Answer
$ f^{-1}(x)=2x+7 $
Domain of $ f^{-1}$ is $(-\infty,\infty)$
Range of $ f^{-1}$ is $(-\infty,\infty)$
Work Step by Step
Given $$ f(x) = \frac{1}{2}x-\frac{7}{2}$$
Find the inverse:
\begin{aligned} y &= \frac{1}{2}x-\frac{7}{2} \\2y &=x-7 \\ x&=2y+7 \\
\text {Switch }& x \ and\ y:\\
y=&2x+7 =f^{-1}(x) \end{aligned}
We see that:
Domain of $ f^{-1}$ is $(-\infty,\infty)$
and
Range of $ f^{-1}$ is $(-\infty,\infty)$
Confirm the inverse:
\begin{aligned} f\left(f^{-1}(x)\right) &=f(2 x+7) \\ &=\frac{1}{2}(2 x+7)-\frac{7}{2} \\ &=x+\frac{7}{2}-\frac{7}{2} \\ &=x \\
f^{-1}(f(x)) &=f^{-1}\left(\frac{1}{2} x-\frac{7}{2}\right) \\ &=2\left(\frac{1}{2} x-\frac{7}{2}\right)+7 \\ &=x-7+7 \\ &=x \end{aligned}
