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

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.

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 $$x$$.

• Try z(x)=$\sqrt[3]x$… Also, additive inverse and inverse function are not the same. Commented Jun 22, 2022 at 3:44

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 $$y = \sqrt{x} \implies f^{-1}(x) = \sqrt{x}$$.
• Under the restricted domain, is there a way to algebraically find the inverse : $z(x)$? It should be $z(x)= f^{-1}(x)$, but how it leads to taking root (square root) is unclear algebraically, though can see it fits by composition. So, is the only way is to try see composition? Or can see the same algebraically too, without fitting into composition? Commented Jun 22, 2022 at 3:54
• You are wrong: $(\sqrt{x})^2 = x$, but you got $1$. Commented Jun 22, 2022 at 4:15
• If $z(x)$ is the inverse of $f(x)$, then for any $x$ in the domain of $z$, you always have: $f(z(x)) = x$. So it won't be $1$ as you might think. It is the fundamental property of the inverse function. So for example, $f(z(1)) = 1$. Commented Jun 22, 2022 at 4:27
• Yes, you don't need to further restrict the domain of $z$ because it always equals to the range of $f$ which is already positive reals in my example. Commented Jun 22, 2022 at 5:13