How do you find the difference between the maximum and minimum values of $f(x)=\cos(2x)+\cos x$?

Question

This is a SAT subject test question. Is there a way to do it without a graphing calculator (I do not own one)? Most SAT questions have multiple approaches so I was hoping there might be a quick shortcut that I haven't been able to figure out yet.

Answer 1

With $c:=\cos x$, $f=2c^2+c-1$, so find the extrema of this function on $[-1, 1]$. Do not assume each extremum is at an endpoint of this range. The quadratic has its global minimum at $c=-\frac14$ in this range, so its maximum is at whichever endpoint is furthe from this value of $c$. So the extrema differ by$$[2c^2+c-1]_{-1/4}^1=[2x^2]_0^{5/4}=\frac{25}{8}.$$

Answer 2

We have that

$$\cos(2x)+\cos x=2\cos^2 x+\cos x-1$$

and

$$-1\le \cos x \le 1$$

Answer 3

To maximize $f(x) = \cos 2x + \cos x $ , differentiate it with respect to $x$.

$$f'(x) = -2\sin2x -\sin x$$

For the function to achieve a minimum value , $f'(x)$ should be $0$.

$$2\sin 2x = -\sin x$$ $$2 . 2\sin x \cos x = -\sin x$$ $$\cos x = \frac{-1}{4}$$

Hence minimum value is $f(x) = 2 \times \frac{1}{16} - \frac14 - 1 \implies f(x) = -\frac 98$

It is easy to see that the function is maximum at $\cos x =1 $ , $f(x) = 2\times1 -1+1 \implies f(x) = 2$.

Hence the difference is $2 - (- \frac98) = \boxed{\color{blue}{\frac{25}{8}}}$