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Maxima & Minima · Question 62 of 80

Question 62

HardOriginal question

What is the greatest value of f(x)=2sin⁡x+sin⁡2xf(x) = 2\sin x + \sin 2x on [0,2π][0, 2\pi]?

Options for question 62

Show the answer and explanation
CorrectNot quiteAnswer C

Correct answer: C

Explanation

  1. f′(x)=2cos⁡x+2cos⁡2x=0f'(x) = 2\cos x + 2\cos 2x = 0 gives 2cos⁡2x+cos⁡x−1=02\cos^{2} x + \cos x - 1 = 0: cos⁡x=12\cos x = \frac{1}{2} or −1-1, so x=π3,π,5π3x = \frac{\pi}{3}, \pi, \frac{5\pi}{3}.

  2. f(π3)=3+32=332f\left(\frac{\pi}{3}\right) = \sqrt{3} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}, f(π)=0f(\pi) = 0, f(5π3)=−332f\left(\frac{5\pi}{3}\right) = -\frac{3\sqrt{3}}{2}, and 0 at both ends.

Worth noting

3 adds the two peaks, 2 and 1, but sin⁡x\sin x and sin⁡2x\sin 2x do not peak at the same xx.

Concept
Extreme values of a trigonometric function
Typical time
120 s
Your time
—
The rule this testsGreatest and least values on an interval
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Ten questions from Maxima & Minima, at your own pace.

Practise this chapter →

Ten questions from Maxima & Minima, at your own pace.