How do you analytically bound $f(x)=-2\cos(2x)+4\cos(x)-3$ to get 0 for upper bound?
If you graph $f(x)$ it looks like:
Clearly from this graph we can see $-9 \leq f(x) \leq 0 $.
But how do you conclude this analytically?
My thinking is
$-2 \leq -2\cos(2x) \leq 2$ and $-4 \leq 4\cos(x) \leq 4 \implies -6 \leq 2\cos(2x)+4\cos(x) \leq 6 $ .
Then subtracting by 3 across the inequality, it becomes
$$-9 \leq -2\cos(2x)+4\cos(x)-3 \leq 3 $$
While the upper bound of 3 is technically correct, the graph shows a tighter upper bound at 0. Why can't I derive that tight upper bound to be 0 analytically? Is there a better way?
Let $$f(x) = 2 \cos 2x + 4 \cos x - 3 = 2 (2 \cos^2 x - 1) + 4 \cos x - 3 = 4 \cos^2 x + 4 \cos x - 5.$$ So as a quadratic in $z = \cos x$, we have $$f(x) = 4z^2 + 4z - 5 = 4z^2 + 4z + 1 - 6 = (2z + 1)^2 - 6.$$ It follows that on the interval $z \in [-1,1]$, the maximum is attained when $(2z+1)^2$ is as large as possible, which occurs when $z = 1$, and the value attained is $3$. The plot you have provided does not match the function you specified. Clearly, when $x = 0$, we have $\cos 2x = \cos x = 1$, hence $$f(0) = 2 + 4 - 3 = 3.$$
After seeing the comments, if we consider instead the function $$g(x) = -2 \cos 2x + 4 \cos x - 3 = -4 \cos^2 x + 4 \cos x - 1 = -(2 \cos x - 1)^2,$$ it becomes immediately obvious that $g$ is at most $0$.