# Inverse of a piecewise continuous function.

I am trying to find the inverse of a two dimensional map $$f\left(\begin{bmatrix} x\\y\end{bmatrix}\right)$$, For example $$f\left(\begin{bmatrix}x\\y \end{bmatrix}\right) = \begin{cases} ax + by, &(x,y)\in D_1 \\ ax^3 + b x^4, & (x,y)\in D_2\end{cases}$$

For this question we have $$f\left(\begin{bmatrix}x\\y\end{bmatrix}\right)$$ is defined as

$$f\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right) = \begin{cases} F_{0}\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right), &yb\end{cases}$$

We know the inverses of $$F_{0}$$ and $$F_{1}$$.

The inverse of the map $$f$$ depends on the point $$(x,y)$$ we give, like for $$y, we will take the inverse of $$f$$ as the inverse of the map $$F_{0}$$ and if the $$y$$ value is greater than $$b$$, then we take the inverse of the map $$T_{1}$$. But when the $$y$$ value lies in between $$a$$ and $$b$$, then we take the inverse of $$\big(1-s(y)\big)F_{0}\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right) + s(y) F_{1}\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right),$$ which I am not sure how to calculate the inverse of seems like applying $$(g.h)^{-1} = h^{-1}.g^{-1}$$ will help in this case, still how to find the inverse of the piece o fthe function defined in $$a?

• $s(y)F_0(v)$ is not a composite function. It is just a product. The rule you have mentioned is true for composite functions. I don't think products follow that rule. – Balakrishnan Rajan Mar 28 at 2:14
• This function doesn't have an inverse if $F_0(x,y) = F_1(u,v)$ for some $(x,y)$ with $y<a$ and $(u,v)$ with $v > b$. – amsmath Mar 28 at 3:13
• @BalakrishnanRajan yup you are right! – BAYMAX Mar 28 at 4:09
• @amsmath nice, but let us suppose this doesnot happen! – BAYMAX Mar 28 at 4:09

A linear combination of invertible functions is not necessarily invertible. A very simple example is $$f_1(x) = x$$ and $$f_2(x) = -x$$. They are each invertible, actually being their own inverses, but $$f(x) = f_1(x) + f_2(x) = 0$$ is not invertible.
In your case, invertibility will depend on what $$\big(1-s(y)\big)F_{0}\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right) + s(y) F_{1}\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right)$$ results in, but even if this is invertible, you're not likely to find any easy formula to determine what it is from the base parts of $$s$$, $$F_0$$ and $$F_1$$.