Chapter 4

$\frac{1}{72\pi }\text{cm/sec},$ or approximately 0.0044 cm/sec

$500\text{ft/sec}$

$\frac{1}{10}\text{rad/sec}$

$\mathrm{-0.61}\text{ft/sec}$

$L\left(x\right)=2+\frac{1}{12}\left(x-8\right);$ 2.00833

$L(x)=\text{−}x+\frac{\pi }{2}$

$L(x)=1+4x$

$dy=2x{e}^{x^2}dx$

$dy=1.6,$ $\text{Δ}y=1.64$

The volume measurement is accurate to within $21.6{\text{cm}}^3.$

7.6%

$x=-\frac{2}{3},$ $x=1$

The absolute maximum is $3$ and it occurs at $x=4.$ The absolute minimum is $\mathrm{-1}$ and it occurs at $x=2.$

$c=2$

$\frac{5}{2\sqrt{2}}$ sec

$f$ has a local minimum at $\mathrm{-2}$ and a local maximum at $3.$

$f$ has no local extrema because $f^{\prime }$ does not change sign at $x=1.$

$f$ is concave up over the interval $\left(\text{−}\infty ,\frac{1}{2}\right)$ and concave down over the interval $\left(\frac{1}{2},\infty \right)$

$f$ has a local maximum at $\mathrm{-2}$ and a local minimum at $3.$

Both limits are $3.$ The line $y=3$ is a horizontal asymptote.

Let $\epsilon >0.$ Let $N=\frac{1}{\sqrt{\epsilon }}.$ Therefore, for all $x>N,$ we have

$\left|3-\frac{1}{x^2}-3\right|=\frac{1}{x^2}<\frac{1}{N^2}=\epsilon$

Therefore, $\underset{x\to \infty }{\text{lim}}\left(3-1\text{/}{x}^2\right)=3.$

Let $M>0.$ Let $N=\sqrt{\frac{M}{3}}.$ Then, for all $x>N,$ we have

$3x^2>3N^2=3{\left(\sqrt{\frac{M}{3}}\right)}^2{}^2=\frac{3M}{3}=M$

$\text{−}\infty$

$\frac{3}{5}$

$\text{±}\sqrt{3}$

$\underset{x\to \infty }{\text{lim}}f\left(x\right)=\frac{3}{5},$ $\underset{x\to \text{−}\infty }{\text{lim}}f\left(x\right)=\mathrm{-2}$

$y=\frac{3}{2}x$

The function $f$ has a cusp at $\left(0,5\right)$ $\underset{x\to 0^{-}}{\text{lim}}{f}^{\prime }\left(x\right)=\infty ,$ $\underset{x\to 0^{+}}{\text{lim}}{f}^{\prime }\left(x\right)=\text{−}\infty .$ For end behavior, $\underset{x\to \text{±}\infty }{\text{lim}}f\left(x\right)=\text{−}\infty .$

The maximum area is $5000{\text{ft}}^2.$

$V\left(x\right)=x\left(20-2x\right)\left(30-2x\right).$ The domain is $\left[0,10\right].$

$T\left(x\right)=\frac{x}{6}+\frac{\sqrt{{\left(15-x\right)}^2+1}}{2.5}$

The company should charge $\text{\$}75$ per car per day.

$A\left(x\right)=4x\sqrt{1-x^2}.$ The domain of consideration is $\left[0,1\right].$

$c\left(x\right)=\frac{259.2}{x}+0.2x^2$ dollars

$1$

$0$

$\underset{x\to 0^{+}}{\text{lim}}\text{cos}x=1.$ Therefore, we cannot apply L’Hôpital’s rule. The limit of the quotient is $\infty$

$1$

$0$

$e$

$1$

The function $2^x$ grows faster than $x^{100}.$

$x_1\approx 0.33333333,x_2\approx 0.347222222$

$x_1=2,x_2=1.75$

$x_1\approx -1.842105263,x_2\approx -1.772826920$

$x_1=6,x_2=8,x_3=\frac{26}{3},x_4=\frac{80}{9},x_5=\frac{242}{27};x*=9$

$\text{−}\text{cos}x+C$

$\frac{d}{dx}\left(x\text{sin}x+\text{cos}x+C\right)=\text{sin}x+x\text{cos}x-\text{sin}x=x\text{cos}x$

$x^4-\frac{5}{3}{x}^3+\frac{1}{2}{x}^2-7x+C$

$y=-\frac{3}{x}+5$

$2.93\text{sec},64.5\text{ft}$

$8$

$\pm \frac{13}{\sqrt{10}}$

$2\sqrt{3}$ ft/sec

The distance is decreasing at $390\text{mi/h}.$

The distance between them shrinks at a rate of $\frac{1320}{13}\approx 101.5\text{mph}.$

$\frac{9}{2}$ ft/sec

It grows at a rate $\frac{4}{9}$ ft/sec

The distance is increasing at $\frac{\left(135\sqrt{26}\right)}{26}$ ft/sec

$-\frac{5}{6}$ m/sec

$240\pi$ m²/sec

$\frac{1}{2\sqrt{\pi }}$ cm

The area is increasing at a rate $\frac{\left(3\sqrt{3}\right)}{8}{\text{ft}}^2\text{/sec.}$

The depth of the water decreases at $\frac{128}{125\pi }$ ft/min.

The volume is decreasing at a rate of $\frac{\left(25\pi \right)}{16}{\text{ft}}^3\text{/min}.$

The water flows out at rate $\frac{\left(2\pi \right)}{5}{\text{m}}^3\text{/min.}$

$\frac{3}{2}$ m/sec

$\frac{25}{19\pi }$ ft/min

$\frac{2}{45\pi }$ ft/min

The angle decreases at $\frac{400}{1681}\text{rad/sec}.$

$100\pi \text{mi/min}$

The angle is changing at a rate of $\frac{11}{25}\text{rad/sec}.$

The distance is increasing at a rate of $62.50$ ft/sec.

The distance is decreasing at a rate of $11.99$ ft/sec.

$f^{\prime }\left(a\right)=0$

The linear approximation exact when $y=f(x)$ is linear or constant.

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

$L(x)=1$

$L(x)=0$

0.02

$1.9996875$

$0.001593$

$1;$ error, $\mathrm{~0.00005}$

$0.97;$ error, $\mathrm{~0.0006}$

$3-\frac{1}{600};$ error, $\mathrm{~4.632}\times {10}^{\mathrm{-7}}$

$dy=\left(\text{cos}x-x\text{sin}x\right)dx$

$dy=\left(\frac{x^2-2x-2}{{(x-1)}^2}\right)dx$

$dy=-\frac{1}{{\left(x+1\right)}^2}dx,$ $-\frac{1}{16}$

$dy=\frac{9x^2+12x-2}{2{\left(x+1\right)}^{3\text{/}2}}dx,$ $\mathrm{-0.1}$

$dy=\left(3x^2+2-\frac{1}{x^2}\right)dx,$ $0.2$

$12xdx$

$4\pi r^{2}dr$

$\mathrm{-1.2}\pi {\text{cm}}^3$

$\mathrm{-100}$ ft³

Answers may vary

Answers will vary

No; answers will vary

Since the absolute maximum is the function (output) value rather than the x value, the answer is no; answers will vary

When $a=0$

Absolute minimum at 3; Absolute maximum at −2.2; local minima at −2, 1; local maxima at −1, 2

Absolute minima at −2, 2; absolute maxima at −2.5, 2.5; minima at –2, 0, and 2 (Note: –2 and 2 are both local and absolute minima); local maxima at −1, 1

Answers may vary.

Answers may vary.

$x=1$

None

$x=0\text{;}x=\pm 2$

None

$x=\mathrm{-1},1$

Absolute maximum: $x=4,$ $y=\frac{33}{2};$ absolute minimum: $x=1,$ $y=3$

Absolute minimum: $x=\frac{1}{2},$ $y=4$

Absolute maximum: $x=2\pi ,$ $y=2\pi ;$ absolute minimum: $x=0,$ $y=0$

Absolute maximum: $x=\mathrm{-3}$ and $y=6;;$ absolute minimum: $\mathrm{-1}\le x\le 1,$ $y=2$

Absolute maximum: $x=\frac{\pi }{4},$ $y=\sqrt{2};$ absolute minimum: $x=\frac{5\pi }{4},$ $y=\text{−}\sqrt{2}$

Absolute minimum: $x=\mathrm{-2},$ $y=1$

Absolute minimum: $x=\mathrm{-3},$ $y=\mathrm{-135};$ local maximum: $x=0,$ $y=0;$ local minimum: $x=1,$ $y=\mathrm{-7}$

Local maximum: $x=1-2\sqrt{2},$ $y=3-4\sqrt{2};$ local minimum: $x=1+2\sqrt{2},$ $y=3+4\sqrt{2}$

Absolute maximum: $x=\frac{\sqrt{2}}{2},$ $y=\frac{3}{2};$ absolute minimum: $x=-\frac{\sqrt{2}}{2},$ $y=-\frac{3}{2}$

Local maximum: $x=\mathrm{-2},$ $y=59;$ local minimum: $x=1,$ $y=\mathrm{-130}$

Absolute maximum: $x=0,$ $y=1;$ absolute minimum: $x=\mathrm{-2},2,$ $y=0$

$h=\frac{9245}{49}\text{m,}$ $t=\frac{300}{49}\text{s}$

The absolute minimum was in 1848, when no gold was produced.

Absolute minima: $x=0,$ $x=2,$ $y=1;$ local maximum at $x=1,$ $y=2$

No maxima/minima if $a$ is odd, minimum at $x=1$ if $a$ is even

One example is $f(x)=\left|x\right|+3,\mathrm{-2}\le x\le 2$

Yes, but the Mean Value Theorem still does not apply

Any closed interval in $(\text{−}\infty ,0) \text{or} (0,\infty )$

Any closed interval in $(\text{−}\infty ,\mathrm{-2}) \text{or} (2,\infty )$

2 points

5 points

$c=\frac{2\sqrt{3}}{3}$

$c=\frac{1}{2},1,\frac{3}{2}$

$c=1$

Not differentiable

Not differentiable

Yes

The Mean Value Theorem does not apply since the function is discontinuous at $x=\frac{1}{4},\frac{3}{4},\frac{5}{4},\frac{7}{4}.$

Yes

The Mean Value Theorem does not apply; discontinuous at $x=0.$

Yes

The Mean Value Theorem does not apply; not differentiable at $x=0.$

$b=\text{±}2\sqrt{c}$

$c=\text{±}\frac{1}{\pi }{\text{cos}}^{\mathrm{-1}}\left(\frac{\sqrt{\pi }}{2}\right),$ $c=\text{±}0.1533$

The Mean Value Theorem does not apply.

$\frac{1}{2\sqrt{c+1}}-\frac{2}{c^3}=\frac{521}{2880};$ $c=3.133,5.867$

Yes

Let $s_1\left(t\right)$ represent the distance of Car 1 from Stoplight 1 at time $t$. Let $s_2\left(t\right)$ represent the distance of Car 2 from Stoplight 1 at time $t$. Then $s_1\left(0\right)=0=s_2\left(0\right)$. Let $t=m$ represent the time that the cars arrive at Stoplight 2. Then $s_1\left(m\right)=d=s_2\left(m\right)$ where $d$ is the distance between the stoplights.

Define a new function $f\left(t\right)=s_1\left(t\right)-s_2\left(t\right)$. Then $f\left(0\right)=0=f\left(m\right)$. By the Mean Value Theorem, there is a time $0<c<m$ such that

At time c,

So, at time c, Car 1 and Car 2 are traveling at the same speed.

It is not a local maximum/minimum because $f^{\prime }$ does not change sign

No

False; for example, $y=\sqrt{x}.$

Increasing for $\mathrm{-2}<x<\mathrm{-1}$ and $x>2;$ decreasing for $x<\mathrm{-2}$ and $\mathrm{-1}<x<2$

Decreasing for $x<1,$ increasing for $x>1$

Decreasing for $\mathrm{-2}<x<\mathrm{-1}$ and $1<x<2;$ increasing for $\mathrm{-1}<x<1$ and $x<\mathrm{-2}$ and $x>2$

a. Increasing over $\mathrm{-2}<x<\mathrm{-1},0<x<1,x>2,$ decreasing over $x<\mathrm{-2},$ $\mathrm{-1}<x<0,1<x<2;$ b. maxima at $x=\mathrm{-1}$ and $x=1,$ minima at $x=\mathrm{-2}$ and $x=0$ and $x=2$

a. Increasing over $x>0,$ decreasing over $x<0;$ b. Minimum at $x=0$

Concave up on all $x,$ no inflection points

Concave up on all $x,$ no inflection points

Concave up for $x<0$ and $x>1,$ concave down for $0<x<1,$ inflection points at $x=0$ and $x=1$

Answers will vary

Answers will vary

a. Increasing over $-\frac{\pi }{2}<x<\frac{\pi }{2},$ decreasing over $x<-\frac{\pi }{2},x>\frac{\pi }{2}$ b. Local maximum at $x=\frac{\pi }{2};$ local minimum at $x=-\frac{\pi }{2}$

a. Concave up for $x>\frac{4}{3},$ concave down for $x<\frac{4}{3}$ b. Inflection point at $x=\frac{4}{3}$

a. Increasing over $x<0$ and $x>4,$ decreasing over $0<x<4$ b. Maximum at $x=0,$ minimum at $x=4$ c. Concave up for $x>2,$ concave down for $x<2$ d. Infection point at $x=2$

a. Increasing over $x<0$ and $x>\frac{60}{11},$ decreasing over $0<x<\frac{60}{11}$ b. Minimum at $x=\frac{60}{11}$, local maximum at x = 0 c. Concave down for $x<\frac{54}{11},$ concave up for $x>\frac{54}{11}$ d. Inflection point at $x=\frac{54}{11}$

a. Increasing over $x>-\frac{1}{2},$ decreasing over $x<-\frac{1}{2}$ b. Minimum at $x=-\frac{1}{2}$ c. Concave up for all $x$ d. No inflection points

a. Increases over $-\frac{1}{4}<x<\frac{3}{4},$ decreases over $x>\frac{3}{4}$ and $x<-\frac{1}{4}$ b. Minimum at $x=-\frac{1}{4},$ maximum at $x=\frac{3}{4}$ c. Concave up for $-\frac{3}{4}<x<\frac{1}{4},$ concave down for $x<-\frac{3}{4}$ and $x>\frac{1}{4}$ d. Inflection points at $x=-\frac{3}{4},x=\frac{1}{4}$

a. Increasing for all $x$ b. No local minimum or maximum c. Concave up for $x>0,$ concave down for $x<0$ d. Inflection point at $x=0$

a. Increasing for all $x$ where defined b. No local minima or maxima c. Concave up for $x<1;$ concave down for $x>1$ d. No inflection points in domain

a. Increasing over $-\frac{\pi }{4}<x<\frac{3\pi }{4},$ decreasing over $x>\frac{3\pi }{4},x<-\frac{\pi }{4}$ b. Minimum at $x=-\frac{\pi }{4},$ maximum at $x=\frac{3\pi }{4}$ c. Concave up for $-\frac{\pi }{2}<x<\frac{\pi }{2},$ concave down for $x<-\frac{\pi }{2},x>\frac{\pi }{2}$ d. Infection points at $x=\text{±}\frac{\pi }{2}$

a. Increasing over $x>4,$ decreasing over $0<x<4$ b. Minimum at $x=4$ c. Concave up for $0<x<8\sqrt[3]{2},$ concave down for $x>8\sqrt[3]{2}$ d. Inflection point at $x=8\sqrt[3]{2}$

$f>0,f^{\prime }>0,f^{″}<0$

$f>0,f^{\prime }<0,f^{″}>0$

$f>0,f^{\prime }>0,f^{″}>0$

True, by the Mean Value Theorem

True, examine derivative

$x=1$

$x=\mathrm{-1},x=2$

$x=0$

Yes, there is a vertical asymptote

Yes, there is vertical asymptote

$0$

$\infty$

$-\frac{1}{7}$

$\mathrm{-2}$

$\mathrm{-4}$

Horizontal: none, vertical: $x=0$

Horizontal: none, vertical: $x=\text{±}2$

Horizontal: none, vertical: none

Horizontal: $y=0,$ vertical: $x=\text{±}1$

Horizontal: $y=0,$ vertical: $x=0$ and $x=\mathrm{-1}$

Horizontal: $y=1,$ vertical: $x=1$

Horizontal: none, vertical: none

Answers will vary, for example: $y=\frac{2x}{x-1}$

Answers will vary, for example: $y=\frac{4x}{x+1}$

$y=0$

$\infty$

$y=3$

$P\left(0\right)\ne 0 \mathrm{and} Q\left(0\right)=0$

$\underset{x\to 1^{-}}{\text{lim}}f\left(x\right)=\mathrm{-\infty }\text{and}\underset{x\to 1^{-}}{\text{lim}}g\left(x\right)=\infty$

The critical points can be the minima, maxima, or neither.

False; $y=x^2$ has a minimum only

$h=\frac{62}{3}$ in.

$1$

$100\text{ft by}100\text{ft}$

$40\text{ft by}40\text{ft}$

$19.73\text{ft}.$