I have this exercise;

For each of the following functions, determine the inverse function.

Here, $\mathbb{R}_{\geq 0}$ denotes the set of all non-negative reals:

$$f : \mathbb{R}_{\geq 0}\to\mathbb{R}, x\mapsto \sqrt{x}$$

But I really don't know where or how to start, could anyone provide some guidance on how to get an inverse function? maybe show some steps? Thank you so much


In general, if you have a function $y = f(x)$ and you want to find the inverse, then you want to rearrange $y = f(x)$ so that $x$ is a function of $y$.

For your example, observe that $y = \sqrt{x}$ so that $y^{2} = x$. Thus your inverse function is $f^{-1}(x) = x^{2}$.

It remains to determine the domain and range. We first note that $f : \mathbb{R}_{\geq 0} \to \mathbb{R}$ has the domain $\mathbb{R}_{\geq 0}$ and the range $\mathbb{R}$. However, the image of $f$ is the set of all elements in the range that are mapped to by $f$ from something in the domain. In this case, the image of $f$ is $\mathbb{R}_{\geq 0}$ (can you see why?). Thus, the domain of $f^{-1}$ will be this set, the image of $f$, and the range of $f^{-1}$ will be the domain of $f$ so that $f^{-1} : \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$. Hence the inverse function is $$ f^{-1} : \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0},\quad f^{-1}(x) = x^{2}. $$

It is important that the domain of $f^{-1}$ is the image of $f$ otherwise, as drhab points out, one may have something like $$ -1\stackrel{f^{-1}}{\to}1\stackrel{f}{\to}1\neq-1, $$ which is not what we want since we should have $f\circ f^{-1} = f^{-1}\circ f = \text{Identity map}$.

  • $\begingroup$ Then $-1\stackrel{f^{-1}}{\to}1\stackrel{f}{\to}1\neq-1$ so no identity (as it should). $\endgroup$ – drhab Aug 22 '18 at 12:13
  • $\begingroup$ @drhab Ah yes, let me fix that! $\endgroup$ – Bill Wallis Aug 22 '18 at 12:15

We have $f(R)=R$ and $f$ is injective.

$y= \sqrt{x} \iff y^2=x$, hence $f^{-1}:R \to R$ is given by $f^{-1}(x)=x^2$.

  • $\begingroup$ What is $R$ here? If it is $\mathbb R$ then it is not correct. We get $-1\stackrel{f^{-1}}{\to}1\stackrel{f}{\to}1\neq1$. $\endgroup$ – drhab Aug 22 '18 at 12:14
  • $\begingroup$ In the original post, the OP wrote $R$ instead of $\mathbb{R}_{\geq 0}$. $\endgroup$ – Fred Aug 22 '18 at 13:05

You can solve the question in the follwoing manner swap x and y then you will get a function $x=f(y)$ in terms of $y$ again swap x and y
$$y=\sqrt{x}$$ $$y^2=x$$ $$x=y^2$$ $$f^{-1}(x)=x^2$$

  • $\begingroup$ Sorry for being stupid but does "f−1" denote that it's an inverse function? $\endgroup$ – ValentineJ Aug 22 '18 at 11:03
  • $\begingroup$ yes $f^{-1}(x)$ is the inverse function of $f(x)$ $\endgroup$ – Deepesh Meena Aug 22 '18 at 11:03
  • $\begingroup$ awesome, great explanation thank you $\endgroup$ – ValentineJ Aug 22 '18 at 11:04

The function $f:\mathbb R_{\geq0}\to\mathbb R$ prescribed by $x\mapsto\sqrt x$ has no inverse.

This because it is not surjective.

The function $g:\mathbb R_{\geq0}\to\mathbb R_{\geq0}$ prescribed by $x\mapsto\sqrt x$ is bijective, hence has an inverse.

It is the function $\mathbb R_{\geq0}\to\mathbb R_{\geq0}$ prescribed by $x\mapsto x^2$.

For finding this function see the other answers.

  • $\begingroup$ At least be so kind to motivate your downvote. As long as you do not I will not take you serious (whoever you are). $\endgroup$ – drhab Aug 23 '18 at 17:12

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