[tex]g(x)=3x+1\to y=3x+1[/tex]
change x to y, and y to x
[tex]x=3y+1[/tex]
solve for y
[tex]3y+1=x\ \ \ \ |-1\\\\3y=x-1\ \ \ \ |:3\\\\y=\dfrac{x-1}{3}[/tex]
therefore:
[tex]g^{-1}(x)=\dfrac{x-1}{3}[/tex]
put the value of x = 1 to the equation:
[tex]g^{-1}(1)=\dfrac{1-1}{3}=\dfrac{0}{3}=0[/tex]
Answer: 0
