Answer
See below for detailed answers.
Work Step by Step
$$f(x)=\frac{1}{5}x+7$$
a) To find $f^{-1}(x)$, we substitute $x$ with $f^{-1}x$ and $f(x)$ with $x$: $$x=\frac{1}{5}f^{-1}x+7$$
Then solve for $f^{-1}(x)$: $$f^{-1}(x)=\frac{x-7}{\frac{1}{5}}=5(x-7)$$
b) The graphs of $f(x)$ and $f^{-1}(x)$ are enclosed below.
c) - Find $df/dx$: $$\frac{df}{dx}=\frac{d}{dx}\Big(\frac{1}{5}x+7\Big)=\frac{1}{5}$$
At $x=-1$, $df/dx=1/5$
- Find $df^{-1}/dx$:
$$\frac{df^{-1}}{dx}=\frac{d}{dx}(5(x-7))=5(1-0)=5$$
At $x=f(-1)=1/5(-1)+7=34/5$, $$df^{-1}/dx=5$$
It can be seen that at these points, $df^{-1}/dx=1/(df/dx)$
