Answer
(a)
Domain of $f$: (-$\infty$, $\infty$)
Range of $f$: (-$\infty$, $\infty$)
(b)
Domain of $f^{-1}$: (-$\infty$, $\infty$)
Range of $f^{-1}$: (-$\infty$, $\infty$)
(d)
$f$'($\frac{1}{2}$) = $\frac{3}{4}$
($f^{-1}$)'($\frac{1}{8}$) = $\frac{4}{3}$
Work Step by Step
Function: $f$(x) = $x^{3}$, Point: ($\frac{1}{2}$,$\frac{1}{8}$)
Function: $f$(x) = $\sqrt[3] x$, Point: ($\frac{1}{8}$,$\frac{1}{2}$)
The domain and range of $x^{3}$ is (-$\infty$, $\infty$), (-$\infty$, $\infty$).
The domain and range of $\sqrt[3] x$ is (-$\infty$, $\infty$), (-$\infty$, $\infty$).
These graphs are inverses of each other.
How to show that $x^{3}$ and $\sqrt[3] x$ are reciprocals at the given points:
1. Evaluate the derivative of $x^{3}$ at $x$ = $\frac{1}{2}$
The derivative of $x^{3}$ is $3x^{2}$
Plug in $\frac{1}{2}$ for $x$, $3($$\frac{1}{2}$)$^{2}$
$\frac{1}{2}$$^{2}$ equals $\frac{1}{4}$, multiply it by 3 and it equals $\frac{3}{4}$
So $f$'($\frac{1}{2}$) = $\frac{3}{4}$
2. Evaluate the derivative of $\sqrt[3] x$ at $x$ = $\frac{1}{8}$
($f^{-1}$)'($x$) = $\frac{1}{f'(f^{-1}(x))}$
($f^{-1}$)'($\frac{1}{8}$) = $\frac{1}{f'(f^{-1}(\frac{1}{8}))}$
$f^{-1}(\frac{1}{8})$ = $\sqrt[3] \frac{1}{8}$ = $\frac{1}{2}$
($f^{-1}$)'($\frac{1}{8}$) = $\frac{1}{f'(\frac{1}{2})}$
Earlier, we learned that $f$'($\frac{1}{2}$) = $\frac{3}{4}$
$\frac{1}{\frac{3}{4}}$ = $\frac{4}{3}$
So ($f^{-1}$)'($\frac{1}{8}$) = $\frac{4}{3}$
They are reciprocals
