Problem Statement
Find the absolute maximum and minimum values of $f(x) = 2\cos 2x - \cos 4x$ in the interval $[0, \pi]$.
Verified Solution & Marking Scheme
Find Stationary Points (f'(x) = 0)
$f'(x) = 2(-\sin 2x)(2) - (-\sin 4x)(4) = -4\sin 2x + 4\sin 4x = 4(\sin 4x - \sin 2x)$
Set $f'(x) = 0 \implies \sin 4x = \sin 2x$.
Using $\sin 4x - \sin 2x = 2\cos 3x \sin x = 0$:
1. $\sin x = 0 \implies x = 0, \pi$.
2. $\cos 3x = 0 \implies 3x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2} \implies x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$.
Evaluate f(x) at All Critical Points and Endpoints
- At $x = 0$: $f(0) = 2\cos 0 - \cos 0 = 2(1) - 1 = 1$.
- At $x = \pi/6$: $f(\pi/6) = 2\cos(\pi/3) - \cos(2\pi/3) = 2(1/2) - (-1/2) = 1 + 1/2 = \frac{3}{2}$.
- At $x = \pi/2$: $f(\pi/2) = 2\cos(\pi) - \cos(2\pi) = 2(-1) - 1 = -3$.
- At $x = 5\pi/6$: $f(5\pi/6) = 2\cos(5\pi/3) - \cos(10\pi/3) = 2(1/2) - (-1/2) = \frac{3}{2}$.
- At $x = \pi$: $f(\pi) = 2\cos(2\pi) - \cos(4\pi) = 2(1) - 1 = 1$.
Identify Absolute Extrema
Absolute Maximum $= \frac{3}{2}$ (at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$).
Absolute Minimum $= -3$ (at $x = \frac{\pi}{2}$).