Curve sketching, question 18

Topic: Curve sketching
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §7 CLP-1 §3.6Key results

On the interval $0 < x < 2\pi$ , where is $f(x) = 2\sin x + \cos(2x)$ increasing?

$\left(0, \tfrac{\pi}{6}\right) \cup \left(\tfrac{5\pi}{6}, 2\pi\right)$
$\left(0, \tfrac{\pi}{2}\right) \cup \left(\tfrac{3\pi}{2}, 2\pi\right)$
$\left(\tfrac{\pi}{6}, \tfrac{\pi}{2}\right) \cup \left(\tfrac{5\pi}{6}, \tfrac{3\pi}{2}\right)$
$\left(0, \tfrac{\pi}{6}\right) \cup \left(\tfrac{\pi}{2}, \tfrac{5\pi}{6}\right) \cup \left(\tfrac{3\pi}{2}, 2\pi\right)$

How sure are you?

SureNot sureGuessing

Theory UBC Math 100 §7 CLP-1 §3.6Key results

Worked solution

Strategy

Differentiate, factor with the double-angle identity, and make a sign chart from the zeros of both factors on the interval.

Working

$f'(x) = 2\cos x - 2\sin(2x) = 2\cos x - 4\sin x\cos x = 2\cos x\,(1 - 2\sin x)$
Chain rule on cos(2x), then sin(2x) = 2 sin x cos x.
$\cos x = 0:\ x = \tfrac{\pi}{2},\ \tfrac{3\pi}{2}; \qquad \sin x = \tfrac{1}{2}:\ x = \tfrac{\pi}{6},\ \tfrac{5\pi}{6}$
Four critical points split the interval into five pieces.
$\begin{array}{c|ccc} & \cos x & 1 - 2\sin x & f'(x) \\ \hline \left(0, \frac{\pi}{6}\right) & + & + & + \\[3pt] \left(\frac{\pi}{6}, \frac{\pi}{2}\right) & + & - & - \\[3pt] \left(\frac{\pi}{2}, \frac{5\pi}{6}\right) & - & - & + \\[3pt] \left(\frac{5\pi}{6}, \frac{3\pi}{2}\right) & - & + & - \\[3pt] \left(\frac{3\pi}{2}, 2\pi\right) & + & + & + \end{array}$
Test one point in each piece.
$\text{increasing on } \left(0, \tfrac{\pi}{6}\right) \cup \left(\tfrac{\pi}{2}, \tfrac{5\pi}{6}\right) \cup \left(\tfrac{3\pi}{2}, 2\pi\right)$
So f has local maxima at π/6 and 5π/6 (value 3/2) and local minima at π/2 (value 1) and 3π/2 (value −3).

Answer

\left(0, \frac{\pi}{6}\right) \cup \left(\frac{\pi}{2}, \frac{5\pi}{6}\right) \cup \left(\frac{3\pi}{2}, 2\pi\right)

Check

f takes the values 1, 3/2, 1, 3/2, −3 and 1 at 0, π/6, π/2, 5π/6, 3π/2 and 2π: up, down, up, down, up.

Where this goes wrong

Cancelling cos x from 2cos x(1 − 2sin x) = 0 and keeping only sin x = 1/2. The zeros of cos x, at π/2 and 3π/2, are critical points too, and f′ changes sign at each.