Answer
$ a.\quad f^{-1}(x)=-\displaystyle \frac{1}{4}x+\frac{5}{4}$
$ b.\quad$ see image
$ c.\quad \displaystyle \left.\frac{df}{dx}\right|_{x=1/2}=-4,\qquad \left.\frac{df^{-1}}{dx}\right|_{x=3}=-\frac{1}{4}$
Work Step by Step
$ a.\quad$
Set $y=f(x)$. Interchange x and y
$y=5-4x$
$x=5-4y$
... solve for y
$x-5=-4y$
$y=-\displaystyle \frac{1}{4}x+\frac{5}{4}$
... replace y with $f^{-1}(x)$
$f^{-1}(x)=-\displaystyle \frac{1}{4}x+\frac{5}{4}$
$ b.\quad$see graphs (attached image)
$ c.\quad$
$f(x)=5-4x$
$\displaystyle \frac{df}{dx}=-4,\qquad \left.\frac{df}{dx}\right|_{x=1/2}=-4$
$a=\displaystyle \frac{1}{2}$
$f(\displaystyle \frac{1}{2})=5-4(\frac{1}{2})=3\quad\Rightarrow\quad f^{-1}(3)=\frac{1}{2}$
$f^{-1}(x)=-\displaystyle \frac{1}{4}x+\frac{5}{4}$
$\displaystyle \frac{df^{-1}}{dx}=-\frac{1}{4},\qquad \left.\frac{df^{-1}}{dx}\right|_{x=3}=-\frac{1}{4}$
