$

Answer: After 5 years, approximately \(\$131,650\).

Problem 74

Step 1 — Set \(P = 100000,\) \(j = 0.035,\) \(t = 5\):

a) Daily (\(n = 365\)):

$$ A = 100000\!\left(1 + \frac{0.035}{365}\right)^{365 \times 5} \approx 100000 \times 1.19096 \approx \$119,096. $$

b) Monthly (\(n = 12\)):

$$ A = 100000\!\left(1 + \frac{0.035}{12}\right)^{60} \approx 100000 \times 1.19064 \approx \$119,064. $$

c) Quarterly (\(n = 4\)):

$$ A = 100000\!\left(1 + \frac{0.035}{4}\right)^{20} \approx 100000 \times 1.18985 \approx \$118,985. $$

d) Yearly (\(n = 1\)):

$$ A = 100000(1.035)^5 \approx 100000 \times 1.18769 \approx \$118,769. $$

Answer: a) \(\approx \$119,096\); b) \(\approx \$119,064\); c) \(\approx \$118,985\); d) \(\approx \$118,769\).

Problem 75

Step 1 — Recall the pH formula: \(\text{pH} = -\log[\text{H}^+]\).

a) Eggs: \([\text{H}^+] = 1.6 \times 10^{-8}\):

$$ \text{pH} = -\log(1.6 \times 10^{-8}) = -({\log 1.6 + \log 10^{-8}}) = -(0.204 - 8) \approx 7.8 \approx 8. $$

Since pH \(> 7\): base.

b) Beer: \([\text{H}^+] = 3.16 \times 10^{-3}\):

$$ \text{pH} = -\log(3.16 \times 10^{-3}) = -(0.500 - 3) = 2.5 \approx 3. $$

Since pH \(< 7\): acid.

c) Tomato Juice: \([\text{H}^+] = 7.94 \times 10^{-5}\):

$$ \text{pH} = -\log(7.94 \times 10^{-5}) = -(0.900 - 5) = 4.1 \approx 4. $$

Since pH \(< 7\): acid.

Answer: a) pH \(\approx 8\) (base); b) pH \(\approx 3\) (acid); c) pH \(\approx 4\) (acid).

Problem 76

Step 1 — Set up the equation for 95% decay:

We want \(Q(t) = 0.05 Q_0\) (only 5% remains):

$$ 0.05 Q_0 = Q_0 e^{-0.08664t} \Rightarrow e^{-0.08664t} = 0.05. $$

Step 2 — Solve for \(t\):

$$ -0.08664t = \ln(0.05) = -2.9957, \quad t = \frac{2.9957}{0.08664} \approx 34.58 \approx 35 \text{ days.} $$

Answer: It takes approximately \(35\) days for 95% of the iodine-131 to decay.

Problem 77

Step 1 — Part a: Population in 2020 (\(t = 7\)):

$$ P(7) = 316 e^{0.0074 \times 7} = 316 e^{0.0518} \approx 316 \times 1.0532 \approx 332.8 \text{ million.} $$

Step 2 — Part b: When does \(P(t) = 2 \times 316 = 632\)?

$$ 632 = 316 e^{0.0074t} \Rightarrow e^{0.0074t} = 2 \Rightarrow 0.0074t = \ln 2 \approx 0.6931 \Rightarrow t \approx 93.7 \text{ years.} $$

So approximately in the year \(2013 + 94 = 2107\).

Answer: a) \(\approx 332.8\) million; b) \(\approx 94\) years after 2013 (year 2107).

Problem 78

Step 1 — Part a: Evaluate at \(t = 5\) and \(t = 10\):

$$ A(5) = 1000(1.04)^5 \approx 1000 \times 1.21665 = \$1216.65. $$ $$ A(10) = 1000(1.04)^{10} \approx 1000 \times 1.48024 = \$1480.24. $$

Step 2 — Part b: Find \(t\) such that \(A(t) = 3000\):

$$ 3 = (1.04)^t \Rightarrow t = \frac{\ln 3}{\ln 1.04} = \frac{1.0986}{0.03922} \approx 28.01 \text{ years.} $$

Answer: a) After 5 years: \(\$1216.65\); after 10 years: \(\$1480.24\). b) Approximately \(28\) years.

Problem 79

Step 1 — Part a: Find \(k\) using the doubling time of 12 hours:

At \(t = 12\), \(Q = 2Q_0\):

$$ 2Q_0 = Q_0 e^{12k} \Rightarrow e^{12k} = 2 \Rightarrow k = \frac{\ln 2}{12} = \frac{0.6931}{12} \approx 0.0578. $$

Step 2 — Part b: Find \(t\) when \(Q = 200000\):

$$ 200000 = 1000 e^{0.0578t} \Rightarrow e^{0.0578t} = 200 \Rightarrow t = \frac{\ln 200}{0.0578} = \frac{5.2983}{0.0578} \approx 91.7 \text{ hours.} $$

Answer: a) \(k \approx 0.0578\); b) approximately \(91.7\) hours.

Problem 80

Step 1 — Part a: Find \(a\) using doubling time of 6 months (\(t = 0.5\) years):

At \(t = 0.5\), \(P = 2 \times 120 = 240\):

$$ 240 = 120 a^{0.5} \Rightarrow a^{0.5} = 2 \Rightarrow a = 4. $$

So \(a = 4.0000\).

Step 2 — Part b: Find \(t\) when \(P = 3500\):

$$ 3500 = 120 \cdot 4^t \Rightarrow 4^t = \frac{3500}{120} = 29.167 \Rightarrow t = \frac{\ln(29.167)}{\ln 4} = \frac{3.3730}{1.3863} \approx 2.43 \text{ years.} $$

Answer: a) \(a = 4.0000\); b) approximately \(2.43\) years.

Problem 81

Step 1 — Apply the Richter scale formula:

$$ R_1 - R_2 = \log_{10}\!\left(\frac{A_1}{A_2}\right), \quad 8.3 - 4.9 = 3.4. $$

Step 2 — Compute the intensity ratio:

$$ \frac{A_1}{A_2} = 10^{3.4} \approx 2512. $$

Answer: The San Francisco earthquake was approximately \(2512\) times more intense than the Japanese earthquake.