compute the inverse function

Question

Assume $h(x)$ is an invertible function.

Let $g(x)=2+8h(4x+1)$. Find the inverse of $g$ in terms of $h^{-1}$

So following the usual steps to get the inverse function, I rearranged to get $h(4x+1)=\frac{g(x)-2}{8}$. Should I now apply $h^{-1}$ to both sides to get $4x+1=h^{-1}\frac{g(x)-2}{8}$ and then isolate $x$ which will be the inverse? I have a feeling I can get a better answer.

Thanks

Answer 1

Your idea is correct. So you have then \begin{align*} x=\frac{1}{4}\left(h^{-1}\left(\frac{g(x)-2}{8}\right)-1\right), \end{align*} and if you replace $x$ by $g^{-1}(x)$ you will get \begin{align*} g^{-1}(x)=\frac{1}{4}\left(h^{-1}\left(\frac{x-2}{8}\right)-1\right). \end{align*}

Answer 2

$$y=2+8h(4x+1)\implies h(4x-1)=\frac{y-2}{8}\implies 4x-1=h^{-1}\left(\frac{y-2}{8}\right)$$$$\implies x=\frac{1}{4}h^{-1}\left(\frac{y-2}{8}\right)-\frac{1}{4}.$$

Therefore, $$g^{-1}(y)=\frac{1}{4}h^{-1}\left(\frac{y-2}{8}\right)-\frac{1}{4}.$$