Find Number of Real roots of $x^{18}+e^{-x}+5x^2-2\cos x=0$
My try:
Let $$f(x)=x^{18}+e^{-x}+5x^2-2\cos x$$
I have used IVT and figured out that there is atleast one root in each of the intervals $(-1,0)$ and $(0,1)$
Since:
$$f(-1)\gt 0$$ $$f(0)\lt 0$$ $$f(1)\gt 0$$
But how to identify how many roots in each interval?
Not an answer but too long for a comment.
I really do not see how to answer the question beside continuing to divide the interval that is to say computing $f(-2^{-n})$ and $f(2^{-n})$. You will find a root between $-0.50$ and $-0.25$ and another one between $0.25$ and $0.50$. But the problem remains the same. For sure, you could repeat the problem changing the $2$ to any number $a>2$ but this is just inspection.
What is "amazing" is that if you make the Taylor series of the monster around $x=0$, you should get $$f(x)\sim -1-x+\frac{13 }{2}x^2+O\left(x^3\right)$$ Ignoring the higher order terms, this would give $$x_1=-\frac{1}{13} \left(3 \sqrt{3}-1\right)\approx -0.322780$$ $$x_2=\frac{1}{13} \left(3 \sqrt{3}+1\right)\approx 0.476627$$ while the "exact" solutions, obtained using Newton method, are $-0.32179$ and $0.480627$.
Pushing the expansion to $O\left(x^4\right)$, you would need to solve a cubic and the solutions would be $-0.321712$ and $0.480162$.