Chapter 7

F

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

Compose $f\left(g\left(x\right)\right)$ and $g\left(f\left(x\right)\right)$ to confirm that each equals $x$.

Compose $f\left(g\left(x\right)\right)$ and $g\left(f\left(x\right)\right)$ to confirm that each equals $x$.

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

$(-\mathrm{\infty },3]$ or $[3,\mathrm{\infty })$

$f^{-1}\left(x\right)={\displaystyle \frac{2x+1}{x-1}}$

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

$0$

$-1/5$

$1/\sqrt{2}$

$7/\sqrt{58}$

$3\pi /4$

$x/2$

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

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

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

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

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

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

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

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

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

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

$f^{\prime }\left(x\right)=-\frac{1}{\sqrt{1-x^2}}$

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

$-\pi /6$

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

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

$\sqrt{91}\approx 9.54$ feet

$(-\mathrm{\infty },\mathrm{\infty })$

$(-\mathrm{\infty },0)\cup (0,\mathrm{\infty })$

$f^{\prime }\left(t\right)=3t^2{e}^{t^3-1}$

$f^{\prime }\left(x\right)={\displaystyle \frac{1-x\ln 5\ln x}{x{5}^x\ln 5}}$

$f^{\prime }\left(x\right)=1$

$h^{\prime }\left(r\right)={\displaystyle \frac{3^r\ln 3}{1+3^{2r}}}$

${\displaystyle \frac{24}{\ln 5}}$

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

$\frac{1}{2}{\sin }^2\left(e^x\right)+C$

${\displaystyle \frac{\ln 245}{\ln 3}}-1$

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

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

$n=-3,2$

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

Tangent line: $y=\left(1-2\ln 2\right)\left(x-1\right)+2$

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

Tangent line: $y=\left(1/4\right)\left(x-1\right)+1/2$

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

Tangent line: $y=1/9\left(x-1\right)+2/3$

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

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

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

Because $\cosh x$ is always positive.

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

$2\cosh 2x$

$2x{\mathrm{sech}}^2\left(x^2\right)$

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

$\frac{-2x}{\left(x^2\right)\sqrt{1-x^4}}$

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

$-\csc x$

$y=x$

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

$y=x$

$\frac{1}{2}\ln \left(\cosh \left(2x\right)\right)+C$

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

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

${\tan }^{-1}\left(e^x\right)+C$

$0$

Using rule #33: $A={\int }_0^{\sinh \theta }\sqrt{1+y^2}-y\coth \theta \mathrm{d}y=\frac{\theta }{2}$.

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

F

derivatives; limits

Answers will vary.

3

$-1$

$5$

$a/b$

$1/2$

$0$

$\mathrm{\infty }$

$0$

$-2$

$0$

$0$

$\mathrm{\infty }$

$\mathrm{\infty }$

$0$

$1$

$1$

$1$

$1$

$1$

$2$

$-\mathrm{\infty }$

$0$

$\sqrt[3]{5}$

Use technology to verify sketch.