1.2 Basic Classes of Functions
Learning Objectives
- Calculate the slope of a linear function and interpret its meaning.
- Recognize the degree of a polynomial.
- Find the roots of a quadratic polynomial.
- Describe the graphs of basic odd and even polynomial functions.
- Identify a rational function.
- Describe the graphs of power and root functions.
- Explain the difference between algebraic and transcendental functions.
- Graph a piecewise-defined function.
- Sketch the graph of a function that has been shifted, stretched, or reflected from its initial graph position.
We have studied the general characteristics of functions, so now let's examine some specific classes of functions. We begin by reviewing the basic properties of linear and quadratic functions, then generalize to include higher-degree polynomials. By combining root functions with polynomials, we can define general algebraic functions and distinguish them from the transcendental functions we examine later in this chapter. We finish the section with piecewise-defined functions and a look at how to sketch the graph of a function that has been shifted, stretched, or reflected from its initial form.
1.2.1 Linear Functions and Slope
Consider line \(L\) passing through points \((x_1, y_1)\) and \((x_2, y_2).\) Let \(\Delta y = y_2 - y_1\) and \(\Delta x = x_2 - x_1\) denote the changes in \(y\) and \(x,\) respectively. The slope of the line is
$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\Delta y}{\Delta x}. $$We now examine the relationship between slope and the formula for a linear function. Consider \(f(x) = ax + b.\) Evaluating at \(x = 0\) gives the point \((0, b);\) evaluating at \(x = 1\) gives \((1, a+b).\) The slope between these two points is
$$ \frac{(a+b) - b}{1 - 0} = a. $$So the coefficient \(a\) is exactly the slope. The formula \(f(x) = ax + b\) describes a line with slope \(a\) and \(y\)-intercept \((0, b).\) Since we often use \(m\) for slope, we write this as
$$ f(x) = mx + b. $$This is the slope-intercept form of a linear function. We can also express a linear function in point-slope form when we know the slope and one point. If the line passes through \((x_1, y_1)\) with slope \(m,\) then any other point \((x, f(x))\) satisfies
$$ m = \frac{f(x) - y_1}{x - x_1}, $$which gives
$$ f(x) - y_1 = m(x - x_1). $$A vertical line cannot be described by either of these forms — it is not a function. Instead, a vertical line is described by \(x = k\) for some constant \(k.\) To handle all lines, including vertical ones, we use the standard form of a line:
$$ ax + by = c, $$where \(a\) and \(b\) are not both zero.
Consider a line passing through the point \((x_1, y_1)\) with slope \(m.\) The equation
$$ y - y_1 = m(x - x_1) $$is the point-slope equation for that line.
Consider a line with slope \(m\) and \(y\)-intercept \((0, b).\) The equation
$$ y = mx + b $$is the slope-intercept form for that line.
The standard form of a line is
$$ ax + by = c, $$where \(a\) and \(b\) are not both zero. This form handles vertical lines \(x = k\) as a special case.
Think of the three line forms as three different ID badges for the same line. Slope-intercept (y = mx + b) is the "quick-glance" badge — slope and \(y\)-intercept right on the front. Point-slope is the "incident-report" badge — you were at point \((x_1, y_1)\) heading in direction \(m.\) Standard form is the "official record" badge — works even for vertical lines that make the other two forms break down.
The easiest type of function to consider is a linear function. Linear functions have the form \(f(x) = ax + b,\) where \(a\) and \(b\) are constants. In Figure 1.15, we see examples of linear functions when \(a\) is positive, negative, and zero. Note that if \(a > 0,\) the graph of the line rises as \(x\) increases — in other words, \(f(x) = ax + b\) is increasing on \((-\infty, \infty).\) If \(a < 0,\) the graph falls as \(x\) increases. If \(a = 0,\) the line is horizontal.
Figure 1.15 — These linear functions are increasing or decreasing on \((-\infty,\infty)\) and one function is a horizontal line.
One of the distinguishing features of a line is its slope. The slope measures both the steepness and the direction of a line. To calculate the slope, we choose any two points \((x_1, y_1)\) and \((x_2, y_2)\) on the line and compute the ratio of the change in \(y\) to the change in \(x.\) As Figure 1.16 shows, this ratio is the same no matter which two points we pick.
Figure 1.16 — For any linear function, the slope \((y_2-y_1)/(x_2-x_1)\) is independent of the choice of points \((x_1,y_1)\) and \((x_2,y_2)\) on the line.
Slope is one of the most important numbers in all of calculus. Every derivative you compute in this course is, at its core, a slope — the slope of the curve at a single point. Before we get there, understanding slope for straight lines gives you the intuition you'll carry everywhere. A positive slope means "climbing left to right," a negative slope means "falling left to right," and zero slope means "flat." Simple as that.
Consider the line passing through points \((-3, 2)\) and \((1, 4).\) Find the slope. Find an equation in point-slope form. Find an equation in slope-intercept form.
Solution
Slope: $$ m = \frac{4 - 2}{1 - (-3)} = \frac{2}{4} = \frac{1}{2}. $$
Point-slope form using \((-3, 2):\) $$ y - 2 = \frac{1}{2}(x + 3). $$
Slope-intercept form: $$ y = \frac{1}{2}x + \frac{3}{2} + 2 = \frac{1}{2}x + \frac{7}{2}. $$
Consider the line passing through the points \((11, -4)\) and \((-4, 5),\) as shown in Figure 1.17.
Figure 1.17 — Finding the equation of a linear function with a graph that is a line between two given points.
- Find the slope of the line.
- Find an equation for this linear function in point-slope form.
- Find an equation for this linear function in slope-intercept form.
Solution
Step 1 — Compute the slope:
$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-4)}{-4 - 11} = \frac{9}{-15} = -\frac{3}{5}. $$Step 2 — Point-slope form: Use \(m = -3/5\) and the point \((11, -4):\)
$$ f(x) + 4 = -\frac{3}{5}(x - 11). $$Step 3 — Slope-intercept form: Solve the equation from Step 2 for \(f(x):\)
$$ f(x) = -\frac{3}{5}x + \frac{33}{5} - 4 = -\frac{3}{5}x + \frac{33 - 20}{5} = -\frac{3}{5}x + \frac{13}{5}. $$Answer: Slope \(m = -3/5,\) point-slope form \(f(x) + 4 = -\frac{3}{5}(x - 11),\) slope-intercept form \(f(x) = -\frac{3}{5}x + \frac{13}{5}.\)
Jessica leaves her house at 5:50 a.m. and goes for a 9-mile run. She returns at 7:08 a.m. Answer the following questions, assuming Jessica runs at a constant pace.
- Describe the distance \(D\) (in miles) Jessica runs as a linear function of her run time \(t\) (in minutes).
- Sketch a graph of \(D.\)
- Interpret the meaning of the slope.
Solution
Step 1 — Identify the two data points. At time \(t = 0,\) Jessica is at her house, so \(D(0) = 0.\) The run lasts from 5:50 a.m. to 7:08 a.m., which is \(78\) minutes, so \(D(78) = 9.\) The slope is
$$ m = \frac{9 - 0}{78 - 0} = \frac{3}{26}. $$The \(y\)-intercept is \((0, 0),\) so the linear function is
$$ D(t) = \frac{3}{26}t. $$Step 2 — Sketch: The graph passes through the origin with slope \(m = 3/26.\)
Step 3 — Interpret the slope: \(m = 3/26 \approx 0.115\) describes the distance (in miles) Jessica runs per minute — her average velocity.
1.2.2 Polynomials
A linear function is a special case of a larger class: polynomial functions. A polynomial function can be written in the form
$$ f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 $$for some integer \(n \ge 0\) and constants \(a_n, a_{n-1}, \ldots, a_0,\) where \(a_n \ne 0.\) The value \(n\) is called the degree of the polynomial; the constant \(a_n\) is called the leading coefficient. A polynomial of degree 1 (if \(m \ne 0\)) is a linear function. A polynomial of degree 0 is a constant function. A polynomial of degree 2 is a quadratic function \(f(x) = ax^2 + bx + c\) with \(a \ne 0.\) A polynomial of degree 3 is a cubic function.
Power Functions
Some polynomial functions are power functions. A power function has the form \(f(x) = ax^b,\) where \(a\) and \(b\) are any real numbers. When the exponent is a positive integer \(n,\) the function \(f(x) = ax^n\) is a polynomial. If \(n\) is even, then \(f(x) = ax^n\) is an even function because \(f(-x) = a(-x)^n = ax^n = f(x).\) If \(n\) is odd, then \(f(x) = ax^n\) is an odd function because \(f(-x) = a(-x)^n = -ax^n = -f(x).\)
Figure 1.18 — (a) For any even integer \(n,\) \(f(x)=ax^n\) is an even function. (b) For any odd integer \(n,\) \(f(x)=ax^n\) is an odd function.
Behavior at Infinity
"What does this function do way out on the number line, far from the origin?" That question — the end behavior question — turns out to matter a lot in calculus. When you're analyzing limits, sketching graphs, or deciding whether an improper integral converges, the end behavior of the function is often the first thing you check. Getting comfortable reading end behavior from the leading term now saves real work later.
To understand what happens to a function \(f\) as the inputs grow without bound, we look at the values \(f(x)\) as \(x\) becomes very large or very small. For some functions, those values approach a finite number — we say the function has a horizontal asymptote. For example, for \(f(x) = 2 + 1/x,\) the values \(1/x\) approach zero as \(x\) grows, so \(f(x) \to 2\) as \(x \to \infty.\) The line \(y = 2\) is a horizontal asymptote.
For a polynomial function, the values don't settle at a finite limit — they grow without bound. We describe what happens to \(f(x)\) as \(x \to \infty\) and as \(x \to -\infty\) as the end behavior of the function, and the leading term determines it.
For a quadratic \(f(x) = ax^2 + bx + c:\)
- If \(a > 0:\) \(f(x) \to \infty\) as \(x \to \pm\infty.\) The parabola opens upward.
- If \(a < 0:\) \(f(x) \to -\infty\) as \(x \to \pm\infty.\) The parabola opens downward.
Figure 1.19 — (a) For a quadratic function, if the leading coefficient \(a>0,\) the parabola opens upward. If \(a<0,\) the parabola opens downward. (b) For a cubic function \(f,\) if the leading coefficient \(a>0,\) the values \(f(x)\to\infty\) as \(x\to\infty\) and \(f(x)\to-\infty\) as \(x\to-\infty.\) If \(a<0,\) the opposite is true.
For a cubic \(f(x) = ax^3 + bx^2 + cx + d:\)
- If \(a > 0:\) \(f(x) \to \infty\) as \(x \to \infty\) and \(f(x) \to -\infty\) as \(x \to -\infty.\)
- If \(a < 0:\) \(f(x) \to -\infty\) as \(x \to \infty\) and \(f(x) \to \infty\) as \(x \to -\infty.\)
The pattern holds for polynomials of any degree: the leading term dominates everything else when \(|x|\) is large.
Zeros of Polynomial Functions
Another key feature of a polynomial graph is where it crosses the \(x\)-axis. To find those zeros (also called roots or \(x\)-intercepts), we solve \(f(x) = 0.\)
For a linear function \(f(x) = mx + b,\) the zero is at \(x = -b/m.\) For a quadratic \(ax^2 + bx + c = 0,\) we can factor (when possible) or use the quadratic formula:
Rule: The Quadratic Formula
For the quadratic equation \(ax^2 + bx + c = 0\) with \(a \ne 0,\) the solutions are
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$- If \(b^2 - 4ac > 0:\) two distinct real solutions.
- If \(b^2 - 4ac = 0:\) one real solution (a repeated root).
- If \(b^2 - 4ac < 0:\) no real solutions.
The quantity \(b^2 - 4ac\) is called the discriminant. Reading its sign tells you right away how many real roots exist, before you do any arithmetic.
Consider the quadratic function \(f(x) = 3x^2 - 6x + 2.\) Find the zeros of \(f.\) Does the parabola open upward or downward?
Solution
Zeros: Apply the quadratic formula with \(a = 3,\) \(b = -6,\) \(c = 2:\)
$$ x = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}. $$Direction: Since \(a = 3 > 0,\) the parabola opens upward.
For each of the following functions, (i) describe the end behavior, (ii) find all zeros, and (iii) sketch a graph.
a. \(f(x) = -2x^2 + 4x - 1\)
b. \(f(x) = x^3 - 3x^2 - 4x\)
Solution
Part a. \(f(x) = -2x^2 + 4x - 1\)
Step 1 — End behavior. The leading coefficient is \(a = -2 < 0.\) Since the degree is even, \(f(x) \to -\infty\) as \(x \to \pm\infty.\)
Step 2 — Zeros. Use the quadratic formula with \(a = -2,\) \(b = 4,\) \(c = -1:\)
$$ x = \frac{-4 \pm \sqrt{16 - 4(-2)(-1)}}{2(-2)} = \frac{-4 \pm \sqrt{16 - 8}}{-4} = \frac{-4 \pm 2\sqrt{2}}{-4} = \frac{2 \pm \sqrt{2}}{2}. $$Step 3 — Sketch. The parabola opens downward, has two \(x\)-intercepts at \(\frac{2 + \sqrt{2}}{2} \approx 1.71\) and \(\frac{2 - \sqrt{2}}{2} \approx 0.29,\) and vertex between them.
Part b. \(f(x) = x^3 - 3x^2 - 4x\)
Step 1 — End behavior. The leading coefficient is \(a = 1 > 0.\) Since the degree is odd, \(f(x) \to \infty\) as \(x \to \infty\) and \(f(x) \to -\infty\) as \(x \to -\infty.\)
Step 2 — Zeros. Factor out \(x:\)
$$ f(x) = x(x^2 - 3x - 4). $$Factor the quadratic:
$$ f(x) = x(x - 4)(x + 1). $$The zeros are \(x = 0, 4, -1.\)
Step 3 — Sketch. The cubic rises from lower-left to upper-right, crossing the \(x\)-axis at \(-1,\) \(0,\) and \(4.\)
Mathematical Models
A mathematical model is a method of simulating real-life situations with mathematical equations. Physicists, engineers, economists, and researchers develop models by combining observation with quantitative data. Models are useful because they help predict future outcomes. The study of population dynamics, weather patterns, and product sales are all examples.
As a concrete example, consider how a company might model its revenue. The revenue \(R\) from selling \(n\) items at price \(p\) dollars each is \(R = p \cdot n.\) Suppose the data in Table 1.2.1 show how many units a company sells at different prices.
| Price per item \(p\) (\$) | Number sold \(n\) (thousands) |
|---|---|
| 1 | 24.86 |
| 5 | 20.79 |
| 8 | 17.07 |
| 10 | 14.99 |
| 13 | 11.97 |
| 17 | 8.27 |
| 20 | 5.34 |
From Figure 1.20, the number of units sold looks approximately linear in price, well approximated by \(n = -1.04p + 26\) for \(0 \le p \le 25.\) Using this, the revenue (in thousands of dollars) is
$$ R(p) = p \cdot (-1.04p + 26) = -1.04p^2 + 26p. $$
Figure 1.20 — The data collected for the number of items sold as a function of price is roughly linear. We use the linear function \(n=-1.04p+26\) to estimate this function.
A company models its revenue \(R\) (in thousands of dollars) as a function of price per item \(p\) by
$$ R(p) = p \cdot (-1.04p + 26) = -1.04p^2 + 26p $$for \(0 \le p \le 25.\)
- Predict the revenue if the company sells the item at \(p = \$5\) and \(p = \$17.\)
- Find the zeros of this function and interpret their meaning.
- Sketch a graph of \(R.\)
- Use the graph to determine the value of \(p\) that maximizes revenue. Find the maximum revenue.
Solution
Step 1 — Evaluate at \(p = 5\) and \(p = 17:\)
$$ R(5) = -1.04(25) + 26(5) = -26 + 130 = 104, \quad \text{so revenue} = \$104{,}000. $$ $$ R(17) = -1.04(289) + 26(17) = -300.56 + 442 = 141.44, \quad \text{so revenue} \approx \$141{,}440. $$Step 2 — Zeros. Solve \(-1.04p^2 + 26p = 0.\) Factor: \(p(-1.04p + 26) = 0.\) Solutions: \(p = 0\) and \(p = 25.\)
Interpretation: at \(p = \$0,\) no revenue because the company gives the item away. At \(p = \$25,\) no revenue because the price is so high that no one buys.
Step 3 — Sketch. The parabola opens downward (leading coefficient \(-1.04 < 0\)). Zeros at \(p = 0\) and \(p = 25,\) so the axis of symmetry is at \(p = 12.5.\)
Step 4 — Maximum. By symmetry, the vertex (maximum) occurs at \(p = 12.5.\) The maximum revenue is
$$ R(12.5) = -1.04(12.5)^2 + 26(12.5) = -1.04(156.25) + 325 = -162.5 + 325 = 162.5. $$Answer: Maximum revenue of \(\$162{,}500\) at a price of \(\$12.50\) per item.
1.2.3 Algebraic Functions
By allowing quotients and fractional powers in polynomial expressions, we create a larger class: algebraic functions. An algebraic function involves addition, subtraction, multiplication, division, rational powers, and roots. The two main types are rational functions and root functions.
Just as rational numbers are quotients of integers, rational functions are quotients of polynomials. Any function of the form \(f(x) = p(x)/q(x),\) where \(p(x)\) and \(q(x)\) are polynomials, is a rational function. For example,
$$ f(x) = \frac{3x - 1}{5x + 2} \qquad \text{and} \qquad g(x) = \frac{4}{x^2 + 1} $$are rational functions. A root function is a power function \(f(x) = x^{1/n}\) where \(n\) is a positive integer greater than one. For example, \(f(x) = \sqrt{x} = x^{1/2}\) is the square-root function, and \(g(x) = \sqrt[3]{x} = x^{1/3}\) is the cube-root function. Compositions of root functions and rational functions are also algebraic — for instance, \(f(x) = \sqrt{4 - x^2}\) is algebraic.
Rational functions show up constantly in applications: cost-per-unit formulas, average-rate expressions, concentration in mixture problems. The key skill is finding the domain — wherever the denominator is zero, the function is undefined. Root functions appear in geometry and physics (distance, velocity, area formulas). Before you can differentiate or integrate any of these, you need to know where they live (domain) and where they can go (range).
Find the domain and range for the function \(f(x) = \dfrac{5x + 2}{2x - 1}.\)
Solution
Domain: We need \(2x - 1 \ne 0,\) so \(x \ne 1/2.\) Domain: \(\{x \mid x \ne 1/2\}.\)
Range: Set \(y = \frac{5x+2}{2x-1}\) and solve for \(x:\) \(y(2x-1) = 5x+2,\) so \(2xy - y = 5x + 2,\) giving \(x(2y - 5) = y + 2.\) If \(y = 5/2,\) no solution. Otherwise \(x = \frac{y+2}{2y-5}.\) Range: \(\{y \mid y \ne 5/2\}.\)
The root functions \(f(x) = x^{1/n}\) have different domains depending on whether \(n\) is odd or even.
- Even \(n \ge 2:\) domain is \([0, \infty)\) (we can't take the square root of a negative number in the reals).
- Odd \(n \ge 1:\) domain is \((-\infty, \infty)\) (any real number has a real odd root).
Since \(x^{1/n} = -(-x)^{1/n}\) for odd \(n,\) the odd root function is an odd function. See Figure 1.21 for graphs of root functions.
![Figure 1.21 — (a) If \(n\) is even, the domain of \(f(x)=\sqrt[n]{x}\) is \([0,\infty).\) (b) If \(n\) is odd, the domain of \(f(x)=\sqrt[n]{x}\) is \((-\infty,\infty)\) and the function \(f(x)=\sqrt[n]{x}\) is an odd function.](../images/figure_1.21_light.png?v=20260528b)
Figure 1.21 — (a) If \(n\) is even, the domain of \(f(x)=\sqrt[n]{x}\) is \([0,\infty).\) (b) If \(n\) is odd, the domain of \(f(x)=\sqrt[n]{x}\) is \((-\infty,\infty)\) and the function is odd.
For each of the following functions, find the domain and range.
- \(f(x) = \dfrac{3x - 1}{5x + 2}\)
- \(f(x) = \sqrt{4 - x^2}\)
Solution
Part 1 — Domain: Division by zero is forbidden, so we need \(5x + 2 \ne 0,\) i.e., \(x \ne -2/5.\) Domain: \(\{x \mid x \ne -2/5\}.\)
Part 1 — Range: We want to find all values \(y\) such that \(y = \frac{3x-1}{5x+2}\) has a real solution for \(x.\) Multiply both sides by \(5x + 2:\)
$$ 5xy + 2y = 3x - 1. $$Rearrange to isolate \(x:\)
$$ 2y + 1 = x(3 - 5y). $$If \(y = 3/5,\) the right side is zero but the left side is \(2(3/5) + 1 = 11/5 \ne 0,\) so no solution exists. For \(y \ne 3/5,\)
$$ x = \frac{2y + 1}{3 - 5y}, $$which always exists. Range: \(\{y \mid y \ne 3/5\}.\)
Part 2 — Domain: We need \(4 - x^2 \ge 0.\) The quadratic equals zero at \(x = \pm 2.\) Solving \((2-x)(2+x) > 0,\) both factors must have the same sign.
Case 1 — Both positive: \(x < 2\) and \(x > -2,\) so \(-2 < x < 2.\) Case 2 — Both negative: \(x > 2\) and \(x < -2,\) which is impossible.
Including the boundary points where \(4 - x^2 = 0:\) domain is \(\{x \mid -2 \le x \le 2\}.\)
Part 2 — Range: For \(-2 \le x \le 2,\) we have \(0 \le 4 - x^2 \le 4,\) so \(0 \le \sqrt{4 - x^2} \le 2.\) Range: \(\{y \mid 0 \le y \le 2\}.\)
Find the domain for each of the following functions: \(f(x) = \dfrac{5 - 2x}{x^2 + 2}\) and \(g(x) = \sqrt{5x - 1}.\)
Solution
For \(f:\) The denominator \(x^2 + 2 \ge 2 > 0\) for all real \(x,\) so the domain is \((-\infty, \infty).\)
For \(g:\) We need \(5x - 1 \ge 0,\) so \(x \ge 1/5.\) Domain: \([1/5, \infty).\)
For each of the following functions, determine the domain.
- \(f(x) = \dfrac{3}{x^2 - 1}\)
- \(f(x) = \dfrac{2x + 5}{3x^2 + 4}\)
- \(f(x) = \sqrt{4 - 3x}\)
- \(f(x) = \sqrt[3]{2x - 1}\)
Solution
1. We need \(x^2 - 1 \ne 0,\) i.e., \(x \ne \pm 1.\) Domain: \(\{x \mid x \ne \pm 1\}.\)
2. Check whether the denominator can be zero: \(3x^2 + 4 \ge 4 > 0\) for all real \(x.\) The denominator is never zero. Domain: \((-\infty, \infty).\)
3. We need \(4 - 3x \ge 0,\) so \(x \le 4/3.\) Domain: \(\{x \mid x \le 4/3\}.\)
4. The cube root is defined for all real numbers. Domain: \((-\infty, \infty).\)
1.2.4 Transcendental Functions
Thus far we have discussed algebraic functions. Some functions cannot be described by basic algebraic operations — these are called transcendental functions because they "transcend," or go beyond, algebra. The most common transcendental functions are trigonometric, exponential, and logarithmic functions.
- A trigonometric function relates the ratios of two sides of a right triangle: \(\sin x,\) \(\cos x,\) \(\tan x,\) \(\cot x,\) \(\sec x,\) and \(\csc x.\)
- An exponential function has the form \(f(x) = b^x,\) where the base \(b > 0,\) \(b \ne 1.\)
- A logarithmic function has the form \(f(x) = \log_b(x)\) for some constant \(b > 0,\) \(b \ne 1,\) where \(\log_b(x) = y\) if and only if \(b^y = x.\)
We examine all three types in detail later in this chapter.
The dividing line between algebraic and transcendental is this: algebraic functions can only involve rational-number exponents — you can add, subtract, multiply, divide, and raise to rational powers. The moment you raise to a variable exponent (\(2^x\)), or introduce a circular relationship (\(\sin x\)), you've left the algebraic world. This matters in calculus because transcendental functions require genuinely new differentiation rules.
Is \(f(x) = x/2\) an algebraic or a transcendental function?
Solution
\(f(x) = x/2\) involves only division by a constant — a basic algebraic operation. It is algebraic (in fact, it is a linear function with slope \(1/2\)).
Classify each of the following functions as algebraic or transcendental.
- \(f(x) = \dfrac{\sqrt{x^3 + 1}}{4x + 2}\)
- \(f(x) = 2^{x^2}\)
- \(f(x) = \sin(2x)\)
Solution
1. This function involves only basic algebraic operations (root, polynomial, quotient), so it is algebraic.
2. The exponent is the variable \(x^2,\) which cannot be expressed using only rational-power algebraic operations. This function is transcendental.
3. This function involves a trigonometric operation, which cannot be written using only algebraic operations. This function is transcendental.
1.2.5 Piecewise-Defined Functions
Sometimes a function is defined by different formulas on different parts of its domain. A function with this property is a piecewise-defined function. The absolute value function is a classic example:
$$ f(x) = |x| = \begin{cases} -x, & x < 0 \\ x, & x \ge 0. \end{cases} $$Other piecewise-defined functions may use entirely different formulas on each piece. To graph a piecewise function, we graph each piece on its respective domain on the same coordinate system, using open circles where a piece does not include its endpoint and closed circles where it does.
Sketch a graph of the function
$$ f(x) = \begin{cases} 2 - x, & x \le 2 \\ x + 2, & x > 2. \end{cases} $$Solution
For \(x \le 2:\) the line \(y = 2 - x.\) At \(x = 2,\) \(f(2) = 0\) — draw a closed circle at \((2, 0).\) For \(x > 2:\) the line \(y = x + 2.\) At \(x = 2,\) this piece gives \(4,\) but it doesn't include \(x = 2,\) so draw an open circle at \((2, 4).\) The two lines meet at the break \(x = 2\) but take different values there, creating a jump discontinuity.
Sketch a graph of the following piecewise-defined function:
$$ f(x) = \begin{cases} x + 3, & x < 1 \\ (x - 2)^2, & x \ge 1. \end{cases} $$Solution
Step 1 — Identify the pieces. For \(x < 1:\) linear piece \(y = x + 3.\) For \(x \ge 1:\) quadratic piece \(y = (x-2)^2.\)
Step 2 — Check the boundary. At \(x = 1,\) the formula \(f(x) = (x-2)^2\) applies, giving \(f(1) = (1-2)^2 = 1.\) Draw a closed circle at \((1, 1).\) The linear piece gives \(x + 3 = 4\) at \(x = 1,\) but that piece is only valid for \(x < 1,\) so draw an open circle at \((1, 4).\)
Figure 1.22 — This piecewise-defined function is linear for \(x<1\) and quadratic for \(x\ge1.\)
The cost of mailing a letter is a function of the weight of the letter. Suppose the cost of mailing a letter is \(49 \text{¢}\) for the first ounce and \(21 \text{¢}\) for each additional ounce. Write a piecewise-defined function describing the cost \(C\) (in cents) as a function of weight \(x\) (in ounces) for \(0 < x \le 3.\)
Solution
For the first ounce, the cost is \(49 \text{¢}.\) Each additional ounce adds \(21 \text{¢},\) so the second ounce gives \(49 + 21 = 70 \text{¢}\) and the third gives \(70 + 21 = 91 \text{¢}.\)
In a big city, drivers are charged variable rates for parking in a garage. They are charged $10 for the first hour or any part of the first hour and an additional $2 for each hour or part thereof, up to a maximum of $30 for the day. The garage is open from 6 a.m. to 12 midnight.
- Write a piecewise-defined function that describes the cost \(C\) to park as a function of hours parked \(x.\)
- Sketch a graph of \(C(x).\)
Solution
Step 1 — Domain and formula. The garage is open for 18 hours, so the domain is \(\{x \mid 0 < x \le 18\}.\) The cost increases in whole-dollar steps:
$$ C(x) = \begin{cases} 10, & 0 < x \le 1 \\ 12, & 1 < x \le 2 \\ 14, & 2 < x \le 3 \\ 16, & 3 < x \le 4 \\ \vdots & \\ 30, & 10 < x \le 18. \end{cases} $$Step 2 — Graph. The graph consists of horizontal line segments, one for each one-hour band. Each segment starts with an open circle on the left (the previous hour's endpoint is not included in this band) and ends with a closed circle on the right.
1.2.6 Transformations of Functions
We have seen several cases where we added, subtracted, or multiplied constants to form variations of simple functions. These adjustments are called transformations of a function. Understanding them lets you sketch the graph of any transformed function from the graph of its base function, without computing a table of values.
Vertical and Horizontal Shifts
A vertical shift adds or subtracts a constant from each output. For \(c > 0:\)
- \(f(x) + c\) shifts the graph up \(c\) units.
- \(f(x) - c\) shifts the graph down \(c\) units.
Figure 1.23 — (a) For \(c>0,\) the graph of \(y=f(x)+c\) is a vertical shift up \(c\) units. (b) For \(c>0,\) the graph of \(y=f(x)-c\) is a vertical shift down \(c\) units.
A horizontal shift adds or subtracts a constant from each input. For \(c > 0:\)
- \(f(x + c)\) shifts the graph left \(c\) units.
- \(f(x - c)\) shifts the graph right \(c\) units.
Why does adding \(c\) inside the argument shift left? Consider \(f(x) = |x + 3|.\) The value \(f(0) = 3\) is the same as \(|x|\) at \(x = 3.\) So the graph of \(|x + 3|\) reaches the same height as \(|x|\) but does so 3 units to the left. Similarly, \(|x - 3|\) shifts 3 units to the right (Figure 1.24).
Figure 1.24 — (a) For \(c>0,\) the graph of \(y=f(x+c)\) is a horizontal shift left \(c\) units. (b) For \(c>0,\) the graph of \(y=f(x-c)\) is a horizontal shift right \(c\) units.
Vertical and Horizontal Scalings
A vertical scaling multiplies all outputs by a positive constant \(c.\) If \(c > 1,\) the graph is stretched vertically; if \(0 < c < 1,\) it is compressed vertically (Figure 1.25).
Figure 1.25 — (a) If \(c>1,\) the graph of \(y=cf(x)\) is a vertical stretch. (b) If \(0<c<1,\) the graph of \(y=cf(x)\) is a vertical compression.
A horizontal scaling multiplies all inputs by a positive constant \(c.\) If \(c > 1,\) the graph is compressed horizontally; if \(0 < c < 1,\) it is stretched horizontally (Figure 1.26). Note that the direction is opposite to vertical scaling: multiplying inputs by a number greater than 1 actually compresses (squishes toward the \(y\)-axis), while multiplying by a fraction stretches.
Figure 1.26 — (a) If \(c>1,\) the graph of \(y=f(cx)\) is a horizontal compression. (b) If \(0<c<1,\) the graph of \(y=f(cx)\) is a horizontal stretch.
Reflections
When the constant \(c\) in a scaling is negative:
- \(-f(x)\): multiply all outputs by \(-1\) → reflection about the \(x\)-axis.
- \(f(-x)\): multiply all inputs by \(-1\) → reflection about the \(y\)-axis.
For example, \(y = -(x^3 + 1)\) is \(y = x^3 + 1\) reflected about the \(x\)-axis, and \(y = (-x)^3 + 1\) is \(y = x^3 + 1\) reflected about the \(y\)-axis (Figure 1.27).
Figure 1.27 — (a) The graph of \(y=-f(x)\) is reflected about the \(x\)-axis. (b) The graph of \(y=f(-x)\) is reflected about the \(y\)-axis.
Order of Transformations
When multiple transformations are combined, the order matters. To graph \(y = cf(a(x + b)) + d\) from the graph of \(y = f(x),\) apply the transformations in this order:
- Horizontal shift: shift left by \(b\) if \(b > 0,\) right if \(b < 0.\)
- Horizontal scaling: scale by \(|a|.\) If \(a < 0,\) also reflect about the \(y\)-axis.
- Vertical scaling: scale by \(|c|.\) If \(c < 0,\) also reflect about the \(x\)-axis.
- Vertical shift: shift up by \(d\) if \(d > 0,\) down if \(d < 0.\)
We can summarize the different transformations in the table below.
| Transformation | Effect on graph |
|---|---|
| \(f(x) + c,\quad c > 0\) | Shift up \(c\) units |
| \(f(x) - c,\quad c > 0\) | Shift down \(c\) units |
| \(f(x + c),\quad c > 0\) | Shift left \(c\) units |
| \(f(x - c),\quad c > 0\) | Shift right \(c\) units |
| \(cf(x),\quad c > 1\) | Vertical stretch by factor \(c\) |
| \(cf(x),\quad 0 < c < 1\) | Vertical compression by factor \(c\) |
| \(f(cx),\quad c > 1\) | Horizontal compression by factor \(c\) |
| \(f(cx),\quad 0 < c < 1\) | Horizontal stretch by factor \(c\) |
| \(-f(x)\) | Reflection about \(x\)-axis |
| \(f(-x)\) | Reflection about \(y\)-axis |
Describe how the function \(f(x) = -(x + 1)^2 - 4\) can be graphed using the graph of \(y = x^2\) and a sequence of transformations.
Solution
Start with \(y = x^2.\)
- Replace \(x\) with \(x + 1\) → shift left 1 unit: \(y = (x+1)^2.\)
- Multiply output by \(-1\) → reflect about \(x\)-axis: \(y = -(x+1)^2.\)
- Subtract \(4\) → shift down 4 units: \(y = -(x+1)^2 - 4.\)
The result is a downward-opening parabola with vertex at \((-1, -4).\)
For each of the following functions, sketch a graph using a sequence of transformations of a well-known function.
- \(f(x) = -|x + 2| - 3\)
- \(f(x) = 3\sqrt{-x} + 1\)
Solution
Part 1 — \(f(x) = -|x + 2| - 3:\)
Start with \(y = |x|.\)
- Step 1: Replace \(x\) with \(x + 2\) → shift left 2 units: \(y = |x + 2|.\) - Step 2: Multiply output by \(-1\) → reflect about \(x\)-axis: \(y = -|x + 2|.\) - Step 3: Subtract \(3\) from output → shift down 3 units: \(y = -|x + 2| - 3.\)
Figure 1.28 — The function \(f(x)=-|x+2|-3\) can be viewed as a sequence of three transformations of \(y=|x|.\)
Part 2 — \(f(x) = 3\sqrt{-x} + 1:\)
Start with \(y = \sqrt{x}.\)
- Step 1: Replace \(x\) with \(-x\) → reflect about \(y\)-axis: \(y = \sqrt{-x}.\) - Step 2: Multiply output by \(3\) → vertical stretch by factor 3: \(y = 3\sqrt{-x}.\) - Step 3: Add \(1\) to output → shift up 1 unit: \(y = 3\sqrt{-x} + 1.\)
Figure 1.29 — The function \(f(x)=3\sqrt{-x}+1\) can be viewed as a sequence of three transformations of \(y=\sqrt{x}.\)
Problem Set 1.2
For the following exercises, for each pair of points, a. find the slope of the line passing through the points and b. indicate whether the line is increasing, decreasing, horizontal, or vertical.
Problem 1. \((-2, 4)\) and \((1, 1)\)
Problem 2. \((-1, 4)\) and \((3, -1)\)
Problem 3. \((3, 5)\) and \((-1, 2)\)
Problem 4. \((6, 4)\) and \((4, -3)\)
Problem 5. \((2, 3)\) and \((5, 7)\)
Problem 6. \((1, 9)\) and \((-8, 5)\)
Problem 7. \((2, 4)\) and \((1, 4)\)
Problem 8. \((1, 4)\) and \((1, 0)\)
Solutions 1–8
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.
For the following exercises, write the equation of the line satisfying the given conditions in slope-intercept form.
Problem 9. Slope \(= -6,\) passes through \((1, 3)\)
Problem 10. Slope \(= 3,\) passes through \((-3, 2)\)
Problem 11. Slope \(= \dfrac{1}{3},\) passes through \((0, 4)\)
Problem 12. Slope \(= \dfrac{2}{5},\) \(x\)-intercept \(= 8\)
Problem 13. Passing through \((2, 1)\) and \((-2, -1)\)
Problem 14. Passing through \((-3, 7)\) and \((1, 2)\)
Problem 15. \(x\)-intercept \(= 5\) and \(y\)-intercept \(= -3\)
Problem 16. \(x\)-intercept \(= -6\) and \(y\)-intercept \(= 9\)
Solutions 9–16
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: \(y = \frac{3}{2}x + 9.\)
For the following exercises, for each linear equation, a. give the slope \(m\) and \(y\)-intercept \(b,\) if any, and b. graph the line.
Problem 17. \(y = 2x - 3\)
Problem 18. \(y = -\dfrac{1}{7}x + 1\)
Problem 19. \(f(x) = -6x\)
Problem 20. \(f(x) = -5x + 4\)
Problem 21. \(4y + 24 = 0\)
Problem 22. \(8x - 4 = 0\)
Problem 23. \(2x + 3y = 6\)
Problem 24. \(6x - 5y + 15 = 0\)
Solutions 17–24
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 \((0, 3).\)
For the following exercises, for each polynomial, a. find the degree; b. find the zeros, if any; c. find the \(y\)-intercept(s), if any; d. use the leading coefficient to determine the graph's end behavior; and e. determine algebraically whether the polynomial is even, odd, or neither.
Problem 25. \(f(x) = 2x^2 - 3x - 5\)
Problem 26. \(f(x) = -3x^2 + 6x\)
Problem 27. \(f(x) = \dfrac{1}{2}x^2 - 1\)
Problem 28. \(f(x) = x^3 + 3x^2 - x - 3\)
Problem 29. \(f(x) = 3x - x^3\)
Solutions 25–29
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.
For the following exercises, use the graph of \(f(x) = x^2\) to graph each transformed function \(g.\)
Problem 30. \(g(x) = x^2 - 1\)
Problem 31. \(g(x) = (x + 3)^2 + 1\)
Solutions 30–31
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 \((-3, 1).\)
For the following exercises, use the graph of \(f(x) = \sqrt{x}\) to graph each transformed function \(g.\)
Problem 32. \(g(x) = \sqrt{x + 2}\)
Problem 33. \(g(x) = -\sqrt{x} - 1\)
Solutions 32–33
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 \(x \ge 0.\)
For the following exercises, use the graph of \(y = f(x)\) to graph each transformed function \(g.\)
Problem 34. \(g(x) = f(x) + 1\)
Problem 35. \(g(x) = f(x - 1) + 2\)
Solutions 34–35
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.
For the following exercises, for each of the piecewise-defined functions, a. evaluate at the given values of the independent variable and b. sketch the graph.
Problem 36. \(f(x) = \begin{cases} 4x + 3, & x \le 0 \\ -x + 1, & x > 0 \end{cases};\quad f(-3);\ f(0);\ f(2)\)
Problem 37. \(f(x) = \begin{cases} x^2 - 3, & x < 0 \\ 4x - 3, & x \ge 0 \end{cases};\quad f(-4);\ f(0);\ f(2)\)
Problem 38. \(h(x) = \begin{cases} x + 1, & x \le 5 \\ 4, & x > 5 \end{cases};\quad h(0);\ h(\pi);\ h(5)\)
Problem 39. \(g(x) = \begin{cases} \dfrac{3}{x - 2}, & x \ne 2 \\ 4, & x = 2 \end{cases};\quad g(0);\ g(-4);\ g(2)\)
Solutions 36–39
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 \((2, 4).\)
For the following exercises, determine whether the statement is true or false. Explain why.
Problem 40. \(f(x) = (4x + 1)/(7x - 2)\) is a transcendental function.
Problem 41. \(g(x) = \sqrt[3]{x}\) is an odd root function.
Problem 42. A logarithmic function is an algebraic function.
Problem 43. A function of the form \(f(x) = x^b,\) where \(b\) is a real-valued constant, is an exponential function.
Problem 44. The domain of an even root function is all real numbers.
Problem 45. [T] A company purchases computer equipment for $20,500. At the end of a 3-year period, the value has decreased linearly to $12,300.
a) Find a function \(y = V(t)\) that determines the value \(V\) of the equipment at the end of \(t\) years.
b) Find and interpret the meaning of the \(x\)- and \(y\)-intercepts for this situation.
c) What is the value of the equipment at the end of 5 years?
d) When will the value of the equipment be $3,000?
Problem 46. [T] Total online shopping during the Christmas holidays has increased dramatically in recent years. In 2012 \((t = 0),\) total online holiday sales were $42.3 billion; in 2013 they were $48.1 billion.
a) Find a linear function \(S\) that estimates the total online holiday sales in year \(t.\)
b) Interpret the slope of the graph of \(S.\)
c) Use part a. to predict the year when online shopping during Christmas will reach $60 billion.
Problem 47. [T] A family bakery makes cupcakes and sells them at local outdoor festivals. For a music festival, there is a fixed cost of $125 to set up a stand. The owner estimates it costs $0.75 to make each cupcake.
a) Find a linear function that relates cost \(C\) to \(x,\) the number of cupcakes made.
b) Find the cost to bake 160 cupcakes.
c) If the owner sells the cupcakes for $1.50 apiece, how many cupcakes does she need to sell to start making profit?
Problem 48. [T] A house purchased for $250,000 is expected to be worth twice its purchase price in 18 years.
a) Find a linear function that models the price \(P\) of the house versus the number of years \(t\) since the original purchase.
b) Interpret the slope of the graph of \(P.\)
c) Find the price of the house 15 years from the original purchase.
Problem 49. [T] A car was purchased for $26,000. The value of the car depreciates by $1,500 per year.
a) Find a linear function that models the value \(V\) of the car after \(t\) years.
b) Find and interpret \(V(4).\)
Problem 50. [T] A condominium in an upscale part of the city was purchased for $432,000. In 35 years it is worth $60,500. Find the rate of depreciation.
Problem 51. [T] The total cost \(C\) to produce a certain item is modeled by the function \(C(x) = 10.50x + 28{,}500,\) where \(x\) is the number of items produced. Determine the cost to produce 175 items.
Problem 52. [T] A professor asks her class to report the amount of time \(t\) they spent writing two assignments. Most students report that it takes about 45 minutes to type a four-page assignment and about 1.5 hours to type a nine-page assignment.
a) Find the linear function \(y = N(t)\) that models this situation, where \(N\) is the number of pages typed and \(t\) is the time in minutes.
b) Use part a. to determine how many pages can be typed in 2 hours.
c) Use part a. to determine how long it takes to type a 20-page assignment.
Problem 53. [T] The output (as a percent of total capacity) of nuclear power plants in the United States can be modeled by the function \(P(t) = 1.8576t + 68.052,\) where \(t\) is time in years and \(t = 0\) corresponds to the beginning of 2000. Use the model to predict the approximate percentage output at the beginning of 2015.
Problem 54. [T] The admissions office at a public university estimates that 65% of the students offered admission to the class of 2019 will actually enroll.
a) Find the linear function \(y = N(x),\) where \(N\) is the number of students that actually enroll and \(x\) is the number of all students offered admission.
b) If the university wants the 2019 freshman class size to be 1,350, determine how many students should be offered admission.
Solutions 40–54
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.
Key Terms
linear function — a function of the form \(f(x) = ax + b\)
slope — the ratio \(\Delta y / \Delta x\) measuring steepness and direction of a line
slope-intercept form — \(y = mx + b\)
point-slope equation — \(y - y_1 = m(x - x_1)\)
standard form of a line — \(ax + by = c\)
polynomial function — a function expressible as a finite sum of terms \(a_k x^k\)
degree — the highest power \(n\) with nonzero coefficient in a polynomial
leading coefficient — the coefficient \(a_n\) of the highest-degree term
constant function — a polynomial of degree 0
quadratic function — a polynomial of degree 2
cubic function — a polynomial of degree 3
power function — a function of the form \(f(x) = ax^b\)
end behavior — the behavior of \(f(x)\) as \(x \to \pm\infty\)
mathematical model — a mathematical description of a real-world situation
algebraic function — a function built from basic arithmetic operations and rational powers
rational function — a quotient of two polynomials
root function — a function of the form \(f(x) = x^{1/n}\)
transcendental function — a function that cannot be expressed using only algebraic operations
logarithmic function — a function of the form \(f(x) = \log_b(x)\)
piecewise-defined function — a function defined by different formulas on different parts of its domain
transformation of a function — a shift, scaling, or reflection that maps the graph of one function to another