Chapter 3.3, Problem 37E

(a)

To determine

To show: that $f$ has a differentiable inverse.

Explanation of Solution

Given information :

Function $f\left(x\right)=\cos x+3x$

Differentiate $f\left(x\right)=\cos x+3x$ with respect to $x$ ,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(\cos x+3x\right)\\ =\frac{d}{dx}\left(\cos x\right)+\frac{d}{dx}\left(3x\right)\\ =-\sin x+3\end{array}$

Here $f^{\prime }\left(x\right)>0$ and $f\left(x\right)$ is injective, hence $f^{-1}\left(x\right)$ exists.

Apply chain rule of derivative,

  $\begin{array}{c}\frac{d}{dx}{f}^{-1}\left(x\right)={\left(f^{-1}\right)}^{\prime }\left(x\right)\cdot \frac{d}{dx}x\\ ={\left(f^{-1}\right)}^{\prime }\left(x\right)\cdot 1\\ ={\left(f^{-1}\right)}^{\prime }\left(x\right)\end{array}$

Hence, $f$ has a differentiable inverse.

(b)

To determine

To find: $f\left(0\right)$ and $f^{\prime }\left(0\right)$ .

Answer to Problem 37E

  $f\left(0\right)=1$ and $f^{\prime }\left(0\right)=3$ .

Explanation of Solution

Given information :

Function $f\left(x\right)=\cos x+3x$ and $f^{\prime }\left(x\right)=-\sin x+3$

Plug $x=0$ in $f\left(x\right)=\cos x+3x$ .

  $\begin{array}{c}f\left(0\right)=\cos 0+3\cdot 0\\ =1+0\\ =1\end{array}$

Plug $x=0$ in $f^{\prime }\left(x\right)=-\sin x+3$ .

  $\begin{array}{c}{f}^{\prime }\left(0\right)=-\sin 0+3\\ =0+3\\ =3\end{array}$

Hence, $f\left(0\right)=1$ and $f^{\prime }\left(0\right)=3$ .

(c)

To determine

To find: $f^{-1}\left(1\right)$ and ${\left(f^{-1}\right)}^{\prime }\left(1\right)$ .

Answer to Problem 37E

  $f^{-1}\left(1\right)=0$ and ${\left(f^{-1}\right)}^{\prime }\left(1\right)=\frac{1}{3}$ .

Explanation of Solution

Given information :

Function $f\left(x\right)=\cos x+3x$ and $f^{\prime }\left(x\right)=-\sin x+3$

Plug $x=0$ in $f\left(x\right)=\cos x+3x$ .

  $\begin{array}{c}f\left(0\right)=\cos 0+3\cdot 0\\ =1+0\\ =1\end{array}$

Take $f^{-1}$ on both side of $f\left(0\right)=1$ .

  $\begin{array}{c}{f}^{-1}\left(f\left(0\right)\right)=f^{-1}\left(1\right)\\ 0=f^{-1}\left(1\right)\\ f^{-1}\left(1\right)=0\end{array}$

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

  $\begin{array}{c}{\left(f^{-1}\right)}^{\prime }\left(1\right)=\frac{1}{f^{\prime }\left(0\right)}\\ =\frac{1}{3}\end{array}$

Hence, $f^{-1}\left(1\right)=0$ and ${\left(f^{-1}\right)}^{\prime }\left(1\right)=\frac{1}{3}$ .