Answer
$f^{-1}(1)=0$
$f^{-1}(3)=4$
Work Step by Step
The equation on the domain [0,4) is $f(x)=\frac{1}{2}x+1$, so to find $f^{-1}(1)$ and $f^{-1}(3)$, one must first find $f^{-1}(x)$
To do so, one must take the original equation, swap $f(x)$ for $x$ and solve for $f(x)$:
$f(x)=\frac{1}{2}x+1$
$x=\frac{1}{2}f^{-1}(x)+1$
$x-1=\frac{1}{2}f^{-1}(x)+1-1$
$2(x-1)=(2)\frac{1}{2}f^{-1}(x)$
$f^{-1}(x)=2x-2$
Now, we can find the values asked:
$f^{-1}(1)=2(1)-2=0$
$f^{-1}(3)=2(3)-2=4$
