I was looking at the extremum points of the function
$$f(x)=\dfrac {2^{2 x + 1} – 3^{2 x + 1} – 6^x \cos(\pi x)}{4^x + 9^x + 2^{x + 1} 3^x \cos(\pi x)},$$
where $x$ is a real number. I couldn't do it because I couldn't take the derivative of this function (because the derivative of this function is very complicated). Even if I took the derivative, it doesn't end there, you have to find the roots of the derivative function which is even more complicated than the previous step.
Because I didn't do it manually, I used GeoGebra to find the extremum points. The extremum points of this function have an interesting pattern, they are approximately $1.97,2.98,3.98,4.98,\dots,17.98,18,98,\dots$ and so on. But these approximate findings are not satisfactory. I want to know the full answer, and I don't know how to find it.
Best Answer
$$f(x)=\dfrac {2^{2 x + 1} - 3^{2 x + 1} - 6^x \cos(π x)}{4^x + 9^x + 2^{x + 1} 3^x \cos(π x)}$$ $$f'(x)=-\frac{5\ 6^x \left(2^{x+1} 3^x \log \left(\frac{3}{2}\right)-\pi \left(4^x-9^x\right) \sin (\pi x)+\left(4^x+9^x\right) \log \left(\frac{3}{2}\right) \cos (\pi x)\right)}{\left(4^x+9^x+2^{x+1} 3^x \cos (\pi x)\right)^2}$$
So, we look first for the zeros of function $$g(x)=2^{x+1} 3^x \log \left(\frac{3}{2}\right)-\pi \left(4^x-9^x\right) \sin (\pi x)+\left(4^x+9^x\right) \log \left(\frac{3}{2}\right) \cos (\pi x)$$ for which $$g'(x)=\log (6) \left(2^{x+1} 3^x \log \left(\frac{3}{2}\right)-\pi \left(4^x-9^x\right) \sin (\pi x)\right)+$$ $$\left(\pi ^2 \left(9^x-4^x\right)+\log \left(\frac{3}{2}\right) \left(4^x \log (4)+9^x \log (9)\right)\right) \cos (\pi x)$$
As you noticed, the solutions are quite close to integer values of $x$. So, use the first iteration of Newton methods with $x_0=n$
and you will have $$x_1=n-\frac{g(n)}{g'(n)}$$
Now, notice that $$g(2m)=\left(3^{2m}+2^{2m}\right)^2 \log \left(\frac{3}{2}\right)$$ $$g(2m+1)=-\left(3^{2 m+1}-2^{2 m+1}\right)^2 \log \left(\frac{3}{2}\right)$$
Simplify $g'(2m)$ and $g'(2m+1)$ in the same manner.
Repeating @Moo's calculations $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ -10 & -9.95477 & -9.95768 \\ -9 & -8.95881 & -8.96124 \\ -8 & -7.95260 & -7.95577 \\ -7 & -6.96159 & -6.96371 \\ -6 & -5.94726 & -5.95114 \\ -5 & -4.96717 & -4.96874 \\ -4 & -3.93239 & -3.93850 \\ -3 & -2.97717 & -2.97793 \\ -2 & -1.87057 & -1.88914 \\ -1 & -0.99178 & -0.99188\\ 1 & 0.99203 & 0.99188 \\ 2 & 1.91158 & 1.88914 \\ 3 & 2.97889 & 2.97793 \\ 4 & 3.94558 & 3.93850 \\ 5 & 4.97063 & 4.96874 \\ 6 & 5.95564 & 5.95114 \\ 7 & 6.96623 & 6.96371 \\ 8 & 7.95948 & 7.95577 \\ 9 & 8.96411 & 8.96124 \\ 10 & 9.96108 & 9.95768 \end{array} \right)$$
Now, you just need to classify the points (just compute the second derivative for $x=n$; this should be more than sufficient)