Answer
$f^{-1}(x)=2x+3$
Work Step by Step
We are given the function:
$f(x)=\dfrac{x-3}{2}$
Determine the inverse of $f$:
$y=\dfrac{x-3}{2}$
Interchange $x$ and $y$:
$x=\dfrac{y-3}{2}$
$2x=y-3$
$y=2x+3$
$f^{-1}(x)=2x+3$
Compute $f\circ f^{-1}$ and $f^{-1}\circ f$:
$(f\circ f^{-1})(x)=f(f^{-1}(x))=f(2x+3)=\dfrac{2x+3-3}{2}=\dfrac{2x}{2}=x$
$(f^{-1}\circ f)(x)=f^{-1}\left(\dfrac{x-3}{2}\right)=2\left(\dfrac{x-3}{2}\right)+3=x-3+3=x$
We got that $f^{-1}\circ f=f\circ f^{-1}=x$; therefore $f$ and $f^{-1}$ are inverses.
