Solutions To Selected Problems Chapter F Chapter H

Chapter G

Exercises G.1

1.

F
2.

Answers will vary.
3.

The point $(10,1)$ lies on the graph of $y=f^{-1}(x)$ (assuming $f$ is invertible).
4.

Restrict the domain of the original function so that it is one to one.
5.
6.
7.

Compose $f(g(x))$ and $g(f(x))$ to confirm that each equals $x$ .
8.

Compose $f(g(x))$ and $g(f(x))$ to confirm that each equals $x$ .
9.

Compose $f(g(x))$ and $g(f(x))$ to confirm that each equals $x$ .
10.

Compose $f(g(x))$ and $g(f(x))$ to confirm that each equals $x$ .
11.

$[-4,0]$ or $[0,4]$
12.

$(-\infty,-4]$ or $[4,\infty)$ .
13.

$(-\infty,3]$ or $[3,\infty)$
14.

This is one-to-one on its domain of $[0,\infty)$ .
15.

$f^{-1}(x)=\dfrac{2x+1}{x-1}$
16.

$f^{-1}(x)=\pm\sqrt{x-4}$
17.

$f^{-1}(x)=\ln(x+2)-3$
18.

$f^{-1}(x)=e^{x-1}+5$
19.

$0$
20.

$\pi/7$
21.

$-1/5$
22.

$\pi/3$
23.

$1/\sqrt{2}$
24.

$5/4$
25.

$7/\sqrt{58}$
26.

$-\pi/3$
27.

$3\pi/4$
28.

$6\pi/7$
29.

$x/2$
30.

$x/2$
31.

$x/\sqrt{x^{2}+25}$
32.

$3/\sqrt{x-3}$
33.
34.
35.
36.
37.

$2\pi/3\sqrt{3}$

Exercises G.2

1.

The point $(10,1)$ lies on the graph of $y=f^{-1}(x)$ (assuming $f$ is invertible) and $(f^{-1})^{\prime}(10)=1/5$ .
2.

This sum is constant (in fact, it is $\pi/2$ ).
3.

$\left(f^{-1}\right)^{\prime}(20)=\frac{1}{f\,^{\prime}(2)}=1/5$
4.

$\left(f^{-1}\right)^{\prime}(7)=\frac{1}{f\,^{\prime}(3)}=1/4$
5.

$\left(f^{-1}\right)^{\prime}(\sqrt{3}/2)=\frac{1}{f\,^{\prime}(\pi/6)}=1$
6.

$\left(f^{-1}\right)^{\prime}(8)=\frac{1}{f\,^{\prime}(1)}=1/6$
7.

$\left(f^{-1}\right)^{\prime}(1/2)=\frac{1}{f\,^{\prime}(1)}=-2$
8.

$\left(f^{-1}\right)^{\prime}(6)=\frac{1}{f\,^{\prime}(0)}=1/6$
9.

$h^{\prime}(t)=\frac{2}{\sqrt{1-4t^{2}}}$
10.

$f\,^{\prime}(t)=\frac{1}{\left\lvert t\right\rvert\sqrt{4t^{2}+1}}$
11.

$g^{\prime}(x)=\frac{2}{1+4x^{2}}$
12.

$f\,^{\prime}(x)=\frac{x}{\sqrt{1-x^{2}}}+\sin^{-1}(x)$
13.

$g^{\prime}(t)=\cos^{-1}(t)\cos(t)-\frac{\sin(t)}{\sqrt{1-t^{2}}}$
14.

$f\,^{\prime}(t)=\frac{e^{t}}{t}+e^{t}\ln t$
15.

$h^{\prime}(x)=\frac{\sin^{-1}x+\cos^{-1}x}{\sqrt{1-x^{2}}(\cos^{-1}x)^{2}}$
16.

$g^{\prime}(x)=\frac{1}{\sqrt{x}(2x+2)}$
17.

$f\,^{\prime}(x)=-\frac{1}{\sqrt{1-x^{2}}}$
18.
- $f(x)=x$ , so $f\,^{\prime}(x)=1$
  
  $f\,^{\prime}(x)=\cos(\sin^{-1}x)\frac{1}{\sqrt{1-x^{2}}}=1$ .
19.
- $f(x)=x$ , so $f\,^{\prime}(x)=1$
  
  $f\,^{\prime}(x)=\cos(\sin^{-1}x)\frac{1}{\sqrt{1-x^{2}}}=1$ .
20.
- $f(x)=x$ , so $f\,^{\prime}(x)=1$
  
  $f\,^{\prime}(x)=\frac{1}{1+\tan^{2}x}\sec^{2}x=1$
21.
- $f(x)=\sqrt{1-x^{2}}$ , so $f\,^{\prime}(x)=\frac{-x}{\sqrt{1-x^{2}}}$
  
  $f\,^{\prime}(x)=\cos(\cos^{-1}x)(\frac{1}{\sqrt{1-x^{2}}})=\frac{-x}{\sqrt{1-x% ^{2}}}$
22.
- $f(x)=\frac{x}{\sqrt{x^{2}+1}}$ , so $f\,^{\prime}(x)=\frac{1}{(x^{2}+1)^{3/2}}$
  
  $f\,^{\prime}(x)=\cos(\tan^{-1}x)(\frac{1}{x^{2}+1})=\frac{1}{\sqrt{x^{2}+1}}% \cdot\frac{1}{x^{2}+1}$
23.

$y=\sqrt{2}(x-\sqrt{2}/2)+\pi/4$
24.

$y=-4(x-\sqrt{3}/4)+\pi/6$
25.

$-\pi/6$
26.

$2\pi/9$
27.

$\frac{1}{2}\left(\sin^{-1}r\right)^{2}+C$
28.

$\frac{1}{8}\tan^{-1}(x^{4}/2)+C$
29.

$\sin^{-1}(e^{t}/\sqrt{10})+C$
30.

$2\tan^{-1}(\sqrt{x})+C$
31.

$\sqrt{91}\approx 9.54$ feet

Exercises G.3

1.

$(-\infty,\infty)$
2.

$(-1,1)$
3.

$(-\infty,0)\cup(0,\infty)$
4.

$(-\infty,0)\cup(0,\infty)$
5.

$f\,^{\prime}(t)=3t^{2}e^{t^{3}-1}$
6.

$g\mkern 1.35mu^{\prime}(r)=2^{r}\log_{2}r+\dfrac{r}{\ln 2}$
7.

$f\,^{\prime}(x)=\dfrac{1-x\ln 5\ln x}{x5^{x}\ln 5}$
8.

$f\,^{\prime}(x)=5x^{4}(4^{x^{5}})\ln 4$
9.

$f\,^{\prime}(x)=1$
10.

$g\mkern 1.35mu^{\prime}(x)=2xe^{x^{2}}\sin(x-\ln x)+3^{x^{2}}\cos(x-\ln x)(1-1% /x)$
11.

$h\mkern 1.35mu^{\prime}(r)=\dfrac{3^{r}\ln 3}{1+3^{2r}}$
12.

$h\mkern 1.35mu^{\prime}(x)=\dfrac{2x}{(x^{2}+1)\ln 10}-\dfrac{4}{x\ln 10}$
13.

$\dfrac{24}{\ln 5}$
14.

$\frac{1}{2}\ln 3$
15.

$\frac{3^{x^{2}-1}}{2ln3}+C$
16.

$\sin(\ln x)+C$
17.

$\frac{1}{2}\sin^{2}(e^{x})+C$
18.

$24-\dfrac{31}{\ln 2}$
19.

$\dfrac{\ln 245}{\ln 3}-1$
20.

$\ln\left\lvert\tan^{-1}x\right\rvert+C$
21.

$\frac{1}{2}\ln^{2}(x)+C$
22.

$\frac{\bigl(\ln x\bigr)^{3}}{3}+C$
23.

$\frac{1}{6}\ln^{2}\left(x^{3}\right)+C$
24.

$\frac{1}{2}\ln\left(\ln\left(x^{2}\right)\right)+C$
25.

$n=-3,2$
26.
- $c=4$
  
  Since $g(0)-f(0)=1-0>0$ and $g(-1)-f(-1)=\frac{1}{2}-1<0$ , the Intermediate Value Theorem implies that there is a number $a$ between $-1$ and $0$ so that $g(a)-f(a)=0$ , or $g(a)=f(a)$ .
  
  $-\frac{25}{32}$
  
  $f(18)=18^{2}=324$ , so the graph of $f$ is 27 feet high.
  $g(18)=2^{18}=262{,}144$ , so the graph of $g$ is approximately $4.14$ miles high.
27.

$y^{\prime}=(1+x)^{1/x}\big(\frac{1}{x(x+1)}-\frac{\ln(1+x)}{x^{2}}\big)$

Tangent line: $y=(1-2\ln 2)(x-1)+2$
28.

$y^{\prime}=(2x)^{x^{2}}\big(2x\ln(2x)+x\big)$

Tangent line: $y=(2+4\ln 2)(x-1)+2$
29.

$y^{\prime}=\frac{x^{x}}{x+1}\big(\ln x+1-\frac{1}{x+1}\big)$

Tangent line: $y=(1/4)(x-1)+1/2$
30.

$y^{\prime}=x^{\sin(x)+2}\big(\cos x\ln x+\frac{\sin x+2}{x}\big)$

Tangent line: $y=(3\pi^{2}/4)(x-\pi/2)+(\pi/2)^{3}$
31.

$y^{\prime}=\frac{x+1}{x+2}\big(\frac{1}{x+1}-\frac{1}{x+2}\big)$

Tangent line: $y=1/9(x-1)+2/3$
32.

$y^{\prime}=\frac{(x+1)(x+2)}{(x+3)(x+4)}\big(\frac{1}{x+1}+\frac{1}{x+2}-\frac% {1}{x+3}-\frac{1}{x+4}\big)$

Tangent line: $y=11/72x+1/6$
33.

$y^{\prime}=x^{e^{x}-1}e^{x}(1+x\ln x)$

Tangent line: $y=ex-e+1$
34.

$y^{\prime}=\sin x(-\cot x)^{\cos x}(\csc^{2}x+\ln(\cot x))$

Tangent line: $y=x-\pi$ .
35.

$r=(\ln 2)/5730$ ; $5730\ln 10/\ln 2\approx 19034.65$ years

Exercises G.4

1.

Because $\cosh x$ is always positive.
2.

The points on the left hand side can be defined as $(-\cosh x,\sinh x)$ .
3.

$\cosh t=13/12$ , etc.
4.

$\sinh t=-3/4$ , $\cosh t=5/4$ , etc.
5.

$\begin{aligned} \coth^{2}x-\operatorname{csch}^{2}x&=\left(\frac{e^{x}+e^{-x}}% {e^{x}-e^{-x}}\right)^{2}-\left(\frac{2}{e^{x}-e^{-x}}\right)^{2}\\ &=\frac{(e^{2x}+2+e^{-2x})-(4)}{e^{2x}-2+e^{-2x}}\\ &=\frac{e^{2x}-2+e^{-2x}}{e^{2x}-2+e^{-2x}}\\ &=1\end{aligned}$
6.

$\begin{aligned} \cosh^{2}x+\sinh^{2}x&=\left(\frac{e^{x}+e^{-x}}{2}\right)^{2}% +\left(\frac{e^{x}-e^{-x}}{2}\right)^{2}\\ &=\frac{e^{2x}+2+e^{-2x}}{4}+\frac{e^{2x}-2+e^{-2x}}{4}\\ &=\frac{2e^{2x}+2e^{-2x}}{4}\\ &=\frac{e^{2x}+e^{-2x}}{2}\\ &=\cosh 2x.\end{aligned}$
7.

$\begin{aligned} \cosh^{2}x&=\left(\frac{e^{x}+e^{-x}}{2}\right)^{2}\\ &=\frac{e^{2x}+2+e^{-2x}}{4}\\ &=\frac{1}{2}\frac{(e^{2x}+e^{-2x})+2}{2}\\ &=\frac{1}{2}\left(\frac{e^{2x}+e^{-2x}}{2}+1\right)\\ &=\frac{\cosh 2x+1}{2}.\end{aligned}$
8.

$\begin{aligned} \sinh^{2}x&=\left(\frac{e^{x}-e^{-x}}{2}\right)^{2}\\ &=\frac{e^{2x}-2+e^{-2x}}{4}\\ &=\frac{1}{2}\frac{(e^{2x}+e^{-2x})-2}{2}\\ &=\frac{1}{2}\left(\frac{e^{2x}+e^{-2x}}{2}-1\right)\\ &=\frac{\cosh 2x-1}{2}.\end{aligned}$
9.

$\begin{aligned} \frac{\operatorname{d}\!}{\operatorname{d}\!x}\left[% \operatorname{sech}x\right]&=\frac{\operatorname{d}\!}{\operatorname{d}\!x}% \left[\frac{2}{e^{x}+e^{-x}}\right]\\ &=\frac{-2(e^{x}-e^{-x})}{(e^{x}+e^{-x})^{2}}\\ &=-\frac{2(e^{x}-e^{-x})}{(e^{x}+e^{-x})(e^{x}+e^{-x})}\\ &=-\frac{2}{e^{x}+e^{-x}}\cdot\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\\ &=-\operatorname{sech}x\tanh x\end{aligned}$
10.

$\begin{aligned} \frac{\operatorname{d}\!}{\operatorname{d}\!x}&\left[\coth x% \right]\\ &=\frac{\operatorname{d}\!}{\operatorname{d}\!x}\left[\frac{e^{x}+e^{-x}}{e^{x% }-e^{-x}}\right]\\ &=\frac{(e^{x}-e^{-x})(e^{x}-e^{-x})-(e^{x}+e^{-x})(e^{x}+e^{-x})}{(e^{x}-e^{-% x})^{2}}\\ &=\frac{e^{2x}+e^{-2x}-2-(e^{2x}+e^{-2x}+2)}{(e^{x}-e^{-x})^{2}}\\ &=-\frac{4}{(e^{x}-e^{-x})^{2}}\\ &=-\operatorname{csch}^{2}x\end{aligned}$
11.

$\displaystyle\int\tanh x\operatorname{d}\!x=\int\frac{\sinh x}{\cosh x}% \operatorname{d}\!x$

Let $u=\cosh x$ ; $\operatorname{d}\!u=(\sinh x)\operatorname{d}\!x$

$\begin{aligned} &=\int\frac{1}{u}\operatorname{d}\!u\\ &=\ln\left\lvert u\right\rvert+C\\ &=\ln(\cosh x)+C.\end{aligned}$
12.

Let $u=\sinh x$ ; $\operatorname{d}\!u=(\cosh x)\operatorname{d}\!x$ .

$\begin{aligned} \int\coth x\operatorname{d}\!x&=\int\frac{\cosh x}{\sinh x}% \operatorname{d}\!x\\ &=\int\frac{1}{u}\operatorname{d}\!u\\ &=\ln\left\lvert u\right\rvert+C\\ &=\ln\left\lvert\sinh x\right\rvert+C.\end{aligned}$
13.

$2\cosh 2x$
14.

$2\cosh x\sinh x$
15.

$2x\operatorname{sech}^{2}(x^{2})$
16.

$\coth x$
17.

$\sinh^{2}x+\cosh^{2}x$
18.

$x\cosh x$
19.

$\frac{-2x}{(x^{2})\sqrt{1-x^{4}}}$
20.

$\frac{3}{\sqrt{9x^{2}+1}}$
21.

$\frac{4x}{\sqrt{4x^{4}-1}}$
22.

$\frac{1}{1-(x+5)^{2}}$
23.

$-\csc x$
24.

$\sec x$
25.

$y=x$
26.

$y=3/4(x-\ln 2)+5/4$
27.

$y=\frac{9}{25}(x+\ln 3)-\frac{4}{5}$
28.

$y=-72/125(x-\ln 3)+9/25$
29.

$y=x$
30.

$y=(x-\sqrt{2})+\cosh^{-1}(\sqrt{2})\approx(x-1.414)+0.881$
31.

$\frac{1}{2}\ln(\cosh(2x))+C$
32.

$\frac{1}{3}\sinh(3x-7)+C$
33.

$\frac{1}{2}\sinh^{2}x+C$ or $1/2\cosh^{2}x+C$
34.

$\begin{cases}\frac{1}{3}\tanh^{-1}\left(\frac{x}{3}\right)+C&x^{2}<9\\ \frac{1}{3}\coth^{-1}\left(\frac{x}{3}\right)+C&9<x^{2}\end{cases}=\frac{1}{2}% \ln\left\lvert x+1\right\rvert-\frac{1}{2}\ln\left\lvert x-1\right\rvert+C$
35.

$\cosh^{-1}(x^{2}/2)+C=\ln(x^{2}+\sqrt{x^{4}-4})+C$
36.

$2/3\sinh^{-1}x^{3/2}+C=2/3\ln(x^{3/2}+\sqrt{x^{3}+1})+C$
37.

$\tan^{-1}(e^{x})+C$
38.

$\tan^{-1}(\sinh x)+C$
39.

$0$
40.

$3/2$
41.

Using rule #32: $A=\int_{0}^{\sinh\theta}\sqrt{1+y^{2}}-y\coth\theta\operatorname{d}\!y=\frac{% \theta}{2}$ .

Exercises G.5

1.

$0/0,\infty/\infty,0\cdot\infty,\infty-\infty,0^{0},1^{\infty},\infty^{0}$
2.

F
3.

F
4.

The base of an expression is approaching 1 while its power is growing without bound.
5.

derivatives; limits
6.

Answers will vary.
7.

Answers will vary.
8.

Answers will vary.
9.

3
10.

$-5/3$
11.

$-1$
12.

$-\sqrt{2}/2$
13.

$5$
14.

$0$
15.

$a/b$
16.

$\infty$
17.

$1/2$
18.

$0$
19.

$0$
20.

$0$
21.

$\infty$
22.

$\infty$
23.

$0$
24.

$2$
25.

$-2$
26.

$0$
27.

$0$
28.

$0$
29.

$0$
30.

$0$
31.

$\infty$
32.

$\infty$
33.

$\infty$
34.

$0$
35.

$0$
36.

$e$
37.

$1$
38.

$1$
39.

$1$
40.

$1$
41.

$1$
42.

$0$
43.

$1$
44.

$1$
45.

$1$
46.

$1$
47.

$2$
48.

$1/2$
49.

$-\infty$
50.

$1$
51.

$0$
52.

$3$
53.

$\sqrt[3]{5}$
54.

$\sqrt{7}$
55.

Use technology to verify sketch.