I have an equation involving a polynomial function with coefficients that are exponentials.

Question

To solve: $$2\left[(2^{x}-1)\left(x^{2}-1\right)+(2^{x^{2}-1}-1)x\right]= 0$$

There are three obvious solutions by observation, $x = -1, 0, 1$, and the graph of the function (red font) and its first derivative (blue font) look like this:

How can I prove that these are the only zeroes of the function?

Answer 1

$$(2^x-1)(x+1)(x-1)$$ has exactly three roots, at $-1,0,1$, by cancelling the three factors, and

$$x(2^{(x+1)(x-1)}-1)$$ as well.

Then it is easy to see that these two functions have the same signs everywhere, and their sum can only be zero at the common roots.

Answer 2

HINT

If you prove the derivative has 2 roots, by Rolle's theorem, you are limited to 3 roots of the original function, say $r<s<t$, with a root of the derivative between each pair, i.e. in $(r,s)$ and another in $(s,t)$.

In other words, if there were a fourth root, $x <r$, say, you would have to have another root of the derivative in $(x,r)$...