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