[tex]g(x)=\dfrac{1}{2}(x-3)+1\\\\y=\dfrac{1}{2}(x-3)+1\\\\\text{replace x with y and y with x}\\\\x=\dfrac{1}{2}(y-3)+1\\\\\text{solve for y}\\\\\dfrac{1}{2}(y-3)+1=x\ \ \ \ |-1\\\\\dfrac{1}{2}(y-3)=x-1\ \ \ \ |\cdot2\\\\y-3=2x-2\ \ \ \ |+3\\\\y=2x+1\\\\Answer:\ \boxed{g^{-1}(x)=2x+1}[/tex]
