How to graph $f(x)= \frac{1}{2} (e^x + e^{-x})$ without graphing app?

Question

The function $f$ is defined by $f(x)= \frac{1}{2} (e^x + e^{-x})$, $x \in \Re$. Show that $f$ has no inverse.

If I am to show that function $f$ has no inverse, I would graph function $f$ and show that it does not pass the horizontal-line test, as a horizontal line passing through the graph would have more than one intersection point.

The problem is, I am not sure how to graph $f(x)= \frac{1}{2} (e^x + e^{-x})$ without using a graphing application. I know how a $y=e^x$ graph looks like, but how do I combine $e^x + e^{-x}$?

I thought of using Piecewise function, but I do not know how to proceed.

Answer 1

We have that

$$\lim_{x\to0^+}\frac{e^x}2=\frac12\;,\;\;\lim_{x\to0^+}\frac{e^{-x}}2=\frac12$$

and then

$$\lim_{x\to\infty}\frac{e^x}2=\infty\;,\;\;\lim_{x\to0^+}\frac{e^{-x}}2=0$$

And all this basically tells you what happens to the graph of the function for $\;x\ge0\;$.

If we add to this that the function is monotonically ascending since

$$f'(x)=\frac12\left(e^x-e^{-x}\right)=\frac{2^{2x}-1}{2e^x}>0\,,$$

Since the exponential is ascending and $\;e^{2x}>e^x\;$, you can see how the positive (and thus also the negative) branch of this function ascend pretty quickly to $\;\infty\;$

Answer 2

Let $f(x) = \frac12(e^x + e^{-x}).$

$f''(x) = \frac12(e^x + e^{-x}) > \frac12(0+0)= 0$ for all $x$, i.e. $f(x)$ is strictly concave.

Also, as pointed out before, $f(x)$ is even: $f(x) = f(-x)$, and this means that the graph can be reflected in the $y-$axis. Concavity along with evenness of $f$ suggests that $f(0)$ is a minimum.

Let's confirm the stationary points of the graph for good measure. $f'(x) = 0 \implies e^x - e^{-x} = 0 \implies e^x = \frac{1}{e^x} \implies e^{2x} = 1 \implies 2x = 0 \implies x=0,$ so $(0,f(0)) = (0,1)$ is the only stationary point and it is a minimum because $f$ is concave.

Also, $\lim_{x \to \infty}f(x) = \lim_{x \to -\infty}f(x) = \infty $, but we are still not done, because we haven't shown that our graph doesn't have asymptotes. But $\lim_{x \to \infty}f'(x) = \lim_{x \to \infty}{ \left(e^x - e^{-x}\right)} = \infty,$ and $\lim_{x \to -\infty}f'(x) = \lim_{x \to -\infty}{ \left(e^x - e^{-x}\right)} = -\infty.$

And now we know what the graph looks like.