Answer
It is a decreasing function, so it has an inverse (it is one-to-one.)
$\displaystyle \frac{df^{-1}}{dx}=-\frac{1}{6}(1-x)^{-2/3}$
Work Step by Step
Take two values of x,
$x_{1} \gt x_{2}\qquad $... the cube function is an increasing function,
$x_{1}^{3} \gt x_{2}^{3}\qquad $... multiply with -8 (inequality changes)
$-8x_{1}^{3} \lt -8x_{2}^{3}\qquad$ ... add 1
$1-8x_{1}^{3} \lt 1-8x_{2}^{3}$
So f is a decreasing funtion.
By Exercise 49, it is one-to-one and has an inverse.
Theorem 1 provides a way to find a formula for $\displaystyle \frac{df^{-1}}{dx}$
$\displaystyle \frac{d}{dx}[f^{-1}(x)]=\frac{1}{f'[f^{-1}(x)]}$
We find $f^{-1}(x)\quad \left[\begin{array}{l}
y=1-8x^{3}\\
x=1-8y^{3}\\
8y^{3}=1-x\\
y=(1-x)/8\\
y=(1-x)^{1/3}/2\\
f^{-1}(x)=(1-x)^{1/3}/2
\end{array}\right]\quad f^{-1}(x)=\displaystyle \frac{1}{2}(1-x)^{1/3}$
We find$ f'(x).$
$f'(x)=-8\cdot 3x^{2}=-24x^{2}$
Calculate $f'[f^{-1}(x)]$
$f'[\displaystyle \frac{1}{2}(1-x)^{1/3}]=-24\cdot[\frac{1}{2}(1-x)^{1/3}]^{2}=-6(1-x)^{2/3}$
Apply the formula:
$\displaystyle \frac{d}{dx}[f^{-1}(x)]=\frac{1}{-6(1-x)^{2/3}}=-\frac{1}{6}(1-x)^{-2/3}$
