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Chapter 6 · Application of Derivatives

Exercise 6.3 · Q12: Greatest and least values of x+sin⁡2xx+\sin 2x on [0,2π][0,2\pi]

Exercise 6.3Absolute extrema on a closed interval · Section 6.4.1
The question

Find the maximum and minimum values of x+sin⁡2xx + \sin 2x on [0,2π][0, 2\pi]⁠.

Exam answer

Let f(x)=x+sin⁡2xf(x) = x + \sin 2x⁠. Then f′(x)=1+2cos⁡2xf'(x) = 1 + 2\cos 2x⁠.

f′(x)=0f'(x) = 0 gives cos⁡2x=−12\cos 2x = -\frac12⁠, so in [0,2π][0, 2\pi]⁠, x=π3x = \frac{\pi}{3}⁠, 2π3\frac{2\pi}{3}⁠, 4π3\frac{4\pi}{3}⁠, 5π3\frac{5\pi}{3}⁠.

f(0)=0f(π3)=π3+32f(2π3)=2π3−32f(4π3)=4π3+32f(5π3)=5π3−32f(2π)=2π\begin{aligned} f(0) &= 0 \\ f\left(\tfrac{\pi}{3}\right) &= \tfrac{\pi}{3} + \tfrac{\sqrt3}{2} \\ f\left(\tfrac{2\pi}{3}\right) &= \tfrac{2\pi}{3} - \tfrac{\sqrt3}{2} \\ f\left(\tfrac{4\pi}{3}\right) &= \tfrac{4\pi}{3} + \tfrac{\sqrt3}{2} \\ f\left(\tfrac{5\pi}{3}\right) &= \tfrac{5\pi}{3} - \tfrac{\sqrt3}{2} \\ f(2\pi) &= 2\pi \end{aligned}

As 0<32<2π30 < \frac{\sqrt3}{2} < \frac{2\pi}{3}⁠, the four middle values lie between 00 and 2π2\pi⁠.

Maximum value f(2π)=2πf(2\pi) = 2\pi⁠; minimum value f(0)=0f(0) = 0⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

On a closed interval, a continuous function has a greatest and a least value, and they can only occur at critical points inside the interval or at its two end points (Theorems 5 and 6, Section 6.4.1). So list those few points, evaluate the function at each, and compare. Here the local peaks inside the interval turn out to be lower than the value at the right-hand end, because the term xx keeps climbing while sin⁡2x\sin 2x only wobbles between −1-1 and 11⁠.

Given, to find, and the plan
Given
f(x)=x+sin⁡2xf(x) = x + \sin 2x on the closed interval [0,2π][0, 2\pi]⁠.
To find
The maximum (greatest) and minimum (least) values of ff on [0,2π][0, 2\pi]⁠.
Plan
Working rule: differentiate (chain rule for sin⁡2x\sin 2x⁠); solve f′(x)=0f'(x) = 0 for every xx in [0,2π][0, 2\pi]⁠, which means every angle 2x2x in [0,4π][0, 4\pi]⁠; evaluate ff at these points and at 00 and 2π2\pi⁠; pick the largest and smallest.

Basics you need

1Working rule on a closed interval

Find the critical points in the interval, take the end points, compute ff at all of them; the largest is the absolute maximum value and the smallest the absolute minimum value (Working Rule, Section 6.4.1). Lesson: absolute extrema on a closed interval

2Chain rule for sin⁡2x\sin 2x

ddxsin⁡2x=cos⁡2x⋅ddx(2x)\frac{d}{dx}\sin 2x = \cos 2x \cdot \frac{d}{dx}(2x)⁠, which is 2cos⁡2x2\cos 2x⁠: outer function sin⁡\sin⁠, inner function 2x2x⁠. Basics: the chain rule

3Where cos⁡θ=−12\cos\theta = -\frac{1}{2}

In [0,2π][0, 2\pi]⁠, cos⁡θ=−12\cos\theta = -\frac{1}{2} exactly at θ=2π3\theta = \frac{2\pi}{3} (second quadrant) and θ=4π3\theta = \frac{4\pi}{3} (third quadrant); cos repeats every 2π2\pi⁠. Basics: signs of sin and cos

Solution

  1. The derivative

    Let f(x)=x+sin⁡2xf(x) = x + \sin 2x for 0≤x≤2π0 \le x \le 2\pi⁠. Then

    f′(x)=1+2cos⁡2x.f'(x) = 1 + 2\cos 2x.
    ddxx=1\frac{d}{dx}x = 1⁠. For sin⁡2x\sin 2x the chain rule gives cos⁡2x\cos 2x times the derivative of the inner function 2x2x⁠, which is 2. ff is differentiable everywhere, so the only critical points are where f′(x)=0f'(x) = 0⁠.
    Chain rule: derivative of the outer function at the inner one, times the derivative of the inner one. Basics: the chain rule

  2. Solve f′(x)=0f'(x) = 0 on the whole interval

    1+2cos⁡2x=0  ⟺  cos⁡2x=−12.1 + 2\cos 2x = 0 \iff \cos 2x = -\frac{1}{2}.

    As xx runs over [0,2π][0, 2\pi]⁠, the angle θ=2x\theta = 2x runs over [0,4π][0, 4\pi]⁠, which is two full turns. In the first turn [0,2π][0, 2\pi]⁠, cos⁡θ=−12\cos\theta = -\frac{1}{2} at θ=2π3\theta = \frac{2\pi}{3} and 4π3\frac{4\pi}{3}⁠; in the second turn add 2π2\pi⁠: θ=8π3\theta = \frac{8\pi}{3} and 10π3\frac{10\pi}{3}⁠. Halving,

    x=π3, 2π3, 4π3, 5π3.x = \frac{\pi}{3},\ \frac{2\pi}{3},\ \frac{4\pi}{3},\ \frac{5\pi}{3}.
    The common slip is to solve only in the first turn and get two points. Because the angle is 2x2x⁠, not xx⁠, it goes round the circle twice while xx goes round once, so there are four solutions. All four lie in [0,2π][0, 2\pi]⁠.
    cos⁡2π3=cos⁡4π3=−12\cos\frac{2\pi}{3} = \cos\frac{4\pi}{3} = -\frac{1}{2} (reference angle π3\frac{\pi}{3}⁠, cos negative in quadrants II and III). Basics: signs of sin and cos

  3. Evaluate ff at the critical points and the end points

    We need sin⁡2x\sin 2x at each point. Using sin⁡2π3=32\sin\frac{2\pi}{3} = \frac{\sqrt3}{2}⁠, sin⁡4π3=−32\sin\frac{4\pi}{3} = -\frac{\sqrt3}{2} and the period 2π2\pi of sin:

    xxsin⁡2x\sin 2xf(x)f(x)decimal
    00sin⁡0=0\sin 0 = 0000
    π3\frac{\pi}{3}sin⁡2π3=32\sin\frac{2\pi}{3} = \frac{\sqrt3}{2}π3+32\frac{\pi}{3} + \frac{\sqrt3}{2}1.9132
    2π3\frac{2\pi}{3}sin⁡4π3=−32\sin\frac{4\pi}{3} = -\frac{\sqrt3}{2}2π3−32\frac{2\pi}{3} - \frac{\sqrt3}{2}1.2284
    4π3\frac{4\pi}{3}sin⁡8π3=32\sin\frac{8\pi}{3} = \frac{\sqrt3}{2}4π3+32\frac{4\pi}{3} + \frac{\sqrt3}{2}5.0548
    5π3\frac{5\pi}{3}sin⁡10π3=−32\sin\frac{10\pi}{3} = -\frac{\sqrt3}{2}5π3−32\frac{5\pi}{3} - \frac{\sqrt3}{2}4.3700
    2π2\pisin⁡4π=0\sin 4\pi = 02π2\pi6.2832
    8π3=2π+2π3\frac{8\pi}{3} = 2\pi + \frac{2\pi}{3} and 10π3=2π+4π3\frac{10\pi}{3} = 2\pi + \frac{4\pi}{3}⁠, so their sines equal sin⁡2π3\sin\frac{2\pi}{3} and sin⁡4π3\sin\frac{4\pi}{3}⁠. Decimals use π≈3.14159\pi \approx 3.14159 and 32≈0.86603\frac{\sqrt3}{2} \approx 0.86603⁠.
    sin is positive in quadrant II (2π3\frac{2\pi}{3}⁠) and negative in quadrant III (4π3\frac{4\pi}{3}⁠). Basics: signs of sin and cos

  4. Compare: greatest and least

    The six values are 00⁠, about 1.911.91⁠, 1.231.23⁠, 5.055.05⁠, 4.374.37 and 6.286.28⁠. The largest is 2π2\pi⁠, at x=2πx = 2\pi⁠; the smallest is 00⁠, at x=0x = 0⁠.

    To be sure the comparisons are exact (not decimal accidents), compare the end value with the higher local peak:

    2π−(4π3+32)=2π3−32>0,\begin{aligned} &2\pi - \left(\frac{4\pi}{3} + \frac{\sqrt3}{2}\right) \\ &= \frac{2\pi}{3} - \frac{\sqrt3}{2} > 0, \end{aligned}

    since 2π3>2>32\frac{2\pi}{3} > 2 > \frac{\sqrt3}{2}⁠. And every value at a critical point is positive, because each critical point is at least π3\frac{\pi}{3}⁠, which exceeds 1>321 > \frac{\sqrt3}{2}⁠; subtracting at most 32\frac{\sqrt3}{2} leaves a positive number.

    π/3 2π/3 π 4π/3 5π/3 2π 2 4 6 x f(x) 1.91 1.23 5.05 4.37 least: f(0) = 0 greatest: f(2π) = 2π y = x
    f(x)=x+sin⁡2xf(x) = x + \sin 2x on [0,2π][0, 2\pi] wiggles about the dashed line y=xy = x⁠. Violet dots: the four critical points (local maxima at π3,4π3\frac{\pi}{3}, \frac{4\pi}{3}⁠, local minima at 2π3,5π3\frac{2\pi}{3}, \frac{5\pi}{3}⁠). The least and greatest values are at the two end points.
    By Theorem 5, ff (continuous on a closed interval) attains a greatest and a least value; by Theorem 6 they are at critical points inside or at end points. We have listed all six candidates, so the largest and smallest of these six are the answers.
    Working Rule, Section 6.4.1. Lesson: absolute extrema

  5. Why the local peaks lose (optional understanding)

    f′′(x)=−4sin⁡2xf''(x) = -4\sin 2x⁠. At x=π3x = \frac{\pi}{3} and 4π3\frac{4\pi}{3}⁠, sin⁡2x=32\sin 2x = \frac{\sqrt3}{2}⁠, so f′′=−23<0f'' = -2\sqrt3 < 0⁠: local maxima. At x=2π3x = \frac{2\pi}{3} and 5π3\frac{5\pi}{3}⁠, f′′=23>0f'' = 2\sqrt3 > 0⁠: local minima. But a local maximum only beats its neighbours. The higher local maximum value, 4π3+32\frac{4\pi}{3} + \frac{\sqrt3}{2}⁠, is still below f(2π)=2πf(2\pi) = 2\pi⁠, as Step 4 showed.

    This explains the shape: ff is a rising line y=xy = x plus a bounded wobble. Over a long enough stretch the line wins, so the far end is the highest point and the near end the lowest.
    Second derivative test, Theorem 4, Section 6.4. Lesson: derivative tests
Result

On [0,2π][0, 2\pi]⁠, the maximum value is 2π2\pi, attained at x=2πx = 2\pi⁠, and the minimum value is 00, attained at x=0x = 0⁠. (Textbook answer: maximum at x=2πx = 2\pi⁠, maximum value =2π= 2\pi⁠; minimum at x=0x = 0⁠, minimum value =0= 0⁠.)

Check it with numbers

f′(π3)=1+2cos⁡2π3f'\left(\frac{\pi}{3}\right) = 1 + 2\cos\frac{2\pi}{3} =1+2(−12)=0= 1 + 2\left(-\frac{1}{2}\right) = 0⁠, as it should be. Near the right end, f(6.2)=6.2+sin⁡12.4f(6.2) = 6.2 + \sin 12.4⁠, about 6.2−0.1656=6.03446.2 - 0.1656 = 6.0344⁠, below 2π≈6.28322\pi \approx 6.2832⁠. A computer search over 200 001 equally spaced points of [0,2π][0, 2\pi] finds the greatest value 6.2832 at x=2πx = 2\pi and the least value 0 at x=0x = 0⁠.

Common mistakes
  1. Finding only two critical points. 2x2x runs over [0,4π][0, 4\pi]⁠, so cos⁡2x=−12\cos 2x = -\frac{1}{2} has four solutions in the interval.
  2. Forgetting the chain rule. ddxsin⁡2x=2cos⁡2x\frac{d}{dx}\sin 2x = 2\cos 2x⁠, not cos⁡2x\cos 2x⁠; the wrong derivative gives cos⁡2x=−1\cos 2x = -1 and wrong points.
  3. Reporting a local maximum as the maximum. 4π3+32≈5.05\frac{4\pi}{3} + \frac{\sqrt3}{2} \approx 5.05 is a local maximum value, but f(2π)=2πf(2\pi) = 2\pi is larger.
  4. Ignoring the end points. Here both answers are at the end points.
What to take away
  1. On a closed interval: critical points inside, plus both ends, then compare. Nothing else can be the answer.
  2. With sin⁡kx\sin kx or cos⁡kx\cos kx⁠, the angle runs over kk times the interval: count the solutions carefully.
  3. Compare with Exercise 6.3, Question 8 (sin⁡2x\sin 2x on [0,2π][0, 2\pi]⁠) and Question 5.

Every fact used, and where it comes from

FactUsed inSource
Chain rule; derivatives of sin⁡\sin and cos⁡\cosSteps 1, 5Chapter 5 (Basics: the chain rule)
cos⁡2π3=cos⁡4π3=−12\cos\frac{2\pi}{3} = \cos\frac{4\pi}{3} = -\frac12⁠; sin⁡2π3=32\sin\frac{2\pi}{3} = \frac{\sqrt3}{2}⁠; period 2π2\piSteps 2, 3Trigonometric values (Basics: signs of sin and cos)
A continuous function on [a,b][a, b] has absolute maximum and minimum valuesStep 4Theorem 5, Section 6.4.1
At an interior absolute extremum, f′=0f' = 0Step 4Theorem 6, Section 6.4.1
Working rule for absolute extremaSteps 3, 4Working Rule, Section 6.4.1 (Lesson: absolute extrema)
Second derivative testStep 5Theorem 4, Section 6.4 (Lesson: derivative tests)
AnswerResultAnswers to Exercise 6.3, Question 12
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