Algebra and Trigonometry 10th Edition

It has been verified that $f^{-1}[f (x)]=f[f^{-1} (x)]$, that is, the function is the inverse function. So, the inverse function is: $\dfrac{x-1}{3}$
We have the inverse function: $f^{-1} (x)=\dfrac{x-1}{3}$ Therefore, $f[f^{-1} (x)]=f(\dfrac{x-1}{3})=3(\dfrac{x-1}{3})+1=x$ and $f^{-1}[f (x)]=f^{-1}(3x+1)=\dfrac{(3x+1)-1}{3}=x$ Therefore, it has been verified that $f^{-1}[f (x)]=f[f^{-1} (x)]$, that is, the function is the inverse function. So, the inverse function is: $\dfrac{x-1}{3}$