Answer
$ a.\quad f^{-1}(x)=\sqrt{\dfrac{x}{2}}$
$ b.\quad$ see image
$ c.\quad \displaystyle \left.\frac{df}{dx}\right|_{x=5}=20,\qquad \left.\frac{df^{-1}}{dx}\right|_{x=50}=\frac{1}{20}$
Work Step by Step
$ a.\quad$
Set $y=f(x)$. Interchange x and y
$y=2x^{2}\qquad $... note that $y\geq 0$
$x=2y^{2}$
... solve for y
$y^{2}=\displaystyle \frac{x}{2}$
$y=\sqrt{\frac{x}{2}}$
... replace y with $f^{-1}(x)$
$f^{-1}(x)=\sqrt{\frac{x}{2}}$
$ b.\quad$see graphs (attached image)
$ c.\quad$
$a=5$
$f(x)=2x^{2}$
$\displaystyle \frac{df}{dx}=4x,\qquad $
$\left.\frac{df}{dx}\right|_{x=5}=4(5)=20$
$a=5$
$f(5)=2\cdot 5^{2}=50\quad\Rightarrow\quad f^{-1}(50)=5$
$f^{-1}(x)=\sqrt{\frac{x}{2}}=2^{-1/2}x^{1/2}$
$\displaystyle \frac{df^{-1}}{dx}=2^{-1/2}(\frac{1}{2}x^{-1/2})=\frac{1}{2\sqrt{2x}}$
$ \displaystyle \left.\frac{df^{-1}}{dx}\right|_{x=50}=\frac{1}{2\sqrt{2(50)}}=\frac{1}{20}$
