# Injection and surjection ( with continuity )

Okay so I'm required to find if the given function is one-one/many-one and onto/into.

The function is : $\mathbb R \to \mathbb R$ and $$f(x) = x\left(\frac{2^x - 2^{-x}}{2^x + 2^{-x}}\right)$$

So as this function is even , it can't be one-one .But I'm facing difficulty in deducing if it's onto or not. My book says "The function is continuous and any even-continuous function cannot have range ‘R’. Hence function is many one into."

Can someone explain how do I prove it continuous just by observation ? Or is there any other way I can show it to be an onto function ?

• The function is the same as $f(x)=x$ for $x\neq 0$. To be a function $\mathbb{R}\to\mathbb{R}$ they need to define it at $x=0$. If they define $f(0)=0$, then $f(x)=x$, which is odd, continuous, one-to-one, and onto. – arts Nov 24 '17 at 15:37

Well, for every $x$ in $\mathbb R ,\ f(x)$ clearly has a value mapped into $\mathbb R$.

Also, since the denominator doesn't go to zero for any $x$, you can say the function doesn't have any point discontinuity.

So, we can say that $f$ is continuous for every $x$ in its domain (just by observation).

• Quite helpful , is it sufficient to check if denominator becomes 0 ? – Tanuj Nov 24 '17 at 15:46
• That's where generally the point of discontinuity occurs in polynomial based functions. In case, the function is a break-up of multiple sub-functions like $f(x) = \begin{cases} |x|, & \text{if$x \ne 0$} \\ 0, & \text{if$x=0$} \end{cases}$ Here, $f$ is continuous( though not differentiable at $x=0$) which you can just check by looking at the end point values. – Your IDE Nov 24 '17 at 16:07

Your function is not onto, because it is non-negative. To see this, examine the cases when $x\ge0$ and $x\le0$ separately. The denominator is always positive and the numerator is positive when $x$ is positive and negative when $x$ is negative. So the function is always non-negative.

Also note that $\lim_{x\rightarrow \infty} f(x) = \lim_{x\rightarrow -\infty} f(x) = \infty$ so by continuity $f$ maps $\mathbb{R}$ onto $[0,\infty)$.

The statement in your book stating that any even, continuous function can't map $\mathbb{R}$ onto $\mathbb{R}$ is incorrect. Since the function $f(x)=x \sin(x)$ is even, continuous and maps $\mathbb{R}$ onto $\mathbb{R}$. To see this, it suffices to see that $f(\frac{\pi}{2} +2n\pi) = \frac{\pi}{2} +2n\pi$ and $f(\frac{3\pi}{2} +2n\pi) = -(\frac{3\pi}{2} +2n\pi)$. then using the Mean value theorem to conclude that the function attains all the values in between.

It's interesting to note that $f(x) = x\left(\frac{2^x - 2^{-x}}{2^x + 2^{-x}}\right) = x \tanh(x\log2 )$.