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