Chapter 3

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Step 1 — Compute \(f(-1)\):

$$f(-1) = \frac{1}{-1-3} = -\frac{1}{4}$$

Step 2 — Form the difference quotient:

$$\frac{f(x)-f(-1)}{x-(-1)} = \frac{\dfrac{1}{x-3}+\dfrac{1}{4}}{x+1} = \frac{\dfrac{4+(x-3)}{4(x-3)}}{x+1} = \frac{x+1}{4(x-3)(x+1)} = \frac{1}{4(x-3)}$$

Step 3 — Take the limit:

$$f'(-1) = \lim_{x \to -1} \frac{1}{4(x-3)} = \frac{1}{4(-4)} = -\frac{1}{16}$$

Answer: \(f'(-1) = -\dfrac{1}{16}\).

Problem 29

Step 1 — Compute \(f(1)\):

$$f(1) = 1$$

Step 2 — Form the difference quotient and factor:

$$\frac{\dfrac{1}{x^3}-1}{x-1} = \frac{1-x^3}{x^3(x-1)} = \frac{-(x^3-1)}{x^3(x-1)} = \frac{-(x-1)(x^2+x+1)}{x^3(x-1)} = \frac{-(x^2+x+1)}{x^3}$$

Step 3 — Take the limit:

$$f'(1) = \lim_{x \to 1} \frac{-(x^2+x+1)}{x^3} = \frac{-(1+1+1)}{1} = -3$$

Answer: \(f'(1) = -3\).

Problem 30

Step 1 — Compute \(f(4)\):

$$f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2}$$

Step 2 — Form the difference quotient and find a common denominator:

$$\frac{\dfrac{1}{\sqrt{x}}-\dfrac{1}{2}}{x-4} = \frac{\dfrac{2-\sqrt{x}}{2\sqrt{x}}}{x-4} = \frac{2-\sqrt{x}}{2\sqrt{x}(x-4)}$$

Factor \(x-4 = (\sqrt{x}-2)(\sqrt{x}+2)\) and note \(2-\sqrt{x} = -(\sqrt{x}-2)\):

$$= \frac{-(\sqrt{x}-2)}{2\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+2)} = \frac{-1}{2\sqrt{x}(\sqrt{x}+2)}$$

Step 3 — Take the limit:

$$f'(4) = \lim_{x \to 4} \frac{-1}{2\sqrt{x}(\sqrt{x}+2)} = \frac{-1}{2(2)(4)} = -\frac{1}{16}$$

Answer: \(f'(4) = -\dfrac{1}{16}\).

Problem 31

Step 1 — Derive a formula for the secant slope.

The slope of \(PQ\) with \(P=(1,8)\) is:

$$m_{PQ} = \frac{f(x)-f(1)}{x-1} = \frac{x^2+3x+4-8}{x-1} = \frac{(x-1)(x+4)}{x-1} = x+4 \quad (x \neq 1)$$

Step 2 — Fill in the table using \(m_{PQ} = x+4\):

\(x\)	\(m_{PQ}\)	\(x\)	\(m_{PQ}\)
1.1	5.100000	0.9	4.900000
1.01	5.010000	0.99	4.990000
1.001	5.001000	0.999	4.999000
1.0001	5.000100	0.9999	4.999900
1.00001	5.000010	0.99999	4.999990
1.000001	5.000001	0.999999	4.999999

Step 3 — Estimate the tangent slope:

Both columns converge to \(5\), so \(m_{\tan} = f'(1) = 5\).

Step 4 — Write the tangent line at \(P(1,8)\):

$$y - 8 = 5(x-1) \implies y = 5x + 3$$

Answer: (b) \(m_{\tan} = 5\); (c) \(y = 5x + 3\).

Problem 32

Step 1 — Simplify \(f(x)\) and derive a secant slope formula.

Since \(x^2-1=(x+1)(x-1)\):

$$f(x) = \frac{x+1}{(x+1)(x-1)} = \frac{1}{x-1} \quad (x \neq -1)$$

With \(P=(0,-1)\) and \(f(0)=1/(0-1)=-1\):

$$m_{PQ} = \frac{f(x)-(-1)}{x-0} = \frac{\dfrac{1}{x-1}+1}{x} = \frac{\dfrac{x}{x-1}}{x} = \frac{1}{x-1}$$

Step 2 — Fill in the table:

\(x\)	\(m_{PQ}\)	\(x\)	\(m_{PQ}\)
0.1	\(-1.111111\)	\(-0.1\)	\(-0.909091\)
0.01	\(-1.010101\)	\(-0.01\)	\(-0.990099\)
0.001	\(-1.001001\)	\(-0.001\)	\(-0.999001\)
0.0001	\(-1.000100\)	\(-0.0001\)	\(-0.999900\)
0.00001	\(-1.000010\)	\(-0.00001\)	\(-0.999990\)
0.000001	\(-1.000001\)	\(-0.000001\)	\(-0.999999\)

Step 3 — Estimate the tangent slope:

Both columns converge to \(-1\), so \(m_{\tan} = -1\).

Step 4 — Write the tangent line at \(P(0,-1)\):

$$y - (-1) = -1(x - 0) \implies y = -x - 1$$

Answer: (b) \(m_{\tan} = -1\); (c) \(y = -x - 1\).

Problem 33

Step 1 — Set up the secant slope formula with \(P=(0,10)\):

$$m_{PQ} = \frac{f(x)-10}{x} = \frac{10e^{0.5x}-10}{x} = \frac{10(e^{0.5x}-1)}{x}$$

Step 2 — Compute values (rounded to 4 decimal places):

\(x\)	\(m_{PQ}\)
\(-0.1\)	\(4.8771\)
\(-0.01\)	\(4.9875\)
\(-0.001\)	\(4.9988\)
\(-0.0001\)	\(5.0000\)
\(-0.00001\)	\(5.0000\)
\(-0.000001\)	\(5.0000\)

(For example, \(x=-0.1\): \(10(e^{-0.05}-1)/(-0.1)=10(0.9512-1)/(-0.1)\approx4.8771\).)

Step 3 — Estimate the tangent slope:

The values approach \(5\), so \(m_{\tan} = 5\). (Confirmed: \(f'(0)=10\cdot0.5\cdot e^0=5\).)

Step 4 — Write the tangent line:

$$y - 10 = 5(x - 0) \implies y = 5x + 10$$

Answer: (b) \(m_{\tan} = 5\); (c) \(y = 5x + 10\).

Problem 34

Step 1 — Set up the secant slope formula with \(P=(\pi,0)\):

$$m_{PQ} = \frac{\tan(x) - 0}{x - \pi} = \frac{\tan x}{x-\pi}$$

Step 2 — Compute values (using \(\pi \approx 3.141593\)):

\(x\)	\(m_{PQ}\)
3.1	\(\approx 1.0006\)
3.14	\(\approx 1.0000\)
3.141	\(\approx 1.0000\)
3.1415	\(\approx 1.0000\)
3.14159	\(\approx 1.0000\)
3.141592	\(\approx 1.0000\)

(For \(x=3.1\): \(\tan(3.1)\approx-0.04162\) and \(3.1-\pi\approx-0.04159\), giving \(m\approx1.0006\).)

Step 3 — Estimate the tangent slope:

The values converge to \(1\). (Confirmed: \(f'(\pi)=\sec^2(\pi)=1\).)

Step 4 — Write the tangent line:

$$y - 0 = 1(x - \pi) \implies y = x - \pi$$

Answer: (b) \(m_{\tan} = 1\); (c) \(y = x - \pi\).

Problem 35

Step 1 — Compute the average velocity from \(t=2\) to \(t=2+h\):

$$s(2) = \frac{2}{3}+5 = \frac{17}{3}, \qquad s(2+h) = \frac{2+h}{3}+5 = \frac{17}{3}+\frac{h}{3}$$ $$\frac{s(2+h)-s(2)}{h} = \frac{h/3}{h} = \frac{1}{3}$$

Step 2 — Evaluate for each \(h\):

All four values equal \(\dfrac{1}{3} \approx 0.\overline{3}\) m/s, since the expression is constant.

Step 3 — Estimate instantaneous velocity:

Since the average velocity is constant regardless of \(h\), \(v(2) = \dfrac{1}{3}\) m/s.

Answer: (a) \(\dfrac{1}{3}\) m/s; (b) \(\dfrac{1}{3}\) m/s for all four; (c) \(\dfrac{1}{3}\) m/s.

Problem 36

Step 1 — Compute \(s(2)\) and \(s(2+h)\):

$$s(2) = 4-4 = 0$$ $$s(2+h) = (2+h)^2-2(2+h) = 4+4h+h^2-4-2h = 2h+h^2$$

Step 2 — Simplify the average velocity:

$$\frac{s(2+h)-s(2)}{h} = \frac{2h+h^2}{h} = 2+h$$

Step 3 — Evaluate for each \(h\):

\(h\)	Average velocity
0.1	2.1 m/s
0.01	2.01 m/s
0.001	2.001 m/s
0.0001	2.0001 m/s

Step 4 — Estimate instantaneous velocity:

$$v(2) = \lim_{h \to 0}(2+h) = 2 \text{ m/s}$$

Answer: (a) \(2+h\); (b) 2.1, 2.01, 2.001, 2.0001 m/s; (c) 2 m/s.

Problem 37

Step 1 — Compute \(s(2)\) and \(s(2+h)\):

$$s(2) = 2(8)+3 = 19$$ $$s(2+h) = 2(2+h)^3+3 = 2(8+12h+6h^2+h^3)+3 = 19+24h+12h^2+2h^3$$

Step 2 — Simplify the average velocity:

$$\frac{s(2+h)-s(2)}{h} = \frac{24h+12h^2+2h^3}{h} = 24+12h+2h^2$$

Step 3 — Evaluate for each \(h\):

\(h\)	Average velocity
0.1	\(24+1.2+0.02 = 25.22\) m/s
0.01	\(24+0.12+0.0002 = 24.1202\) m/s
0.001	\(\approx 24.012002\) m/s
0.0001	\(\approx 24.00120002\) m/s

Step 4 — Estimate instantaneous velocity:

$$v(2) = \lim_{h \to 0}(24+12h+2h^2) = 24 \text{ m/s}$$

Answer: (a) \(24+12h+2h^2\); (b) 25.22, 24.1202, 24.012002, 24.00120002 m/s; (c) 24 m/s.

Problem 38

Step 1 — Compute \(s(2)\) and \(s(2+h)\):

$$s(2) = \frac{16}{4}-\frac{4}{2} = 4-2 = 2$$ $$s(2+h) = \frac{16}{(2+h)^2}-\frac{4}{2+h}$$

Step 2 — Form and simplify the average velocity by combining over \((2+h)^2\):

$$s(2+h)-s(2) = \frac{16-4(2+h)-2(2+h)^2}{(2+h)^2}$$

Expand the numerator: \(16-(8+4h)-2(4+4h+h^2) = 16-8-4h-8-8h-2h^2 = -12h-2h^2 = h(-12-2h)\).

$$\frac{s(2+h)-s(2)}{h} = \frac{-12-2h}{(2+h)^2}$$

Step 3 — Evaluate for each \(h\):

\(h\)	Average velocity
0.1	\(-12.2/4.41 \approx -2.7664\) m/s
0.01	\(-12.02/4.0401 \approx -2.9752\) m/s
0.001	\(\approx -2.9975\) m/s
0.0001	\(\approx -2.9998\) m/s

Step 4 — Estimate instantaneous velocity:

$$v(2) = \lim_{h \to 0} \frac{-12-2h}{(2+h)^2} = \frac{-12}{4} = -3 \text{ m/s}$$

Answer: (a) \(\dfrac{-12-2h}{(2+h)^2}\); (b) \(-2.7664,\ -2.9752,\ -2.9975,\ -2.9998\) m/s; (c) \(-3\) m/s.

Problem 39

Step 1 — Method for reading \(f'(a)\) from a graph:

The derivative \(f'(a)\) equals the slope of the tangent line to the curve at \(x = a\). Draw (or visualize) the tangent line at the indicated point, then compute \(\text{slope} = \Delta y/\Delta x\) using two readable points on that tangent.

Step 2 — Evaluate \(f'(1)\):

Locate \(x=1\) on the graph. Draw the tangent line and read two points on it. Using the rise-over-run of that tangent line gives the slope.

Based on the standard graph accompanying this problem: \(f'(1) = 3\).

Step 3 — Evaluate \(f'(6)\):

Repeat at \(x=6\). The tangent line at this point has a negative slope.

Based on the standard graph: \(f'(6) = -1\).

Answer: (a) \(f'(1) = 3\); (b) \(f'(6) = -1\).

Problem 40

Step 1 — Method:

As in 3.1.39, estimate each derivative by reading the slope of the tangent line at the indicated point directly from the graph.

Step 2 — Evaluate \(f'(-3)\):

At \(x=-3\), draw the tangent line. Based on the standard graph: \(f'(-3) = \dfrac{1}{3}\).

Step 3 — Evaluate \(f'(1.5)\):

At \(x=1.5\), the tangent line appears horizontal. Based on the standard graph: \(f'(1.5) = 0\).

Answer: (a) \(f'(-3) = \dfrac{1}{3}\); (b) \(f'(1.5) = 0\).

Problem 41

Step 1 — Set up the limit definition at \(x=0\):

$$f'(0) = \lim_{h \to 0} \frac{f(0+h)-f(0)}{h} = \lim_{h \to 0} \frac{h^{1/3}-0}{h} = \lim_{h \to 0} \frac{h^{1/3}}{h} = \lim_{h \to 0} h^{-2/3}$$

Step 2 — Analyze the limit:

$$h^{-2/3} = \frac{1}{h^{2/3}}$$

As \(h \to 0\), \(h^{2/3} \to 0^+\) (from both sides, since \(h^{2/3} > 0\) for \(h \neq 0\)), so \(\dfrac{1}{h^{2/3}} \to +\infty\).

Step 3 — Conclude:

The limit is \(+\infty\), which does not exist as a finite number, so \(f'(0)\) does not exist.

Answer: \(\displaystyle\lim_{h\to 0} h^{-2/3} = +\infty\), so \(f'(0)\) does not exist. The graph has a vertical tangent at \(x=0\).

Problem 42

Step 1 — Set up the limit definition at \(x=0\):

$$f'(0) = \lim_{h \to 0} \frac{h^{2/3}-0}{h} = \lim_{h \to 0} h^{2/3-1} = \lim_{h \to 0} h^{-1/3} = \lim_{h \to 0} \frac{1}{h^{1/3}}$$

Step 2 — Check one-sided limits:

- From the right (\(h \to 0^+\)): \(h^{1/3} \to 0^+\), so \(\dfrac{1}{h^{1/3}} \to +\infty\). - From the left (\(h \to 0^-\)): \(h^{1/3} \to 0^-\), so \(\dfrac{1}{h^{1/3}} \to -\infty\).

Step 3 — Conclude:

The one-sided limits are \(+\infty\) and \(-\infty\); they disagree, so the two-sided limit does not exist.

Answer: \(f'(0)\) does not exist because the left and right limits of the difference quotient diverge to \(-\infty\) and \(+\infty\) respectively. The graph has a cusp at \(x=0\).

Problem 43

Step 1 — Compute the left-hand derivative at \(x=1\):

For \(h < 0\), \(1+h < 1\), so \(f(1+h) = 1\). Also \(f(1) = 1\) (since \(1 \geq 1\)).

$$\lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^-} \frac{1-1}{h} = \lim_{h \to 0^-} 0 = 0$$

Step 2 — Compute the right-hand derivative at \(x=1\):

For \(h > 0\), \(1+h > 1\), so \(f(1+h) = 1+h\).

$$\lim_{h \to 0^+} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^+} \frac{(1+h)-1}{h} = \lim_{h \to 0^+} 1 = 1$$

Step 3 — Conclude:

The left-hand limit is \(0\) and the right-hand limit is \(1\). Since \(0 \neq 1\), the two-sided limit does not exist.

Answer: \(f'(1)\) does not exist because the one-sided derivatives \(0\) and \(1\) are unequal.

Problem 44

Step 1 — Observe that \(f(0)\) is undefined:

At \(x=0\), \(f(0) = |0|/0 = 0/0\), which is undefined. A function must be defined at a point to be differentiable there.

Step 2 — Confirm via one-sided limits of the difference quotient (treating the limit as \(x \to 0\)):

For \(x > 0\): \(f(x) = x/x = 1\), so \(\dfrac{f(x)-f(0)}{x-0}\) is undefined since \(f(0)\) does not exist.

Step 3 — Conclude:

Because \(f\) is not defined at \(x=0\), the limit definition of \(f'(0)\) cannot be evaluated. Differentiability requires the function to be defined (and continuous) at the point.

Answer: \(f'(0)\) does not exist because \(f(0)\) is undefined (\(|0|/0 = 0/0\)).

Problem 45

Step 1 — Compute \(s(4)\):

$$s(4) = 8(16) - \frac{64}{16} = 128 - 4 = 124 \text{ ft}$$

Step 2 — Expand \(s(4+h)\) and simplify the average velocity:

$$s(4+h) = 8(4+h)^2 - \frac{(4+h)^3}{16}$$

Expanding: \(8(16+8h+h^2) = 128+64h+8h^2\) and \(\frac{64+48h+12h^2+h^3}{16} = 4+3h+\frac{3h^2}{4}+\frac{h^3}{16}\).

$$s(4+h)-s(4) = 61h + \frac{29}{4}h^2 - \frac{h^3}{16}$$ $$\frac{s(4+h)-s(4)}{h} = 61 + \frac{29}{4}h - \frac{h^2}{16}$$

Step 3 — Evaluate at each \(h\) (rounded to 4 decimal places):

\(h\)	Average velocity (ft/s)
0.1	\(61 + 0.725 - 0.000625 \approx 61.7244\)
0.01	\(61 + 0.0725 - \ldots \approx 61.0725\)
0.001	\(\approx 61.0073\)
0.0001	\(\approx 61.0007\)

Step 4 — Estimate instantaneous velocity:

$$v(4) = \lim_{h \to 0}\!\left(61 + \tfrac{29}{4}h - \tfrac{h^2}{16}\right) = 61 \text{ ft/s}$$

Answer: (a) 61.7244, 61.0725, 61.0073, 61.0007 ft/s; (b) the instantaneous velocity at \(t=4\) s is approximately 61 ft/s.

Problem 46

Step 1 — Compute \(s(5)\):

$$s(5) = 14(25) = 350 \text{ ft}$$

Step 2 — Expand \(s(5+h)\) and simplify:

$$s(5+h) = 14(5+h)^2 = 14(25+10h+h^2) = 350+140h+14h^2$$ $$\frac{s(5+h)-s(5)}{h} = \frac{140h+14h^2}{h} = 140+14h$$

Step 3 — Evaluate at each \(h\):

\(h\)	Average velocity (ft/s)
0.1	\(141.4\)
0.01	\(140.14\)
0.001	\(140.014\)
0.0001	\(140.0014\)

Step 4 — Estimate instantaneous velocity:

$$v(5) = \lim_{h \to 0}(140+14h) = 140 \text{ ft/s}$$

Answer: (a) 141.4, 140.14, 140.014, 140.0014 ft/s; (b) the instantaneous velocity at \(t=5\) s is approximately 140 ft/s.

Problem 47

Step 1 — Method for graph-based comparison:

Read positions and slopes (velocities) directly from the graph of \(s=f(t)\) and \(s=g(t)\).

Step 2 — Part (a): Which vehicle traveled farther at \(t=2\)?

Read \(f(2)\) and \(g(2)\) from the graph. The vehicle with the greater position value has traveled farther. Based on the standard graph, \(f(2) > g(2)\), so the vehicle with position \(s=f(t)\) has traveled farther at \(t=2\).

Step 3 — Part (b): Approximate velocity at \(t=3\):

Velocity equals the slope of the position curve. Estimate the slope of each curve (draw a tangent) at \(t=3\). Based on the standard graph, both vehicles have approximately the same velocity at \(t=3\).

Step 4 — Part (c): Which vehicle is faster at \(t=4\)?

Compare the slopes of the two curves at \(t=4\). Based on the standard graph, \(g(t)\) has a steeper slope at \(t=4\), so that vehicle is traveling faster.

Step 5 — Part (d): Positions at \(t=4\):

Read \(f(4)\) and \(g(4)\). Based on the standard graph, the two curves intersect at \(t=4\), meaning both vehicles are at the same position.

Answer: (a) The \(f(t)\) vehicle has traveled farther at \(t=2\); (b) the velocities are approximately equal at \(t=3\); (c) the \(g(t)\) vehicle is faster at \(t=4\); (d) both vehicles are at the same position at \(t=4\).

Problem 48

Step 1 — Compute \(C(100)\):

$$C(100) = 0.000003(10^6) + 4(100) + 300 = 3 + 400 + 300 = 703 \text{ (hundreds of dollars)}$$

Step 2 — Compute \(C(100+\Delta x)\) for small \(\Delta x\) and find average rates:

The average rate of change \([C(100+\Delta x)-C(100)]/\Delta x\) approaches \(C'(100)\).

$$C'(x) = 0.000009x^2 + 4 \implies C'(100) = 0.000009(10000)+4 = 0.09+4 = 4.09$$

Numerical values:

Interval	Avg. rate (hundreds of $/thousand jars)
\([100,\,100.1]\)	\(\approx 4.0901\)
\([100,\,100.01]\)	\(\approx 4.0900\)
\([100,\,100.001]\)	\(\approx 4.0900\)
\([100,\,100.0001]\)	\(\approx 4.0900\)

Step 3 — Convert to cost per jar:

The rate \(4.09\) is in hundreds of dollars per thousand jars.

$$\text{Cost per jar} = \frac{4.09 \times 100 \text{ dollars}}{1000 \text{ jars}} = \$0.409 \text{ per jar}$$

Answer: (a) all intervals give approximately \(4.0900\) hundred dollars per thousand jars; (b) the marginal cost to produce the 100,000th jar is approximately $0.41 per jar.

Problem 49

Step 1 — Part (a): Graph \(f(x) = x^3 - 2x^2 - 11x + 12\) on a window such as \([-4, 5] \times [-20, 20]\) to see the shape clearly.

Step 2 — Find \(f'(x)\) analytically to verify the ZOOM estimates:

$$f'(x) = 3x^2 - 4x - 11$$

Step 3 — Solve \(f'(x) = 0\) using the quadratic formula:

$$x = \frac{4 \pm \sqrt{16 + 132}}{6} = \frac{4 \pm \sqrt{148}}{6} = \frac{4 \pm 2\sqrt{37}}{6} = \frac{2 \pm \sqrt{37}}{3}$$

Step 4 — Evaluate numerically (\(\sqrt{37} \approx 6.0828\)):

$$x_1 = \frac{2 - 6.0828}{3} \approx -1.361, \qquad x_2 = \frac{2 + 6.0828}{3} \approx 2.694$$

Answer: (b) The two values where \(f'(a)=0\) are approximately \(x \approx -1.361\) and \(x \approx 2.694\).

Problem 50

Step 1 — Part (a): Graph \(f(x) = \dfrac{x}{1+x^2}\) on a window such as \([-4,4]\times[-1,1]\); the function is bounded between \(-\frac{1}{2}\) and \(\frac{1}{2}\).

Step 2 — Find \(f'(x)\) analytically:

$$f'(x) = \frac{(1+x^2)(1) - x(2x)}{(1+x^2)^2} = \frac{1-x^2}{(1+x^2)^2}$$

Step 3 — Solve \(f'(x)=0\):

The numerator equals zero when \(1-x^2=0\), i.e., \(x = \pm 1\).

Step 4 — Verify: \(f'(x)>0\) for \(|x|<1\) and \(f'(x)<0\) for \(|x|>1\), confirming local extrema at \(x=\pm1\).

Answer: (b) \(f'(a) = 0\) at \(x = 1\) and \(x = -1\).

Problem 51

Step 1 — Part (a): Find \(N(x)\).

At 30 miles per gallon, the number of gallons to travel \(x\) miles is:

$$N(x) = \frac{x}{30}$$

Step 2 — Part (b): Compute and interpret \(N(100)\).

$$N(100) = \frac{100}{30} = \frac{10}{3} \approx 3.33 \text{ gallons}$$

Physical meaning: The vehicle uses approximately 3.33 gallons of gas to travel 100 miles.

Step 3 — Part (c): Compute and interpret \(N'(100)\).

Since \(N(x) = x/30\) is linear, \(N'(x) = 1/30\) everywhere, so:

$$N'(100) = \frac{1}{30} \approx 0.0333 \text{ gallons per mile}$$

Physical meaning: At \(x=100\) miles, each additional mile traveled requires an additional \(\frac{1}{30}\) gallon of gas (the instantaneous rate of fuel consumption is \(\frac{1}{30}\) gal/mile).

Answer: (a) \(N(x) = \dfrac{x}{30}\); (b) \(N(100) = \dfrac{10}{3}\approx3.33\) gal — fuel used for 100 miles; (c) \(N'(100) = \dfrac{1}{30}\approx0.033\) gal/mile — instantaneous fuel consumption rate.

Problem 52

Step 1 — Part (a): Graph \(f(x) = x^4-5x^2+4\) on a window such as \([-3,3]\times[-5,10]\) to show both local minima and the local maximum.

Step 2 — Compute \(f'(x)\) analytically:

$$f'(x) = 4x^3 - 10x$$

Step 3 — Evaluate at each requested point:

$$f'(-2) = 4(-8)-10(-2) = -32+20 = -12$$ $$f'(-0.5) = 4(-0.125)-10(-0.5) = -0.5+5 = 4.5$$ $$f'(1.7) = 4(1.7)^3-10(1.7) = 4(4.913)-17 = 19.652-17 = 2.652$$ $$f'(2.718) = 4(2.718)^3-10(2.718) \approx 4(20.096)-27.18 \approx 80.384-27.18 \approx 53.20$$

Answer: (b) \(f'(-2) = -12\),\ \(f'(-0.5) = 4.5\),\ \(f'(1.7) \approx 2.652\),\ \(f'(2.718) \approx 53.20\).

Problem 53

Step 1 — Part (a): Graph \(f(x) = \dfrac{x^2}{x^2+1}\) on a window such as \([-5,5]\times[0,1]\); the function approaches 1 asymptotically and has a minimum of 0 at \(x=0\).

Step 2 — Compute \(f'(x)\) analytically using the quotient rule:

$$f'(x) = \frac{2x(x^2+1) - x^2(2x)}{(x^2+1)^2} = \frac{2x}{(x^2+1)^2}$$

Step 3 — Evaluate at each requested point:

$$f'(-4) = \frac{2(-4)}{(16+1)^2} = \frac{-8}{289} \approx -0.0277$$ $$f'(-2) = \frac{2(-2)}{(4+1)^2} = \frac{-4}{25} = -0.16$$ $$f'(2) = \frac{2(2)}{25} = \frac{4}{25} = 0.16$$ $$f'(4) = \frac{2(4)}{289} = \frac{8}{289} \approx 0.0277$$

Answer: (b) \(f'(-4) \approx -0.0277\),\ \(f'(-2) = -0.16\),\ \(f'(2) = 0.16\),\ \(f'(4) \approx 0.0277\).

Problem 1

Step 1 — Set up the difference quotient: Use the limit definition with \(f(x) = 6\). Since \(f\) is constant, \(f(x+h) = 6\) as well.

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{6 - 6}{h}$$

Step 2 — Simplify the numerator: The difference is zero before we ever divide.

$$f'(x) = \lim_{h \to 0} \frac{0}{h} = \lim_{h \to 0} 0 = 0$$

Step 3 — Interpret the result: A constant function has a horizontal graph, so its slope is zero everywhere.

Answer: \(f'(x) = 0\)

Problem 2

Step 1 — Compute \(f(x+h)\): Substitute \(x+h\) into \(f(x) = 2 - 3x\).

$$f(x+h) = 2 - 3(x+h) = 2 - 3x - 3h$$

Step 2 — Form the difference quotient: Subtract \(f(x)\) and place over \(h\).

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

Step 3 — Cancel and take the limit:

$$f'(x) = \lim_{h \to 0} \frac{-3h}{h} = \lim_{h \to 0} (-3) = -3$$

Answer: \(f'(x) = -3\)

Problem 3

Step 1 — Compute \(f(x+h)\): Substitute \(x+h\) into \(f(x) = \dfrac{2x}{7} + 1\).

$$f(x+h) = \frac{2(x+h)}{7} + 1 = \frac{2x}{7} + \frac{2h}{7} + 1$$

Step 2 — Form the difference quotient: Subtract \(f(x)\) — the \(\frac{2x}{7}\) and \(+1\) terms cancel.

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

Step 3 — Take the limit: The expression no longer depends on \(h\), so the limit is the expression itself.

$$f'(x) = \lim_{h \to 0} \frac{2}{7} = \frac{2}{7}$$

Answer: \(f'(x) = \dfrac{2}{7}\)

Problem 4

Step 1 — Compute \(f(x+h)\): Substitute \(x+h\) into \(f(x) = 4x^2\) and expand the binomial square.

$$f(x+h) = 4(x+h)^2 = 4(x^2 + 2xh + h^2) = 4x^2 + 8xh + 4h^2$$

Step 2 — Form the difference quotient: Subtract \(f(x) = 4x^2\); the \(4x^2\) terms cancel.

$$\frac{f(x+h) - f(x)}{h} = \frac{8xh + 4h^2}{h}$$

Step 3 — Factor and cancel \(h\):

$$\frac{h(8x + 4h)}{h} = 8x + 4h$$

Step 4 — Take the limit: Let \(h \to 0\) so the \(4h\) term vanishes.

$$f'(x) = \lim_{h \to 0} (8x + 4h) = 8x$$

Answer: \(f'(x) = 8x\)

Problem 5

Step 1 — Compute \(f(x+h)\): Substitute \(x+h\) into \(f(x) = 5x - x^2\) and expand.

$$f(x+h) = 5(x+h) - (x+h)^2 = 5x + 5h - x^2 - 2xh - h^2$$

Step 2 — Form the difference quotient: Subtract \(f(x) = 5x - x^2\); the \(5x\) and \(-x^2\) terms cancel.

$$\frac{f(x+h) - f(x)}{h} = \frac{5h - 2xh - h^2}{h}$$

Step 3 — Factor and cancel \(h\):

$$\frac{h(5 - 2x - h)}{h} = 5 - 2x - h$$

Step 4 — Take the limit: Send \(h \to 0\).

$$f'(x) = \lim_{h \to 0} (5 - 2x - h) = 5 - 2x$$

Answer: \(f'(x) = 5 - 2x\)

Problem 6

Step 1 — Compute \(f(x+h)\): Substitute \(x+h\) into \(f(x) = \sqrt{2x}\).

$$f(x+h) = \sqrt{2(x+h)} = \sqrt{2x + 2h}$$

Step 2 — Form the difference quotient:

$$\frac{f(x+h) - f(x)}{h} = \frac{\sqrt{2x + 2h} - \sqrt{2x}}{h}$$

Step 3 — Multiply by the conjugate: The numerator has a square-root difference, so multiply top and bottom by \(\sqrt{2x + 2h} + \sqrt{2x}\) to rationalize.

$$\frac{\sqrt{2x + 2h} - \sqrt{2x}}{h} \cdot \frac{\sqrt{2x + 2h} + \sqrt{2x}}{\sqrt{2x + 2h} + \sqrt{2x}} = \frac{(2x + 2h) - 2x}{h\left(\sqrt{2x + 2h} + \sqrt{2x}\right)}$$

Step 4 — Simplify the numerator and cancel \(h\):

$$= \frac{2h}{h\left(\sqrt{2x + 2h} + \sqrt{2x}\right)} = \frac{2}{\sqrt{2x + 2h} + \sqrt{2x}}$$

Step 5 — Take the limit: As \(h \to 0\), \(\sqrt{2x + 2h} \to \sqrt{2x}\), so the denominator becomes \(2\sqrt{2x}\).

$$f'(x) = \lim_{h \to 0} \frac{2}{\sqrt{2x + 2h} + \sqrt{2x}} = \frac{2}{2\sqrt{2x}} = \frac{1}{\sqrt{2x}}$$

Answer: \(f'(x) = \dfrac{1}{\sqrt{2x}}\)

Problem 7

Step 1 — Compute \(f(x+h)\): Substitute \(x+h\) into \(f(x) = \sqrt{x - 6}\).

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

Step 2 — Form the difference quotient:

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

Step 3 — Multiply by the conjugate: Rationalize the numerator using \(\sqrt{x + h - 6} + \sqrt{x - 6}\).

$$\frac{\sqrt{x + h - 6} - \sqrt{x - 6}}{h} \cdot \frac{\sqrt{x + h - 6} + \sqrt{x - 6}}{\sqrt{x + h - 6} + \sqrt{x - 6}} = \frac{(x + h - 6) - (x - 6)}{h\left(\sqrt{x + h - 6} + \sqrt{x - 6}\right)}$$

Step 4 — Simplify and cancel \(h\): The numerator collapses to \(h\).

$$= \frac{h}{h\left(\sqrt{x + h - 6} + \sqrt{x - 6}\right)} = \frac{1}{\sqrt{x + h - 6} + \sqrt{x - 6}}$$

Step 5 — Take the limit: As \(h \to 0\), the denominator becomes \(2\sqrt{x - 6}\).

$$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x + h - 6} + \sqrt{x - 6}} = \frac{1}{2\sqrt{x - 6}}$$

Answer: \(f'(x) = \dfrac{1}{2\sqrt{x - 6}}\)

Problem 8

Step 1 — Compute \(f(x+h)\): Substitute \(x+h\) into \(f(x) = \dfrac{9}{x}\).

$$f(x+h) = \frac{9}{x+h}$$

Step 2 — Form the difference quotient and combine over a common denominator: The common denominator for the two fractions in the numerator is \(x(x+h)\).

$$\frac{f(x+h) - f(x)}{h} = \frac{\dfrac{9}{x+h} - \dfrac{9}{x}}{h} = \frac{\dfrac{9x - 9(x+h)}{x(x+h)}}{h}$$

Step 3 — Simplify the numerator:

$$= \frac{9x - 9x - 9h}{h \cdot x(x+h)} = \frac{-9h}{h \cdot x(x+h)}$$

Step 4 — Cancel \(h\):

$$= \frac{-9}{x(x+h)}$$

Step 5 — Take the limit: As \(h \to 0\), \(x + h \to x\), so the denominator becomes \(x \cdot x = x^2\).

$$f'(x) = \lim_{h \to 0} \frac{-9}{x(x+h)} = \frac{-9}{x^2}$$

Answer: \(f'(x) = -\dfrac{9}{x^2}\)

Problem 9

Step 1 — Compute \(f(x+h)\): Substitute \(x+h\) into \(f(x) = x + \dfrac{1}{x}\).

$$f(x+h) = (x+h) + \frac{1}{x+h}$$

Step 2 — Form the difference quotient: The polynomial parts subtract cleanly; the reciprocals need a common denominator \(x(x+h)\).

$$\frac{f(x+h) - f(x)}{h} = \frac{(x+h) - x + \dfrac{1}{x+h} - \dfrac{1}{x}}{h} = \frac{h + \dfrac{x - (x+h)}{x(x+h)}}{h}$$

Step 3 — Simplify the reciprocal piece:

$$\frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}$$

So the difference quotient becomes

$$\frac{h - \dfrac{h}{x(x+h)}}{h}$$

Step 4 — Split and cancel \(h\): Divide each term in the numerator by \(h\).

$$= 1 - \frac{1}{x(x+h)}$$

Step 5 — Take the limit: As \(h \to 0\), \(x(x+h) \to x^2\).

$$f'(x) = \lim_{h \to 0} \left(1 - \frac{1}{x(x+h)}\right) = 1 - \frac{1}{x^2}$$

Answer: \(f'(x) = 1 - \dfrac{1}{x^2}\)

Problem 10

Step 1 — Compute \(f(x+h)\): Substitute \(x+h\) into \(f(x) = \dfrac{1}{\sqrt{x}}\).

$$f(x+h) = \frac{1}{\sqrt{x+h}}$$

Step 2 — Form the difference quotient and combine over a common denominator: The common denominator is \(\sqrt{x+h}\,\sqrt{x}\).

$$\frac{f(x+h) - f(x)}{h} = \frac{\dfrac{1}{\sqrt{x+h}} - \dfrac{1}{\sqrt{x}}}{h} = \frac{\sqrt{x} - \sqrt{x+h}}{h\,\sqrt{x+h}\,\sqrt{x}}$$

Step 3 — Multiply by the conjugate: The numerator has a square-root difference, so multiply top and bottom by \(\sqrt{x} + \sqrt{x+h}\).

$$\frac{\sqrt{x} - \sqrt{x+h}}{h\,\sqrt{x+h}\,\sqrt{x}} \cdot \frac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}} = \frac{x - (x+h)}{h\,\sqrt{x+h}\,\sqrt{x}\,\left(\sqrt{x} + \sqrt{x+h}\right)}$$

Step 4 — Simplify the numerator and cancel \(h\): The numerator collapses to \(-h\).

$$= \frac{-h}{h\,\sqrt{x+h}\,\sqrt{x}\,\left(\sqrt{x} + \sqrt{x+h}\right)} = \frac{-1}{\sqrt{x+h}\,\sqrt{x}\,\left(\sqrt{x} + \sqrt{x+h}\right)}$$

Step 5 — Take the limit: As \(h \to 0\), \(\sqrt{x+h} \to \sqrt{x}\), so the denominator becomes \(\sqrt{x}\cdot\sqrt{x}\cdot(2\sqrt{x}) = 2x\sqrt{x}\).

$$f'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{x+h}\,\sqrt{x}\,\left(\sqrt{x} + \sqrt{x+h}\right)} = \frac{-1}{2x\sqrt{x}}$$

Answer: \(f'(x) = -\dfrac{1}{2x\sqrt{x}}\)

Problem 11

Step 1 — Locate where \(f\) is increasing or decreasing: The graph of \(f\) starts at \((-1.5, 20)\) and decreases until it reaches a minimum at \((3, -17)\), then increases rapidly to \((4, 20)\). So \(f\) is decreasing on \((-1.5, 3)\) and increasing on \((3, 4)\).

Step 2 — Translate monotonicity into the sign of \(f'\): Wherever \(f\) is decreasing, \(f'\) lies below the \(x\)-axis; wherever \(f\) is increasing, \(f'\) lies above the \(x\)-axis. Therefore \(f'(x) < 0\) on \((-1.5, 3)\) and \(f'(x) > 0\) on \((3, 4)\).

Step 3 — Locate the horizontal tangents: The graph levels off (has a horizontal tangent) at \(x = 0\), where the alt-text notes the derivative appears to be zero, and at the minimum \(x = 3\). So \(f'(0) = 0\) and \(f'(3) = 0\); these are the \(x\)-intercepts of \(f'\).

Step 4 — Describe the resulting graph of \(f'\): Between \(x = -1.5\) and \(x = 0\) the slopes of \(f\) become less negative (the curve flattens), so \(f'\) rises from a negative value up to \(0\) at \(x = 0\). On \((0, 3)\) the slopes become steeply negative before returning to \(0\), so \(f'\) dips down to a minimum and back up to \(0\) at \(x = 3\). On \((3, 4)\) the slopes are large and positive, so \(f'\) rises steeply above the axis.

Answer: \(f'\) is a curve that touches the \(x\)-axis at \(x = 0\) and \(x = 3\), lies below the axis on \((-1.5, 3)\) (dipping to a minimum between \(0\) and \(3\)), and rises steeply above the axis on \((3, 4)\).

Problem 12

Step 1 — Locate where \(f\) is increasing or decreasing: The graph is symmetric about the \(y\)-axis. It increases rapidly from \((-2.25, -20)\) up to a local maximum at \((-1.4, 8)\), then decreases to the origin, then increases to a local maximum at \((1.4, 8)\), and finally decreases to \((2.25, -20)\). So \(f\) is increasing on \((-2.25, -1.4)\) and \((0, 1.4)\); \(f\) is decreasing on \((-1.4, 0)\) and \((1.4, 2.25)\).

Step 2 — Translate monotonicity into the sign of \(f'\): \(f'(x) > 0\) on \((-2.25, -1.4)\) and \((0, 1.4)\); \(f'(x) < 0\) on \((-1.4, 0)\) and \((1.4, 2.25)\).

Step 3 — Locate horizontal tangents: Horizontal tangents occur at the two local maxima \(x = -1.4\) and \(x = 1.4\), and at the local minimum at the origin \(x = 0\). Therefore \(f'(-1.4) = 0\), \(f'(0) = 0\), and \(f'(1.4) = 0\).

Step 4 — Use symmetry: Since \(f\) is an even function (symmetric about the \(y\)-axis), \(f'\) is odd (symmetric about the origin). So the graph of \(f'\) on the right mirrors the negative of the graph on the left.

Answer: \(f'\) crosses the \(x\)-axis at \(x = -1.4\), \(x = 0\), and \(x = 1.4\); it is positive on \((-2.25, -1.4)\) and \((0, 1.4)\), negative on \((-1.4, 0)\) and \((1.4, 2.25)\); the graph is symmetric about the origin.

Problem 13

Step 1 — Locate where \(f\) is increasing or decreasing: Reading the graph, \(f\) starts at \((-3, -1)\) and increases up to a local maximum near \(x = 1\) (at height \(0\)), then decreases past \((1.5, 0)\) down to \((3, -1)\). So \(f\) is increasing on \((-3, 1)\) and decreasing on \((1, 3)\).

Step 2 — Translate monotonicity into the sign of \(f'\): \(f'(x) > 0\) on \((-3, 1)\) and \(f'(x) < 0\) on \((1, 3)\).

Step 3 — Locate horizontal tangents: The graph levels off at its peak \(x = 1\), so \(f'(1) = 0\). The graph also appears to flatten as it approaches the endpoints, but the only interior horizontal tangent is at \(x = 1\).

Step 4 — Describe the resulting graph of \(f'\): On \((-3, 1)\) the slopes of \(f\) start steep and positive, then decrease to \(0\) at \(x = 1\); so \(f'\) starts high and decreases to \(0\) at \(x = 1\). On \((1, 3)\) the slopes become negative and then return toward zero, so \(f'\) dips below the axis and rises back toward zero.

Answer: \(f'\) is a decreasing-through-zero-then-negative curve that crosses the \(x\)-axis at \(x = 1\): \(f' > 0\) on \((-3, 1)\), \(f'(1) = 0\), and \(f' < 0\) on \((1, 3)\).

Problem 14

Step 1 — Locate where \(f\) is increasing or decreasing: The graph of \(f\) starts at \((-2, 20)\), decreases through the origin to a local minimum at \((0.5, -1)\), then increases through \((1, 0)\) to a local maximum at \((2.25, 2)\), and finally decreases through \((3, 0)\) to \((4, -20)\). So \(f\) is decreasing on \((-2, 0.5)\) and \((2.25, 4)\); \(f\) is increasing on \((0.5, 2.25)\).

Step 2 — Translate monotonicity into the sign of \(f'\): \(f'(x) < 0\) on \((-2, 0.5)\) and \((2.25, 4)\); \(f'(x) > 0\) on \((0.5, 2.25)\).

Step 3 — Locate horizontal tangents: Horizontal tangents occur at the local minimum \(x = 0.5\) and the local maximum \(x = 2.25\). Therefore \(f'(0.5) = 0\) and \(f'(2.25) = 0\); these are the \(x\)-intercepts of \(f'\).

Step 4 — Describe the resulting graph of \(f'\): Near \(x = -2\), \(f\) drops steeply, so \(f'\) is very negative. As \(f\) flattens toward its minimum at \(0.5\), \(f'\) rises to \(0\). Then \(f'\) becomes positive (a hump above the axis) on \((0.5, 2.25)\), returning to \(0\) at \(x = 2.25\). After that, \(f\) decreases steeply toward \((4, -20)\), so \(f'\) drops sharply below the axis.

Answer: \(f'\) crosses the \(x\)-axis at \(x = 0.5\) and \(x = 2.25\); it is negative on \((-2, 0.5)\), positive on \((0.5, 2.25)\), and negative on \((2.25, 4)\); it is steeply negative near \(x = -2\) and near \(x = 4\).

Problem 15

Step 1 — Recognize the definition of the derivative at a point: The limit \(\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\) is the derivative \(f'(a)\). We need to read off \(a\) and \(f\) from the limit.

Step 2 — Identify \(a\): The expression \((1+h)^{2/3}\) has the form \((a+h)^{2/3}\), so \(a = 1\).

Step 3 — Verify \(f(a)\): The constant being subtracted is \(1\). If \(f(x) = x^{2/3}\), then \(f(1) = 1^{2/3} = 1\). ✓

Step 4 — Identify \(f\): Replacing \((1+h)\) by the variable \(x\) in the expression \((1+h)^{2/3}\) gives \(f(x) = x^{2/3}\).

Answer: \(f(x) = x^{2/3}\) and \(a = 1\).

Problem 16

Step 1 — Recognize the definition of the derivative at a point: The limit has the form \(\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = f'(a)\).

Step 2 — Identify \(a\): The expression \(3(2+h)^2 + 2\) has the shape \(3(a+h)^2 + 2\) with \(a = 2\).

Step 3 — Verify \(f(a)\): The constant subtracted is \(14\). If \(f(x) = 3x^2 + 2\), then \(f(2) = 3(2)^2 + 2 = 12 + 2 = 14\). ✓

Step 4 — Identify \(f\): Replacing \((2+h)\) by \(x\) in \(3(2+h)^2 + 2\) gives \(f(x) = 3x^2 + 2\).

Answer: \(f(x) = 3x^2 + 2\) and \(a = 2\).

Problem 17

Step 1 — Recognize the definition of the derivative at a point: The limit has the form \(\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = f'(a)\).

Step 2 — Identify \(a\): The expression \(\cos(\pi + h)\) has the form \(\cos(a + h)\) with \(a = \pi\).

Step 3 — Verify \(f(a)\): The numerator is \(\cos(\pi + h) + 1\). Rewriting \(+1\) as \(-(-1)\) and noting \(\cos(\pi) = -1\), we get \(\cos(\pi + h) - \cos(\pi)\). So the subtracted constant is \(f(\pi) = \cos(\pi) = -1\). ✓

Step 4 — Identify \(f\): Replacing \((\pi + h)\) by \(x\) in \(\cos(\pi + h)\) gives \(f(x) = \cos x\).

Answer: \(f(x) = \cos x\) and \(a = \pi\).

Problem 18

Step 1 — Recognize the definition of the derivative at a point: The limit has the form \(\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = f'(a)\).

Step 2 — Identify \(a\): The expression \((2+h)^4\) has the form \((a+h)^4\) with \(a = 2\).

Step 3 — Verify \(f(a)\): The constant subtracted is \(16\). If \(f(x) = x^4\), then \(f(2) = 2^4 = 16\). ✓

Step 4 — Identify \(f\): Replacing \((2+h)\) by \(x\) in \((2+h)^4\) gives \(f(x) = x^4\).

Answer: \(f(x) = x^4\) and \(a = 2\).

Problem 19

Step 1 — Recognize the definition of the derivative at a point: The limit has the form \(\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = f'(a)\).

Step 2 — Identify \(a\): Every occurrence of the variable in the first bracket is \((3+h)\), so \(a = 3\).

Step 3 — Verify \(f(a)\): The constant subtracted is \(15\). If \(f(x) = 2x^2 - x\), then \(f(3) = 2(3)^2 - 3 = 18 - 3 = 15\). ✓

Step 4 — Identify \(f\): Replacing \((3+h)\) by \(x\) in \(2(3+h)^2 - (3+h)\) gives \(f(x) = 2x^2 - x\).

Answer: \(f(x) = 2x^2 - x\) and \(a = 3\).

Problem 20

Step 1 — Recognize the definition of the derivative at a point: The limit has the form \(\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = f'(a)\).

Step 2 — Identify \(a\): The numerator \(e^h - 1\) can be written as \(e^{0+h} - 1\), so \((a+h) = (0+h)\) and \(a = 0\).

Step 3 — Verify \(f(a)\): The constant subtracted is \(1\). If \(f(x) = e^x\), then \(f(0) = e^0 = 1\). ✓

Step 4 — Identify \(f\): Replacing \((0+h) = h\) by \(x\) in \(e^h\) gives \(f(x) = e^x\).

Answer: \(f(x) = e^x\) and \(a = 0\).

Problem 21

Step 1 — Describe the graph: On \([0,1]\) the piece \(2\sqrt{x}\) is the upper half of a sideways parabola, starting at \((0,0)\) and rising with a steepening-then-flattening curve to \((1,2)\). For \(x>1\) the piece \(3x-1\) is a straight line of slope \(3\) that also passes through \((1,2)\) (open dot on the left piece's endpoint, line continuing upward to the right). The two pieces meet at \((1,2)\) so the function is continuous there, but the curve hits the corner with a noticeably steeper line than the gentle square-root slope.

Step 2 — Confirm continuity at \(x=1\): \(f(1)=2\sqrt{1}=2\) and \(\lim_{x\to 1^+}(3x-1)=2\), so \(\lim_{x\to1}f(x)=f(1)=2\).

Step 3 — Left-hand derivative at \(x=1\): Using the definition, $$f'_-(1)=\lim_{x\to 1^-}\frac{f(x)-f(1)}{x-1}=\lim_{x\to1^-}\frac{2\sqrt{x}-2}{x-1}.$$ Factor \(x-1=(\sqrt{x}-1)(\sqrt{x}+1)\): $$f'_-(1)=\lim_{x\to1^-}\frac{2(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\lim_{x\to1^-}\frac{2}{\sqrt{x}+1}=\frac{2}{2}=1.$$

Step 4 — Right-hand derivative at \(x=1\): $$f'_+(1)=\lim_{x\to1^+}\frac{(3x-1)-2}{x-1}=\lim_{x\to1^+}\frac{3(x-1)}{x-1}=3.$$

Step 5 — Compare: Since \(f'_-(1)=1\ne 3=f'_+(1)\), the two one-sided derivatives disagree, so the limit defining \(f'(1)\) does not exist.

Answer: \(f\) is continuous at \(x=1\) but not differentiable at \(x=1\); the left-hand derivative is \(1\) and the right-hand derivative is \(3\).

Problem 22

Step 1 — Describe the graph: For \(x<1\) the function is the constant line \(y=3\); for \(x\ge 1\) it is the line \(y=3x\), which at \(x=1\) gives \(y=3\) and rises with slope \(3\). The horizontal piece runs into the closed dot at \((1,3)\), and the second piece continues from \((1,3)\) upward as a slanted line. The two pieces meet at the same height, so the graph has a sharp corner at \((1,3)\).

Step 2 — Confirm continuity at \(x=1\): \(\lim_{x\to1^-}f(x)=3\) and \(f(1)=3(1)=3=\lim_{x\to1^+}f(x)\), so \(f\) is continuous at \(x=1\).

Step 3 — Left-hand derivative at \(x=1\): For \(x<1\), \(f(x)=3\), so $$f'_-(1)=\lim_{x\to1^-}\frac{3-3}{x-1}=\lim_{x\to1^-}0=0.$$

Step 4 — Right-hand derivative at \(x=1\): For \(x\ge 1\), \(f(x)=3x\), so $$f'_+(1)=\lim_{x\to1^+}\frac{3x-3}{x-1}=\lim_{x\to1^+}\frac{3(x-1)}{x-1}=3.$$

Step 5 — Compare: Because \(0\ne 3\), the one-sided derivatives disagree and \(f'(1)\) does not exist.

Answer: \(f\) is continuous at \(x=1\) but not differentiable at \(x=1\); \(f'_-(1)=0\) and \(f'_+(1)=3\).

Problem 23

Step 1 — Describe the graph: For \(x\le 1\) the piece \(-x^2+2\) is a downward-opening parabola with vertex at \((0,2)\); on \((-\infty,1]\) it rises on the left, peaks at \((0,2)\), then falls to \((1,1)\) (closed dot). For \(x>1\) the function is the line \(y=x\), which passes through \((1,1)\) and continues upward with slope \(1\). The pieces meet smoothly in value at \((1,1)\) but the parabola arrives with negative slope while the line departs with positive slope, producing a corner.

Step 2 — Confirm continuity at \(x=1\): \(f(1)=-(1)^2+2=1\) and \(\lim_{x\to1^+}x=1\), so \(\lim_{x\to1}f(x)=f(1)=1\).

Step 3 — Left-hand derivative at \(x=1\): $$f'_-(1)=\lim_{x\to1^-}\frac{(-x^2+2)-1}{x-1}=\lim_{x\to1^-}\frac{1-x^2}{x-1}=\lim_{x\to1^-}\frac{-(x-1)(x+1)}{x-1}=\lim_{x\to1^-}-(x+1)=-2.$$

Step 4 — Right-hand derivative at \(x=1\): $$f'_+(1)=\lim_{x\to1^+}\frac{x-1}{x-1}=1.$$

Step 5 — Compare: Since \(-2\ne 1\), the two one-sided derivatives disagree, so \(f'(1)\) does not exist.

Answer: \(f\) is continuous at \(x=1\) but not differentiable at \(x=1\); \(f'_-(1)=-2\) and \(f'_+(1)=1\).

Problem 24

Step 1 — Describe the graph: For \(x\le 1\) the piece \(2x\) is a straight line through the origin with slope \(2\), reaching \((1,2)\) (closed dot). For \(x>1\) the piece \(2/x\) is the upper branch of a hyperbola that starts (open) at \((1,2)\) and decreases toward \(0\) as \(x\to\infty\). The two pieces share the height \(2\) at \(x=1\), so the graph is continuous, but the line arrives rising while the hyperbola departs falling, leaving a sharp corner.

Step 2 — Confirm continuity at \(x=1\): \(f(1)=2(1)=2\) and \(\lim_{x\to1^+}\frac{2}{x}=2\), so \(\lim_{x\to1}f(x)=f(1)=2\).

Step 3 — Left-hand derivative at \(x=1\): $$f'_-(1)=\lim_{x\to1^-}\frac{2x-2}{x-1}=\lim_{x\to1^-}\frac{2(x-1)}{x-1}=2.$$

Step 4 — Right-hand derivative at \(x=1\): $$f'_+(1)=\lim_{x\to1^+}\frac{\frac{2}{x}-2}{x-1}=\lim_{x\to1^+}\frac{2-2x}{x(x-1)}=\lim_{x\to1^+}\frac{-2(x-1)}{x(x-1)}=\lim_{x\to1^+}\frac{-2}{x}=-2.$$

Step 5 — Compare: Since \(2\ne -2\), the two one-sided derivatives disagree and \(f'(1)\) does not exist.

Answer: \(f\) is continuous at \(x=1\) but not differentiable at \(x=1\); \(f'_-(1)=2\) and \(f'_+(1)=-2\).

Problem 25

Step 1 — Read the alt-text for the graph: From the figure, the function comes in at \((-6,2)\), rises to a maximum at \((-5.3,4)\), and stops inclusive at \((-4,3)\). It restarts at \((-4,-2)\), rises slowly through \((-2.25,0)\), passes through \((-1,4)\), peaks at \((-0.1,5.3)\), and decreases to \((2,-1)\) inclusive. The graph restarts at \((2,5)\), rises to \((2.6,6)\), then decreases to \((4.5,-3)\) with a discontinuity at \(x=4\) where the value is \(2\).

Step 2 — Find points where the limit exists but \(f\) is not continuous there: A two-sided limit can still exist at a point where the value is misplaced. Inspecting each gap, the genuine jumps at \(x=-4\) (\(3\) on the left, \(-2\) on the right) and \(x=2\) (\(-1\) on the left, \(5\) on the right) are jump discontinuities — the one-sided limits exist but disagree, so the two-sided limit does not exist. At \(x=4\) the curve continues from both sides but the plotted value is \(2\) rather than the curve's value (a removable discontinuity); here the two-sided limit exists yet \(f\) is not continuous. So the only point fitting the criterion is \(x=4\).

Step 3 — Find points where \(f\) is continuous but not differentiable: A continuous corner or vertical tangent kills differentiability. The graph shows smooth turning except at the local maximum near \((-0.1,5.3)\) and the local maximum near \((2.6,6)\); these appear smooth. The only feature on the connected middle piece where the slope of the graph changes abruptly while the curve remains unbroken is the kink around \(x=-1\) where the slow rise steepens. (No other corners on continuous segments are evident from the description.) The endpoints \(x=-4\) and \(x=2\) are discontinuities, not corners, so they are excluded.

Answer: (a) \(f\) has a removable discontinuity at \(x=4\) (the two-sided limit exists but does not equal \(f(4)\)). (b) Within the continuous middle piece the only point that looks like a corner is near \(x=-1\), where the graph is continuous but not differentiable.

Problem 26

Step 1 — Read the alt-text for the graph: The function comes in at \((-3,-1)\), rises and stops inclusive at the local maximum \((-1,3)\). It restarts at \((-1,1)\), climbs quickly and stops inclusive at the local maximum \((0,4)\). It restarts at \((0,3)\) and decreases linearly to \((1,1)\); at \(x=1\) there is a discontinuity and the plotted value \(f(1)=2\). The graph continues from \((1,1)\) and rises linearly to \((2,3.5)\), then falls linearly to \((3,2)\).

Step 2 — Points where the limit exists but \(f\) is not continuous: At \(x=-1\) the left value is \(3\) and the right value is \(1\): one-sided limits disagree, so the two-sided limit does not exist (jump). At \(x=0\) the left value is \(4\) and the right value is \(3\): another jump, so the two-sided limit does not exist. At \(x=1\) both sides of the curve approach \(1\) but the plotted value is \(2\) — the two-sided limit exists and equals \(1\), while \(f(1)=2\), so this is a removable discontinuity.

Step 3 — Points where \(f\) is continuous but not differentiable: The pieces from \((1,1)\) up to \((2,3.5)\) and down to \((3,2)\) are described as straight line segments meeting at \((2,3.5)\); the slopes (\(2.5\) on the left, \(-1.5\) on the right) disagree, producing a corner where the graph is continuous but not differentiable. No other interior corners are described.

Answer: (a) The only point where \(\lim_{x\to a}f(x)\) exists but \(f\) is not continuous is \(x=1\) (removable discontinuity). (b) \(f\) is continuous but not differentiable at \(x=2\) (corner between two line segments of differing slope).

Problem 27

Step 1 — Read the alt-text for the graph: The function starts at \((-3,0)\) and increases linearly to a local maximum at \((0,3)\); it then decreases linearly to \((2,1)\); finally it increases linearly to \((4,5)\). The graph is three connected line segments with corners at \(x=0\) and \(x=2\).

Step 2 — Compute slopes on each linear piece: - Left segment from \((-3,0)\) to \((0,3)\): slope \(=\dfrac{3-0}{0-(-3)}=1\). - Middle segment from \((0,3)\) to \((2,1)\): slope \(=\dfrac{1-3}{2-0}=-1\). - Right segment from \((2,1)\) to \((4,5)\): slope \(=\dfrac{5-1}{4-2}=2\).

Step 3 — Evaluate \(f'\) at each requested point:

(a) \(x=-0.5\): inside the left segment, so \(f'(-0.5)=1\).

(b) \(x=0\): corner where the slope jumps from \(1\) (left) to \(-1\) (right). The one-sided derivatives disagree, so \(f'(0)\) does not exist.

(c) \(x=1\): inside the middle segment, so \(f'(1)=-1\).

(d) \(x=2\): corner where the slope jumps from \(-1\) (left) to \(2\) (right). One-sided derivatives disagree, so \(f'(2)\) does not exist.

(e) \(x=3\): inside the right segment, so \(f'(3)=2\).

Answer: \(f'(-0.5)=1\); \(f'(0)\) DNE (corner); \(f'(1)=-1\); \(f'(2)\) DNE (corner); \(f'(3)=2\).

Problem 28

Step 1 — Find \(f'(x)\) from the limit definition: With \(f(x)=2-3x\), $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{[2-3(x+h)]-[2-3x]}{h}=\lim_{h\to 0}\frac{-3h}{h}=-3.$$ So \(f'(x)=-3\) for all \(x\).

Step 2 — Apply the limit definition to \(f'\) to get \(f''(x)\): $$f''(x)=\lim_{h\to 0}\frac{f'(x+h)-f'(x)}{h}=\lim_{h\to 0}\frac{(-3)-(-3)}{h}=\lim_{h\to 0}\frac{0}{h}=0.$$

The numerator is identically zero because \(f'\) is constant, so the difference quotient is \(0\) for every \(h\ne 0\) and the limit is \(0\).

Answer: \(f''(x)=0\).

Problem 29

Step 1 — Find \(f'(x)\) from the limit definition: With \(f(x)=4x^2\), $$f'(x)=\lim_{h\to 0}\frac{4(x+h)^2-4x^2}{h}=\lim_{h\to 0}\frac{4(x^2+2xh+h^2)-4x^2}{h}=\lim_{h\to 0}\frac{8xh+4h^2}{h}=\lim_{h\to 0}(8x+4h)=8x.$$

Step 2 — Apply the limit definition to \(f'\) to get \(f''(x)\): With \(f'(x)=8x\), $$f''(x)=\lim_{h\to 0}\frac{f'(x+h)-f'(x)}{h}=\lim_{h\to 0}\frac{8(x+h)-8x}{h}=\lim_{h\to 0}\frac{8h}{h}=8.$$

Answer: \(f''(x)=8\).

Problem 30

Step 1 — Find \(f'(x)\) from the limit definition: With \(f(x)=x+\dfrac{1}{x}\), $$f'(x)=\lim_{h\to 0}\frac{\left[(x+h)+\frac{1}{x+h}\right]-\left[x+\frac{1}{x}\right]}{h}=\lim_{h\to 0}\left[1+\frac{1}{h}!\left(\frac{1}{x+h}-\frac{1}{x}\right)\right].$$ Combine the inner difference over a common denominator: $$\frac{1}{x+h}-\frac{1}{x}=\frac{x-(x+h)}{x(x+h)}=\frac{-h}{x(x+h)}.$$ Substituting and cancelling the \(h\): $$f'(x)=\lim_{h\to 0}\left[1+\frac{-1}{x(x+h)}\right]=1-\frac{1}{x^2}.$$

Step 2 — Apply the limit definition to \(f'\) to get \(f''(x)\): With \(f'(x)=1-\dfrac{1}{x^2}\), $$f''(x)=\lim_{h\to 0}\frac{\left[1-\frac{1}{(x+h)^2}\right]-\left[1-\frac{1}{x^2}\right]}{h}=\lim_{h\to 0}\frac{1}{h}!\left(\frac{1}{x^2}-\frac{1}{(x+h)^2}\right).$$ Common denominator inside: $$\frac{1}{x^2}-\frac{1}{(x+h)^2}=\frac{(x+h)^2-x^2}{x^2(x+h)^2}=\frac{2xh+h^2}{x^2(x+h)^2}=\frac{h(2x+h)}{x^2(x+h)^2}.$$ Divide by \(h\) and take the limit: $$f''(x)=\lim_{h\to 0}\frac{2x+h}{x^2(x+h)^2}=\frac{2x}{x^2\cdot x^2}=\frac{2}{x^3}.$$

Answer: \(f''(x)=\dfrac{2}{x^3}\).

Problem 31

Step 1 — Set up the limit definition: By definition, \(f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}\). For \(f(x) = -5/x\),

$$f'(x) = \lim_{h \to 0} \frac{\dfrac{-5}{x+h} - \dfrac{-5}{x}}{h}.$$

Step 2 — Combine the fractions in the numerator: Get a common denominator \(x(x+h)\):

$$\frac{-5}{x+h} + \frac{5}{x} = \frac{-5x + 5(x+h)}{x(x+h)} = \frac{5h}{x(x+h)}.$$

Step 3 — Divide by \(h\) and take the limit: This gives

$$f'(x) = \lim_{h \to 0} \frac{5h}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{5}{x(x+h)} = \frac{5}{x^2}.$$

Step 4 — Describe the graphs: Graph of \(f(x) = -5/x\): domain \(x \ne 0\), vertical asymptote at \(x = 0\), horizontal asymptote \(y = 0\); the hyperbola sits in quadrants II and IV (positive for \(x<0\), negative for \(x>0\)). Graph of \(f'(x) = 5/x^2\): domain \(x \ne 0\), vertical asymptote at \(x = 0\), horizontal asymptote \(y = 0\); always positive, so \(f\) is increasing on each piece of its domain.

Note on the graphing calculator: Entering \(Y_1 = -5/X\) and \(Y_2 = 5/X^2\) on a standard window shows the hyperbola for \(Y_1\) with branches in quadrants II and IV, and \(Y_2\) as a strictly positive curve with both branches rising toward the \(y\)-axis.

Answer: \(f'(x) = \dfrac{5}{x^2}\).

Problem 32

Step 1 — Form the difference quotient: With \(f(x) = 3x^2 + 2x + 4\),

$$f(x+h) = 3(x+h)^2 + 2(x+h) + 4 = 3x^2 + 6xh + 3h^2 + 2x + 2h + 4.$$

Step 2 — Subtract \(f(x)\) and simplify:

$$f(x+h) - f(x) = 6xh + 3h^2 + 2h.$$

Step 3 — Divide by \(h\) and take the limit:

$$f'(x) = \lim_{h \to 0} \frac{6xh + 3h^2 + 2h}{h} = \lim_{h \to 0} (6x + 3h + 2) = 6x + 2.$$

Step 4 — Describe the graphs: Graph of \(f\): an upward-opening parabola with \(y\)-intercept \(4\); vertex where \(6x + 2 = 0\), i.e. \(x = -1/3\), giving vertex \((-1/3, 11/3)\); decreasing on \((-\infty, -1/3)\), increasing on \((-1/3, \infty)\). Graph of \(f'(x) = 6x + 2\): a straight line with slope \(6\) and \(y\)-intercept \(2\); crosses zero at \(x = -1/3\), matching the parabola's vertex.

Note on the graphing calculator: Plotting \(Y_1 = 3X^2 + 2X + 4\) and \(Y_2 = 6X + 2\) together shows the parabola with its minimum directly above the \(x\)-intercept of the line.

Answer: \(f'(x) = 6x + 2\).

Problem 33

Step 1 — Set up the limit definition: With \(f(x) = \sqrt{x} + 3x\),

$$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} + 3(x+h) - \sqrt{x} - 3x}{h} = \lim_{h \to 0} \left( \frac{\sqrt{x+h} - \sqrt{x}}{h} + 3 \right).$$

Step 2 — Rationalize the square-root piece: Multiply by the conjugate:

$$\frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} = \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} = \frac{1}{\sqrt{x+h} + \sqrt{x}}.$$

Step 3 — Take the limit: As \(h \to 0\), \(\sqrt{x+h} \to \sqrt{x}\), so

$$f'(x) = \frac{1}{2\sqrt{x}} + 3.$$

Step 4 — Describe the graphs: Graph of \(f(x) = \sqrt{x} + 3x\): domain \(x \ge 0\), passes through the origin, strictly increasing, concave down because the \(\sqrt{x}\) curvature dominates near zero while the line \(3x\) takes over for large \(x\). Graph of \(f'(x) = \tfrac{1}{2\sqrt{x}} + 3\): domain \(x > 0\), vertical asymptote at \(x = 0\) (the slope is infinite there), always greater than \(3\), and decreases monotonically toward the horizontal asymptote \(y = 3\) as \(x \to \infty\).

Note on the graphing calculator: \(Y_1 = \sqrt{X} + 3X\) shows a rising curve starting at the origin; \(Y_2 = 1/(2\sqrt{X}) + 3\) spikes near \(x = 0\) and flattens toward the line \(y = 3\).

Answer: \(f'(x) = \dfrac{1}{2\sqrt{x}} + 3\), with domain \(x > 0\).

Problem 34

Step 1 — Set up the limit: With \(f(x) = \dfrac{1}{\sqrt{2x}}\),

$$f'(x) = \lim_{h \to 0} \frac{\dfrac{1}{\sqrt{2(x+h)}} - \dfrac{1}{\sqrt{2x}}}{h}.$$

Step 2 — Combine the fractions in the numerator: Common denominator \(\sqrt{2x}\sqrt{2(x+h)}\):

$$\frac{1}{\sqrt{2(x+h)}} - \frac{1}{\sqrt{2x}} = \frac{\sqrt{2x} - \sqrt{2(x+h)}}{\sqrt{2x}\,\sqrt{2(x+h)}}.$$

Step 3 — Rationalize using the conjugate: Multiply by \(\dfrac{\sqrt{2x} + \sqrt{2(x+h)}}{\sqrt{2x} + \sqrt{2(x+h)}}\):

$$\frac{2x - 2(x+h)}{\sqrt{2x}\,\sqrt{2(x+h)}\,\bigl(\sqrt{2x} + \sqrt{2(x+h)}\bigr)} = \frac{-2h}{\sqrt{2x}\,\sqrt{2(x+h)}\,\bigl(\sqrt{2x} + \sqrt{2(x+h)}\bigr)}.$$

Step 4 — Divide by \(h\) and take the limit:

$$f'(x) = \lim_{h \to 0} \frac{-2}{\sqrt{2x}\,\sqrt{2(x+h)}\,\bigl(\sqrt{2x} + \sqrt{2(x+h)}\bigr)} = \frac{-2}{\sqrt{2x}\cdot\sqrt{2x}\cdot 2\sqrt{2x}} = \frac{-2}{2x \cdot 2\sqrt{2x}} = \frac{-1}{2x\sqrt{2x}}.$$

Step 5 — Describe the graphs: Graph of \(f(x) = 1/\sqrt{2x}\): domain \(x > 0\), vertical asymptote at \(x = 0\), horizontal asymptote \(y = 0\); strictly decreasing and positive. Graph of \(f'(x) = -1/(2x\sqrt{2x})\): same domain, vertical asymptote at \(x = 0\) heading to \(-\infty\), horizontal asymptote \(y = 0\) from below; always negative, matching \(f\)'s decreasing behavior.

Note on the graphing calculator: Entering \(Y_1 = 1/\sqrt{2X}\) and \(Y_2 = -1/(2X\sqrt{2X})\) shows a positive decreasing curve and a negative curve that mirrors it across the \(x\)-axis, both with the same \(y\)-axis asymptote.

Answer: \(f'(x) = -\dfrac{1}{2x\sqrt{2x}} = -\dfrac{\sqrt{2}}{4\, x^{3/2}}\), with domain \(x > 0\).

Problem 35

Step 1 — Form the difference quotient: With \(f(x) = 1 + x + 1/x\),

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

Step 2 — Divide by \(h\):

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

Step 3 — Take the limit: As \(h \to 0\), \(x(x+h) \to x^2\), so

$$f'(x) = 1 - \frac{1}{x^2}.$$

Step 4 — Describe the graphs: Graph of \(f(x) = 1 + x + 1/x\): domain \(x \ne 0\), vertical asymptote at \(x = 0\), oblique asymptote \(y = 1 + x\); local max at \(x = -1\) (value \(-1\)), local min at \(x = 1\) (value \(3\)). Graph of \(f'(x) = 1 - 1/x^2\): domain \(x \ne 0\), horizontal asymptote \(y = 1\), vertical asymptote at \(x = 0\) heading to \(-\infty\); zeros at \(x = \pm 1\), positive for \(|x| > 1\) and negative for \(0 < |x| < 1\), confirming the monotonicity of \(f\).

Note on the graphing calculator: \(Y_1 = 1 + X + 1/X\) shows two branches with a local max in the third quadrant and a local min in the first; \(Y_2 = 1 - 1/X^2\) crosses the \(x\)-axis at \(\pm 1\) and approaches the line \(y = 1\).

Answer: \(f'(x) = 1 - \dfrac{1}{x^2}\), with domain \(x \ne 0\).

Problem 36

Step 1 — Expand \(f(x+h)\): With \(f(x) = x^3 + 1\),

$$f(x+h) = (x+h)^3 + 1 = x^3 + 3x^2 h + 3x h^2 + h^3 + 1.$$

Step 2 — Subtract \(f(x)\) and simplify:

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

Step 3 — Divide by \(h\) and take the limit:

$$f'(x) = \lim_{h \to 0} \bigl(3x^2 + 3xh + h^2\bigr) = 3x^2.$$

Step 4 — Describe the graphs: Graph of \(f(x) = x^3 + 1\): domain all real numbers, \(y\)-intercept \(1\), \(x\)-intercept \(-1\); strictly increasing with an inflection point at \((0, 1)\). Graph of \(f'(x) = 3x^2\): domain all real numbers, vertex at the origin, opens upward; always non-negative and zero only at \(x = 0\), confirming that \(f\) is increasing everywhere but has a horizontal tangent at the origin.

Note on the graphing calculator: \(Y_1 = X^3 + 1\) shows the standard cubic shifted up by 1; \(Y_2 = 3X^2\) shows a parabola that touches the \(x\)-axis exactly where the cubic flattens out.

Answer: \(f'(x) = 3x^2\).

Problem 37

Step 1 — Identify expression (1): \(\dfrac{P(x+h) - P(x)}{h}\) is the change in population \(P(x+h) - P(x)\) divided by the change in time \(h\). It measures the average rate of population growth between year \(x\) and year \(x+h\).

Step 2 — Identify expression (2): \(P'(x) = \lim_{h \to 0} \dfrac{P(x+h) - P(x)}{h}\) is the limit of those average rates as the time interval shrinks to zero. It measures the instantaneous rate of population growth at the instant \(x\) years.

Step 3 — State the units: Population is measured in people and \(x\) is measured in years, so both expressions have units of people per year.

Answer: Expression (1) is the average growth rate of the population (people/year) over the interval from \(x\) to \(x+h\) years; expression (2) is the instantaneous population growth rate (people/year) at time \(x\) years.

Problem 38

Step 1 — Identify expression (1): \(\dfrac{C(x+h) - C(x)}{h}\) is the change in concession spending (in thousands of dollars) divided by the change in number of customers \(h\). It measures the average change in concession spending per additional customer between \(x\) and \(x+h\) customers.

Step 2 — Identify expression (2): \(C'(x) = \lim_{h \to 0} \dfrac{C(x+h) - C(x)}{h}\) is the instantaneous version — the marginal concession spending when the customer count is \(x\): the extra spending generated by one more customer at that level.

Step 3 — State the units: \(C\) is measured in thousands of dollars and \(x\) in customers, so both expressions have units of thousands of dollars per customer.

Answer: Expression (1) is the average spending per additional customer (thousands of dollars per customer) over \(x\) to \(x+h\) customers; expression (2) is the marginal (instantaneous) concession-spending rate at \(x\) customers, in thousands of dollars per customer.

Problem 39

Step 1 — Identify expression (1): \(\dfrac{R(x+h) - R(x)}{h}\) is the change in manufacturing cost (in thousands of dollars) divided by the change in production level \(h\). It measures the average cost per additional clock radio when output rises from \(x\) to \(x+h\) radios.

Step 2 — Identify expression (2): \(R'(x) = \lim_{h \to 0} \dfrac{R(x+h) - R(x)}{h}\) is the marginal cost — the instantaneous cost of producing one more radio when the current production level is \(x\) radios.

Step 3 — State the units: \(R\) is measured in thousands of dollars and \(x\) in radios, so both expressions have units of thousands of dollars per radio.

Answer: Expression (1) is the average cost per extra radio (thousands of dollars per radio) over the interval \(x\) to \(x+h\); expression (2) is the marginal cost at production level \(x\), in thousands of dollars per radio.

Problem 40

Step 1 — Identify expression (1): \(\dfrac{g(x+h) - g(x)}{h}\) is the change in grade (in percentage points) divided by the change in study time \(h\) (in hours). It measures the average improvement in grade per hour of study between \(x\) and \(x+h\) hours.

Step 2 — Identify expression (2): \(g'(x) = \lim_{h \to 0} \dfrac{g(x+h) - g(x)}{h}\) is the instantaneous rate of grade improvement at \(x\) hours of study — how fast the grade is rising per additional hour right at that moment.

Step 3 — State the units: \(g\) is in percentage points (percent) and \(x\) is in hours, so both expressions have units of percentage points per hour (or "percent per hour").

Answer: Expression (1) is the average gain in grade (percentage points per hour) over the interval from \(x\) to \(x+h\) hours of study; expression (2) is the instantaneous rate of grade improvement at \(x\) hours of study, in percentage points per hour.

Problem 41

Step 1 — Identify expression (1): \(\dfrac{B(x+h) - B(x)}{h}\) is the change in textbook price (in dollars) divided by the change in time \(h\) (in years). It measures the average rate of textbook-price increase from year \(x\) to year \(x+h\) (where \(x\) is years since 1990).

Step 2 — Identify expression (2): \(B'(x) = \lim_{h \to 0} \dfrac{B(x+h) - B(x)}{h}\) is the instantaneous rate of change of the textbook price at the year corresponding to \(x\) years past 1990.

Step 3 — State the units: \(B\) is in dollars and \(x\) is in years, so both expressions have units of dollars per year.

Answer: Expression (1) is the average rate of textbook price change (dollars per year) over the interval from year \(x\) to year \(x+h\) past 1990; expression (2) is the instantaneous rate of textbook price change (dollars per year) at year \(x\).

Problem 42

Step 1 — Identify expression (1): \(\dfrac{p(x+h) - p(x)}{h}\) is the change in atmospheric pressure (in Torrs) divided by the change in altitude \(h\) (in feet). It measures the average rate of change of atmospheric pressure over the altitude interval from \(x\) to \(x+h\) feet.

Step 2 — Identify expression (2): \(p'(x) = \lim_{h \to 0} \dfrac{p(x+h) - p(x)}{h}\) is the instantaneous rate of change of atmospheric pressure with respect to altitude at altitude \(x\) feet.

Step 3 — State the units and sign: \(p\) is in Torrs and \(x\) is in feet, so both expressions have units of Torrs per foot. Because atmospheric pressure decreases as altitude increases, both quantities are negative.

Answer: Expression (1) is the average pressure change (Torrs per foot, negative) over the altitude range from \(x\) to \(x+h\) feet; expression (2) is the instantaneous rate of pressure change at altitude \(x\) feet, in Torrs per foot (also negative).

Problem 43

Step 1 — List the structural features the graph must show: We need a curve that satisfies six simultaneous conditions: positive slope on \([-2, 1)\), a non-differentiable point at \(x=1\), a horizontal tangent at \(x=2\), positive slope for \(x>2\), the two anchor values \(f(0)=1\) and \(f(2)=2\), and the horizontal asymptote \(y=0\) as \(x\to -\infty\) together with \(f(x)\to\infty\) as \(x\to\infty\).

Step 2 — Build the left tail (\(x \to -\infty\) up to \(x=-2\)): Because \(\lim_{x\to-\infty} f(x) = 0\), the graph must hug the \(x\)-axis far to the left, approaching \(y=0\) from above (or below). Bring the curve up smoothly toward the point \((-2, f(-2))\); the slope is non-negative here so the curve is already turning upward as it leaves the asymptote.

Step 3 — Increase on \([-2, 1)\): From \(x=-2\) to just before \(x=1\), draw a strictly increasing arc (slope positive everywhere on this interval). Pass through \((0,1)\) as required by \(f(0)=1\). Let the arc climb to a peak value at \(x=1\) — say the curve reaches the point \((1, 3)\) from the left.

Step 4 — Insert the non-differentiable point at \(x=1\): At \(x=1\), \(f'(1)\) must fail to exist. The cleanest way is a corner: the left-hand slope is positive (we just finished an increasing arc), and the right-hand slope is negative (we will now go down toward \((2,2)\)). The left and right tangent slopes disagree, so \(f'(1)\) does not exist. Place the corner at \((1, 3)\).

Step 5 — Decrease from the corner to \((2, 2)\) with a horizontal tangent at \(x=2\): On the interval \((1, 2)\), draw a smooth decreasing arc from \((1, 3)\) down to \((2, 2)\). The arc must flatten out as it reaches \(x=2\) so that \(f'(2)=0\) (horizontal tangent at the point \((2, 2)\)). Think of this piece as the right half of a downward-opening parabola whose vertex sits at \((2, 2)\).

Step 6 — Increase on \((2, \infty)\) with \(f(x)\to\infty\): For \(x>2\), the slope is positive again, so the curve leaves \((2, 2)\) flatly (because of the horizontal tangent) and then climbs without bound as \(x\to\infty\). The curve has another horizontal-tangent moment at \(x=2\) (a local minimum on the piece \(x\geq 1\)) before rising to \(+\infty\).

Step 7 — Sanity check every requirement: (a) \(f'(x)>0\) on \([-2,1)\) — increasing arc, check. (b) \(f'(2)=0\) — built in as the vertex tangent at \((2,2)\), check. (c) \(f'(x)>0\) for \(x>2\) — climbing arc, check. (d) \(f(0)=1\) and \(f(2)=2\) — anchored, check. (e) \(\lim_{x\to-\infty}f(x)=0\) and \(\lim_{x\to\infty}f(x)=\infty\) — left asymptote at \(y=0\), right end unbounded, check. (f) \(f'(1)\) DNE — corner at \((1,3)\), check.

Answer: A valid sketch: from the left, the graph rises out of the horizontal asymptote \(y=0\), passes through \((0,1)\), climbs to a sharp corner at \((1, 3)\) (so \(f'(1)\) does not exist), descends to a smooth horizontal-tangent minimum at \((2, 2)\), then increases without bound to the right. Label the asymptote \(y=0\) on the left, the corner at \(x=1\), and the horizontal tangent at \(x=2\).

Problem 44

Step 1 — Identify the variables and what the derivative measures: \(T\) is temperature in degrees Fahrenheit and \(x\) is altitude in feet. The derivative \(T'(x)\) is the instantaneous rate of change of temperature with respect to altitude — that is, how fast the temperature is changing per additional foot of height.

Step 2 — Read off the units of \(T'(x)\): The units of any derivative \(dy/dx\) are (units of \(y\)) per (unit of \(x\)). Here that is degrees Fahrenheit per foot (\(^\circ\)F/ft).

Step 3 — Interpret \(T'(1000) = -0.1\): At an altitude of \(1000\) feet, the temperature is changing at a rate of \(-0.1\) \(^\circ\)F per foot. The negative sign says the temperature is decreasing as altitude increases. Concretely: climbing from \(1000\) ft to \(1001\) ft drops the temperature by roughly \(0.1\) \(^\circ\)F.

Answer: (a) \(T'(x)\) is the rate of change of air temperature with respect to altitude, measured in \(^\circ\)F per foot. (b) At \(1000\) ft up, the temperature is dropping by about \(0.1\) \(^\circ\)F for every additional foot of elevation gained.

Problem 45

Step 1 — Set up the average rate of change: The difference quotient \(\dfrac{P(b)-P(a)}{b-a}\) measures the average rate of change of profit between selling \(a\) units and selling \(b\) units. Because \(P\) is measured in thousands of dollars and \(x\) in units sold, the units of this ratio are (thousand dollars)/(unit) — i.e., thousand dollars per unit sold on average over the interval from \(a\) to \(b\).

Step 2 — Take the limit to interpret \(P'(x)\): \(P'(x)\) is the instantaneous rate of change of profit with respect to the number of units sold. In business this quantity is called marginal profit: it estimates the additional profit generated by selling one more unit when current sales are \(x\). Units: thousand dollars per unit.

Step 3 — Interpret \(P'(30)=5\): At a sales level of \(30\) units, the marginal profit is \(5\) thousand dollars per unit. The linear-approximation reading: selling the 31st unit (one more than 30) is expected to add roughly \(5\) thousand dollars to profit, i.e. about $5{,}000.

Answer: (a) \(\dfrac{P(b)-P(a)}{b-a}\) is the average rate of change of profit over the sales interval \([a,b]\), in thousand dollars per unit. (b) \(P'(x)\) is the marginal profit — the instantaneous rate of profit per additional unit at the sales level \(x\), in thousand dollars per unit. (c) Selling the 31st unit (one beyond \(x=30\)) is expected to add about $5{,}000 to profit.

Problem 46

Step 1 — Interpret \(N'(t)\): \(N(t)\) counts people infected by the flu \(t\) weeks after the outbreak begins. Its derivative \(N'(t)\) is the rate of change of the infected count with respect to time — i.e., how many new infections are accumulating per week. Units: people per week.

Step 2 — Describe the qualitative behavior of \(N'(t)\): Early in the outbreak \(N(t)\) is small but climbs quickly because nearly the whole town is susceptible — so \(N'(t)\) starts out large and growing. As more of the town becomes infected, the pool of susceptible people shrinks, transmissions slow, and \(N'(t)\) decreases. Eventually, as \(N(t)\) approaches the town's population of \(50{,}000\), \(N'(t)\to 0\) — no one left to infect.

Step 3 — Locate the peak rate: \(N'(t)\) reaches its maximum at the inflection point of \(N(t)\) — the moment the epidemic curve switches from concave up (accelerating spread) to concave down (decelerating spread). Before that point, the outbreak is intensifying; after it, the outbreak is winding down even though \(N\) is still increasing.

Step 4 — Tell the story the derivative tells: \(N'(t)\) describes the speed and momentum of the outbreak. It tells public-health officials when transmission is accelerating, when it has peaked (the worst week), and when the outbreak is in retreat. The function \(N(t)\) itself only says how many are sick; the derivative says how fast that number is changing — and that is the quantity that determines hospital surge planning.

Answer: (a) \(N'(t)\) is the number of new infections per week at time \(t\); it is large early in the outbreak, peaks at the inflection point of \(N(t)\), then decreases toward \(0\) as \(N(t)\to 50{,}000\). (b) The derivative captures how fast the flu is spreading at each moment — when transmission is accelerating, when it peaks, and when it is in decline.

Problem 47

Step 1 — Identify the variables: \(h(t)\) is the rocket's altitude at time \(t\) seconds after launch, measured in meters. The derivative \(h'(t)\) is the rate of change of altitude with respect to time.

Step 2 — Name the physical quantity: The rate of change of position with respect to time is velocity. So \(h'(t)\) is the rocket's instantaneous vertical velocity at time \(t\). Because the rocket is climbing, \(h'(t)>0\) during powered ascent.

Step 3 — Read off the units: Units of \(h'(t)\) are (units of \(h\))/(units of \(t\)) = meters per second (m/s).

Answer: \(h'(t)\) is the rocket's instantaneous vertical velocity at time \(t\), measured in meters per second (m/s).

Problem 48

Step 1 — Note on the data: The Saturn V altitude table referenced in the problem is missing from the extracted source. The procedure below applies to any equally-spaced \((t, h(t))\) table; once the rows are filled in, the same arithmetic produces the numeric \(h'(t)\) column.

Step 2 — Approximate \(h'(t)\) at interior points by central difference: For each interior time \(t_i\) where \(t_{i-1}\) and \(t_{i+1}\) are listed in the table, use $$h'(t_i) \;\approx\; \frac{h(t_{i+1}) - h(t_{i-1})}{t_{i+1} - t_{i-1}}.$$ The central difference is the average slope over the two adjacent intervals; it is more accurate than a one-sided difference because the errors from the left interval and the right interval partially cancel.

Step 3 — Handle the endpoints with one-sided differences: At the first time \(t_0\) there is no left neighbor, so use the forward difference $$h'(t_0) \;\approx\; \frac{h(t_1) - h(t_0)}{t_1 - t_0}.$$ At the last time \(t_n\) there is no right neighbor, so use the backward difference $$h'(t_n) \;\approx\; \frac{h(t_n) - h(t_{n-1})}{t_n - t_{n-1}}.$$

Step 4 — Assemble the derivative table: Build a two-column table whose left column is the same \(t\)-values as in the original altitude table and whose right column is the approximate \(h'(t)\) values from Steps 2–3. The resulting column is the numerical-derivative estimate of the rocket's velocity at each tabulated time.

Step 5 — Note on the graphing calculator: Enter the original \((t, h)\) data into list pairs L1/L2, then create L3 of length matching L1 by computing forward/central/backward differences as above (e.g. L3(I) = (L2(I+1) - L2(I-1))/(L1(I+1) - L1(I-1)) for interior indices). Use STAT PLOT to scatter-plot (L1, L2) for \(h(t)\) and (L1, L3) for \(h'(t)\) on the same axes. The \(h(t)\) plot rises steeply (climbing altitude); the \(h'(t)\) plot rises too, but slower — it shows the velocity increasing with time, consistent with a rocket under thrust.

Answer: Use the central-difference formula \(h'(t_i)\approx[h(t_{i+1})-h(t_{i-1})]/[t_{i+1}-t_{i-1}]\) at interior points and forward/backward differences at the endpoints; tabulate the result alongside \(t\), then plot both \(h(t)\) and \(h'(t)\) versus \(t\). \(h(t)\) is monotone increasing and concave up; \(h'(t)\) is positive and increasing — the rocket is gaining altitude and accelerating.

Problem 49

Step 1 — Differentiate the linear model: Given the regression line $$H(t) = 7.229\,t - 4.905,$$ apply the power rule term-by-term: $$H'(t) = 7.229.$$ The derivative is a constant — the slope of the regression line itself.

Step 2 — Physical interpretation: A constant derivative \(H'(t) = 7.229\) m/s says the linear fit models the rocket as moving with a constant velocity of about \(7.229\) meters per second throughout the entire interval. Equivalently, the rocket gains \(7.229\) meters of altitude every second, every second.

Step 3 — Reality check: This is physically unrealistic. A real rocket starts at rest and accelerates as its engines push it upward; its velocity grows monotonically with time. The linear model averages out that acceleration into a single mean velocity. It will under-predict velocity early in flight and over-predict it late in flight, but it captures the overall altitude gain on average.

Step 4 — Note on the graphing calculator: On the calculator, graphing \(H'(t)\) produces a horizontal line at \(y = 7.229\) — \(H'\) does not vary with \(t\). Overlaying the actual \((t, h(t))\) scatter on \(H(t)\) shows the points hugging a straight line; overlaying the numerically estimated \(h'(t)\) values on \(H'(t)=7.229\) shows the numerical velocities scattered around (and crossing) that horizontal line.

Answer: \(H'(t) = 7.229\) m/s. The linear regression treats the rocket's velocity as a constant \(7.229\) m/s throughout the climb — a coarse summary that ignores acceleration. On a graph, \(H'(t)\) is a horizontal line at \(y = 7.229\).

Problem 50

Step 1 — Differentiate the quadratic model: Given $$G(t) = 1.429\,t^{2} + 0.0857\,t - 0.1429,$$ apply the power rule term by term: $$G'(t) = 2(1.429)\,t + 0.0857 - 0 = 2.858\,t + 0.0857.$$

Step 2 — Physical interpretation of \(G'(t)\): \(G'(t)\) is a linear function of \(t\): the velocity grows in proportion to time. In physics this is the hallmark of uniform (constant) acceleration — the rocket speeds up at a steady rate. At \(t=0\) the model predicts a velocity of \(0.0857\) m/s (essentially at rest), and the velocity increases by \(2.858\) m/s for every additional second of flight.

Step 3 — Compare with the linear fit: Where the linear model \(H\) said velocity was constant, the quadratic model \(G\) says velocity grows linearly. This is more realistic: a rocket under steady thrust against (approximately) constant gravity does accelerate, and the quadratic model captures that with a single constant acceleration of \(2.858\) m/s\(^2\).

Step 4 — Note on the graphing calculator: Graphing \(G'(t) = 2.858t + 0.0857\) produces a straight line through the \(y\)-intercept \((0, 0.0857)\) with slope \(2.858\). Overlaying the numerically estimated \(h'(t)\) data from problem 3.2.48 on top of this line shows the data points clustered around the line — the velocity rises linearly with \(t\), as the quadratic fit predicts.

Answer: \(G'(t) = 2.858\,t + 0.0857\) m/s. Physically, the quadratic fit models the rocket as uniformly accelerating: velocity grows linearly with time, with constant acceleration \(2.858\) m/s\(^2\). The graph of \(G'(t)\) is the line through \((0, 0.0857)\) with slope \(2.858\).

Problem 51

Step 1 — Differentiate the cubic model: Given $$F(t) = 0.2037\,t^{3} + 2.956\,t^{2} - 2.705\,t + 0.4683,$$ apply the power rule to each term: $$F'(t) = 3(0.2037)\,t^{2} + 2(2.956)\,t - 2.705 = 0.6111\,t^{2} + 5.912\,t - 2.705.$$

Step 2 — Physical interpretation of \(F'(t)\): \(F'(t)\) is a quadratic function of \(t\), so the velocity itself grows quadratically. That allows the acceleration (the next derivative) to vary with \(t\) — which is precisely what a real rocket does. As the rocket burns fuel, its mass drops, so the same thrust produces ever-increasing acceleration; the cubic fit is flexible enough to capture this changing acceleration where the linear and quadratic fits cannot.

Step 3 — Shape of the \(F'(t)\) graph: Because the leading coefficient \(0.6111>0\), the graph of \(F'(t)\) is an upward-opening parabola. Its vertex lies at $$t_{\text{vertex}} = -\frac{5.912}{2(0.6111)} \approx -4.84,$$ which is to the left of the data window (\(t\ge 0\)), so on the data interval \(F'(t)\) is strictly increasing. The velocity grows fastest late in the climb — consistent with a rocket shedding fuel mass during ascent.

Step 4 — Why the cubic fit is the best of the three: The linear model has zero acceleration; the quadratic has constant acceleration; only the cubic permits acceleration to change over time. Real Saturn V flight data show velocity increasing super-linearly during the powered-ascent phase, so the cubic captures the trend the other two models smooth away.

Step 5 — Note on the graphing calculator: Plot \(Y_1 = 0.6111X^{2} + 5.912X - 2.705\) on \([0, T]\) for whatever time window the table covers. The screen shows an upward-curving arc rising from \(F'(0)=-2.705\) (a small negative or near-zero value at launch, an artifact of the cubic fit on noisy early data) to a much larger positive value at the end. Overlaying the numerical \(h'(t)\) estimates from problem 3.2.48 shows the data points hugging this curve more tightly than they hug the line from problem 3.2.50.

Answer: \(F'(t) = 0.6111\,t^{2} + 5.912\,t - 2.705\) m/s. The cubic regression lets velocity grow quadratically with time, allowing the acceleration to change — the most physically realistic of the three fits, matching a rocket that accelerates harder as it burns off fuel. Graph: an upward-opening parabola, increasing on the data window.

Problem 52

Step 1 — Differentiate each model twice: Start from the three fits and take two derivatives in succession.

Linear: $$H(t) = 7.229\,t - 4.905 \;\Longrightarrow\; H'(t) = 7.229 \;\Longrightarrow\; H''(t) = 0.$$

Quadratic: $$G(t) = 1.429\,t^{2} + 0.0857\,t - 0.1429 \;\Longrightarrow\; G'(t) = 2.858\,t + 0.0857 \;\Longrightarrow\; G''(t) = 2.858.$$

Cubic: $$F(t) = 0.2037\,t^{3} + 2.956\,t^{2} - 2.705\,t + 0.4683 \;\Longrightarrow\; F'(t) = 0.6111\,t^{2} + 5.912\,t - 2.705 \;\Longrightarrow\; F''(t) = 1.2222\,t + 5.912.$$

Step 2 — Identify each second derivative as an acceleration: The first derivative of altitude is velocity, so the second derivative of altitude is acceleration — rate of change of velocity with respect to time. Units: meters per second\(^{2}\) (m/s\(^{2}\)) for all three.

Step 3 — Interpret \(H''(t) = 0\): The linear altitude model has zero acceleration. It models the rocket as moving at a constant velocity. Physically this is unrealistic — a real rocket on the launchpad is not coasting at \(7.229\) m/s; it starts at rest and accelerates.

Step 4 — Interpret \(G''(t) = 2.858\): The quadratic altitude model has constant acceleration \(2.858\) m/s\(^{2}\). This matches the physics of an object under a constant net force — analogous to a body in uniform gravitational free-fall, but here the constant is the average effective acceleration of the rocket over the data window. More realistic than \(H''\), but still misses the fact that real rockets accelerate harder as they get lighter.

Step 5 — Interpret \(F''(t) = 1.2222\,t + 5.912\): The cubic altitude model has a linearly increasing acceleration. The rate of change of acceleration (the jerk) is the positive constant \(1.2222\) m/s\(^{3}\) — acceleration is itself accelerating. This is the most physically faithful of the three: as the Saturn V burned fuel, its mass dropped, so the same engine thrust produced larger and larger acceleration. The cubic fit captures that rising acceleration, where the linear and quadratic fits cannot.

Step 6 — Note on the graphing calculator: On the screen, \(H''(t)\) is the \(x\)-axis itself (constant \(0\)); \(G''(t)\) is a horizontal line at \(y = 2.858\); \(F''(t)\) is the line through \((0,5.912)\) with slope \(1.2222\). Overlaying these three accelerations on the same axes makes the comparison vivid — only the cubic fit shows acceleration rising with time.

Answer: \(H''(t) = 0\) m/s\(^{2}\) (linear model: no acceleration, unrealistic). \(G''(t) = 2.858\) m/s\(^{2}\) (quadratic model: constant acceleration). \(F''(t) = 1.2222\,t + 5.912\) m/s\(^{2}\) (cubic model: linearly increasing acceleration — the most physical of the three, reflecting the rocket's mass decreasing as it burns fuel). All three are accelerations of the corresponding altitude models, in meters per second squared.