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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

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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 $f\circ f^{-1} = f^{-1}\circ f = \text{Identity map}$.

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  • $\begingroup$ 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
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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 $f^{-1}(x)=x^2$.

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  • $\begingroup$ 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
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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
$$y=\sqrt{x}$$ $$y^2=x$$ $$x=y^2$$ $$f^{-1}(x)=x^2$$

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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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