Any hints to shortly prove that $f(x) = (x^2-1)^2 \cosh2 x-(x^2+1)^2$ crosses the $x$ axis only once?

Question

I want to show that the function $f(x) = (x^2-1)^2 \cosh2 x-(x^2+1)^2\;$ for $x>0$ has only one root. Dealing with the first and second derivatives cannot help much since as can be seen in the picture, the function is not monotone. Is there any hope to prove it in a short and simple way? I appreciate any hints/comments.

Answer 1

There's clearly a min somewhere around $x = 1.1$; you could show that $x$ is strictly decreasing below that min, and strictly increasing above it, and then you'd be done.

Answer 2

Partial answer: for $x>1$ it is not hard to see that $$g(x)=\cosh2x$$ is increasing and $$h(x)=\Bigl(\frac{x^2+1}{x^2-1}\Bigr)^2$$ is decreasing. Also $h(x)\to\infty$ as $x\to1^+$. So $$\cosh2x=\Bigl(\frac{x^2+1}{x^2-1}\Bigr)^2$$ has exactly one solution for $x>1$.

Answer 3

To find the minimum without graphing or numerical method.

Since $f'(1)=-8$, expand the derivative as a series around $x=1$; it gives $$f'(x)=-8+4 \left(-4+\frac{1}{e^2}+e^2\right)(x-1)+$$ $$\left(-12-\frac{6}{e^2}+18 e^2\right)(x-1)^2+\left(-4+\frac{2}{e^2}+34 e^2\right)(x-1)^3+O\left((x-1)^4\right)$$

Using the hyperbolic method, we can solve exactly the cubic equation in $(x-1)$ and obtain as an approximation, $x=1.18153$ while the exact solution is $x=1.17752$.

Using one more term in the expansion would give $x=1.17802$ (solving a quartic equation).

Answer 4

$\cosh2x = \frac{1}{2}(e^{2x} + e^{-2x})$. $e^{2x}$ is exponentially increasing for $x > 0$ and so is $e^{-2x}$ for $x < 0$. This means $\frac{1}{2}(e^{2x} + e^{-2x})$ behaves as a parabolic function with minimum at $(0, 1)$.

This is closely similar to the behaviour of the parabolic function $(x^2 + 1)^2$, and with a common minimum of $(0, 1)$.

Hence, $$\cosh(2x) \approx (x^2 + 1)^2$$

$$(x^2 - 1)^2 \cdot \cosh{2x} - (x^2 + 1)^2 \approx (x^2 - 1)^2(x^2 + 1)^2 - (x^2 + 1)^2$$

Solving we see, $$(x^2 + 1)^2\big((x^2 - 1)^2 - 1\big) = 0$$

$$\text{Since, } (x^2 + 1)^2 > 0 \text{, only } (x^2 - 1)^2 - 1 = 0$$

Expanding, gives

$$x^4 - 2x^2 = 0$$

$$x^2(x^2 - 2) = 0$$

Other than $x^2 = 0$, $x^2 - 2$ has only $1$ positive root, as required.