Inverse of $f(x)= x^2$, under $\mathbb R\rightarrow \mathbb R$.

Question

Find Inverse of $f(x)= x^2$ under the operation of function composition $\circ$, with domain and codomain being $\mathbb R$, i.e. $\mathbb R\rightarrow \mathbb R$.

The given function fails to have single value of inverse, under the horizontal line test. Though, am confused why under addition operation have inverse for given function= $-x^2$.

Though, would like to see what is inverse of $f(x)= x^3$ under composition operation. As, now have a strictly increasing function and inverse should be possible.

Let inverse be $z(x)$, on the same co-domain as its domain & vice-versa.

Need find inverse, s.t. $f(z(x))= 1$, but cannot think further.

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Edit :

How to algebraically show that inverse of function $f(x)= x^n$ is $g(x)= \sqrt[n]{x}$, under function composition $\circ$, with suitable part of $\mathbb R$ as domain, codomain?

By suitable domain mean that for function $f(x)$ if $2\mid n$, then need restrict domain to $x\geq 0$, as then there is bijective map $x\rightarrow x^n$. Though, for inverse function, can choose domain as either non-negative reals, or negative reals. Anyway, $f^{-1}(x) = \sqrt[n]{x}.$

So, is it enough to state $f(g(x))=(\sqrt[n]{x})^n=x$. This has assumption that identity is the original element $x$.

Answer 1

The problem with your function is that it is not one-to-one, therefore it has no inverse. To make it a one-to-one function, you simply make the domain smaller than $\mathbb{R}$. For example, take the domain to be $(0,\infty)$, then there is an inverse function, namely $f^{-1}(x) = \sqrt{x}$. To show that this is indeed the inverse function of $f$, set $y = x^2$ and switch $x$ and $y$ in the equation yields $x = y^2$. Then finally solve for $y$: $y = \pm\sqrt{x}$. But you take $y = \sqrt{x}$ because the range of $f^{-1}$ is the domain of $f$ which is the positive reals. So $y = \sqrt{x} \implies f^{-1}(x) = \sqrt{x}$.