Finding solutions for equations with exponential

Question

I have no idea on how to prove that $$ x(e^x-e^{-x})-e^x=0 $$ has at least 2 real solutions in R? How would you generally go about proving solutions exist if you can't isolate the x or use the quadratic formula?

Answer 1

Multiply through by $e^x$, whihc you can do because $e^x$ is never zero for finite real $x$; so you are looking for solutions to $$ f(x) = x(e^{2x}-1) - e^{2x} = 0$$

You know that the $f(x)$ is continuous because the product of two continuous functions is continuous, as is the sum of two continuous functions, and $e^{2x}$ is continuous.

You also know that at $x=-30$, $$f(-30) = 30(1-e^{-60}) - e^{-60} > 0 $$ since $e^{-60}$ very small compared with 30. And at $x=0$, $$f(0) = - e^{0} = -1 < 0 $$ So by the intermediate value theorem, $f(x)$ is zero somewhere on the interval $(-30,0)$.

And you also know that at $x=2$ $$ f(2) = 2(e^4-1)-e^4 = e^4 -1 > 0 $$ So by the intermediate value theorem, $f(x)$ is also zero somewhere on the interval $(0,2)$.

The harder problem would be to show there are only two real zeros.

Answer 2

Let's set $f(x)=x(e^x-e^{-x})-e^x$. First, look at $f(0)$: $$f(0)=0(e^0+e^{-0})-e^0=0-1=-1$$ This means that there are solutions where $f(x)<0$.

Let's take two limits: $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(xe^x-xe^{-x}-e^x)=\lim_{x\to\infty}(x-1)e^x-\lim_{x\to\infty}\frac{x}{e^x}=?$$ $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(xe^x-xe^{-x}-e^x)=\lim_{x\to-\infty}(x-1)e^x-\lim_{x\to-\infty}\frac{x}{e^x}=?$$ If both of these are greater than $0$, then - given that $f(x)$ is continuous - there are at least two real solutions. Can you tell if this is the case?

Note that this would not guarantee that there are only two solutions, merely that at least two solutions must exist.