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Prove that the function $f\colon\mathbb{R}\to\mathbb{R}$ $$ f(x)= x |x| +1 $$ is bijective and find the inverse.


This is what I did to prove that the function is one to one.

Suppose $f(x_1)=f(x_2)$

Then $x_1 |x_1| +1=x_2 |x_2| +1$ or $x_1 |x_1| =x_2 |x_2|$

Case 1: $x_1,x_2 \ge0$ or $x_1,x_2\le0$

Then $x_1^2= x_2^2$.

Thus $x_1=x_2$ (As both $x_1,x_2$ are of the same sign in this case)

Case 2: Suppose $x_1>0$ and $x_2<0$:

Then $x_1 \cdot x_1=-k\cdot k$ where $x_2 =-k$ and $k\in\mathbb{R}^+$

Then $x_1^2 =-k^2$. This is not possible. So is the case with x$_2$>0 and x$_1$<0.

Hence either $x_1,x_2\ge0$ or $x_1,x_2\le0$. In that case $x_1 =x_2$. Hence $f$ is one to one.

$y= x |x| +1$.

If $x\ge0$:

$y-1=x^2$

$x=\sqrt{y-1}$

If $x<0$:

$y-1=-x^2$

$x=\sqrt{1-y}$

Thus $$f^{-1}= \begin{cases} \sqrt{1-x}, & \text{if $x<0$} \\ \sqrt{x-1}, & \text{if $x \ge0$} \\ \end{cases} $$

Is this the correct inverse? How to show that $f$ is surjective? And is my proof of one to one correct?

Any help is appreciated.

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  • $\begingroup$ Your proof looks good. To prove surjectivity, we need to show that for all $y$ in the co-domain, there exists an $x$ in the domain such that $f(x)=y$. With the inverse that you have found with range = $\mathbb{R}$ you have essentially showed surjectivity. $\endgroup$ – Kelvin Soh Dec 12 '13 at 14:17
  • $\begingroup$ Oh. And spotted two slight mistakes. For the case $x<0$, $x$ should be $-\sqrt{1-y}$ instead. Next, for your inverse function, notice after making $x$ the subject we are interchanging $x$ and $y$ as dummy variables. In that case, the domain for your inverse function should be $x<1$ and $x\geq 1$ instead of $x<0$ and $x \geq 0$. $\endgroup$ – Kelvin Soh Dec 12 '13 at 14:18
  • $\begingroup$ When x<0 it why should there be a minus sign?Is it because square root is positive and to get a negative value(as x<0), we have to have a minus sign?And can you explain how it becomes x<1 and x>=1. $\endgroup$ – clarkson Dec 12 '13 at 15:48
  • $\begingroup$ Why does it say that if the function is one to one then an inverse exists. Shouldn't our function be a bijection in order to have a inverse?So if I have to prove a function is surjective is it enough to find the inverse and if the function's co-domain and the inverse function's domain are equal,can I say that the function is surjective. $\endgroup$ – clarkson Dec 12 '13 at 15:56
  • $\begingroup$ (About the minus sign) Yes, the convention is for the square root symbol to denote the positive square root so we take the negative value as $x<0$. Notice that $y = x|x|+1$ so $y<1$ if $x<0$. (Similarly $y\geq 1$ if $x\geq0$). When we define $f^{-1}(x)$ the "$x$" is actually behaving as a dummy for "$y$". $\endgroup$ – Kelvin Soh Dec 12 '13 at 16:36
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Shortcut:

Function $g:\mathbb{R}\rightarrow\mathbb{R}$ defined by $x\mapsto x\left|x\right|$ has an inverse in $h:\mathbb{R}\rightarrow\mathbb{R}$ defined by $y\mapsto\sqrt{y}$ if $y\geq0$ and $y\mapsto\sqrt{-y}$ otherwise. The existence of an inverse garantees the bijectivity of $g$ and consequently of $f\left(x\right)=g\left(x\right)+1$.

If you are asked to prove bijectivity and to find an inverse then in principle it is enough to do the second task only.

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The proof of injectivity is good, albeit a bit involved.

From $x_1|x_1|=x_2|x_2|$ you immediately deduce that $x_1$ and $x_2$ are both positive, both negative or both $0$. So you can get $|x_1|^2=|x_2|^2$ from which $|x_1|=|x_2|$ and the conclusion is immediate.

For surjectivity, you have to work a bit harder. If $y=x|x|+1$, then $x|x|=y-1$.

So, for $y>1$ you have to find $x>0$, for $y=1$ you can take $x=0$ and for $y<1$ you have to take $x<0$.

If $y>1$, the condition becomes “$x^2=y-1$ and $x>0$”, so $x=\sqrt{y-1}$.

If $y<1$, …

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