Chapter 6 · Application of Derivatives

Miscellaneous · Q11: Absolute extrema of $\cos^2 x+\sin x$ on $[0,\pi]$

Miscellaneous ExerciseAbsolute maxima and minima · Section 6.4.1 · Miscellaneous Exercise

The question

Find the absolute maximum and minimum values of the function $f$ given by

f(x) = \cos^2 x + \sin x,\quad x \in [0, \pi].

Exam answer

\begin{aligned} f'(x) &= -2\cos x\sin x + \cos x \\ &= \cos x\,(1 - 2\sin x) \end{aligned}

$f'(x) = 0 \Rightarrow \cos x = 0$ or $\sin x = \frac12$ ⁠. So in $[0, \pi]$ the critical points are $x = \frac{\pi}{6}$ ⁠, $\frac{\pi}{2}$ ⁠, $\frac{5\pi}{6}$ ⁠.

\begin{aligned} f(0) &= 1,\quad f(\pi) = 1 \\ f\left(\tfrac{\pi}{6}\right) &= \tfrac34 + \tfrac12 = \tfrac54 \\ f\left(\tfrac{\pi}{2}\right) &= 0 + 1 = 1 \\ f\left(\tfrac{5\pi}{6}\right) &= \tfrac34 + \tfrac12 = \tfrac54 \end{aligned}

Absolute maximum value $= \frac54$ ⁠, at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ ⁠.

Absolute minimum value $= 1$ ⁠, at $x = 0$ ⁠, $\dfrac{\pi}{2}$ and $\pi$ ⁠.

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 attains a greatest and a least value, and they can only occur at critical points inside the interval or at its two ends. So list those few points, compute $f$ at each, and pick the largest and smallest. Nothing else needs testing.

Given, to find, and the plan

Given: $f(x) = \cos^2 x + \sin x$ on the closed interval $[0, \pi]$ ⁠.
To find: The absolute maximum value and the absolute minimum value of $f$ on $[0, \pi]$ ⁠.
Plan: Follow the book's Working Rule (Section 6.4.1): (1) critical points in the interval; (2) the end points; (3) $f$ at all of them; (4) the largest and smallest of these values.

Basics you need

1Chain rule for $\cos^2 x$

$\cos^2 x = (\cos x)^2$ ⁠, so its derivative is $2\cos x\cdot(-\sin x) = -2\sin x\cos x$ ⁠. Basics: the chain rule

2Values of sin and cos on $[0, \pi]$

$\sin x \ge 0$ on $[0, \pi]$ ⁠; $\cos x$ is positive on $\left[0, \frac{\pi}{2}\right)$ ⁠, zero at $\frac{\pi}{2}$ ⁠, negative on $\left(\frac{\pi}{2}, \pi\right]$ ⁠. $\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}$ and $\frac{5\pi}{6}$ only. Basics: signs of sin and cos

3Absolute extrema on a closed interval (Section 6.4.1)

Theorem 5: a continuous $f$ on $[a, b]$ attains an absolute maximum and an absolute minimum. Theorem 6: at an interior point where this happens, $f' = 0$ ⁠. Hence the Working Rule: compare $f$ at the critical points and the ends. Lesson: absolute extrema on a closed interval

Solution

Differentiate and factor
$\begin{aligned} f'(x) &= 2\cos x\,(-\sin x) + \cos x \\ &= -2\sin x\cos x + \cos x \\ &= \cos x\,(1-2\sin x). \end{aligned}$
The first term uses the chain rule on $(\cos x)^2$ ⁠: outer function $u^2$ with derivative $2u$ ⁠, inner function $\cos x$ with derivative $-\sin x$ ⁠. Both terms contain $\cos x$ ⁠, so we take it out; a product is much easier to set equal to zero than a sum.

$\frac{d}{dx}\sin x = \cos x$ ⁠, $\frac{d}{dx}\cos x = -\sin x$ ⁠; chain rule Theorem 4, Section 5.3.1. Basics: the chain rule
Working Rule, Step 1: critical points in $[0, \pi]$

$f$ is differentiable everywhere, so the critical points are where $f'(x) = 0$ ⁠:
- $\cos x = 0$ ⁠: in $[0, \pi]$ only at $x = \frac{\pi}{2}$ ⁠;
- $1 - 2\sin x = 0$ ⁠, that is $\sin x = \frac{1}{2}$ ⁠: in $[0, \pi]$ at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ ⁠.
So the critical points are $\frac{\pi}{6}$ ⁠, $\frac{\pi}{2}$ ⁠, $\frac{5\pi}{6}$ ⁠.

$\sin x = \frac{1}{2}$ has two solutions in $[0, \pi]$ ⁠, because sine rises from 0 to 1 on $\left[0, \frac{\pi}{2}\right]$ and falls back on $\left[\frac{\pi}{2}, \pi\right]$ ⁠, passing $\frac{1}{2}$ once each way. $\frac{5\pi}{6} = \pi - \frac{\pi}{6}$ is the second one, and it is easy to miss.

$\sin(\pi - x) = \sin x$ ⁠, so $\sin\frac{5\pi}{6} = \sin\frac{\pi}{6} = \frac{1}{2}$ ⁠. Basics: signs of sin and cos

Working Rule, Steps 2 and 3: values at all candidates

Add the end points 0 and $\pi$ and compute $f$ at all five points:

$x$	$\cos^2 x$	$\sin x$	$f(x)$
$0$	$1$	$0$	$1$
$\frac{\pi}{6}$	$\frac{3}{4}$	$\frac{1}{2}$	$\frac{5}{4}$
$\frac{\pi}{2}$	$0$	$1$	$1$
$\frac{5\pi}{6}$	$\frac{3}{4}$	$\frac{1}{2}$	$\frac{5}{4}$
$\pi$	$1$	$0$	$1$

\cos\frac{\pi}{6} = \frac{\sqrt3}{2}

and

\cos\frac{5\pi}{6} = -\frac{\sqrt3}{2}

⁠; squaring removes the sign, giving

\frac{3}{4}

both times. At

\pi

⁠,

\cos\pi = -1

and its square is 1.

Working Rule, Steps 2 and 3, Section 6.4.1: take the end points and calculate

f

at every listed point. Lesson: absolute extrema

Working Rule, Step 4: pick the largest and smallest

The largest value in the table is $\frac{5}{4}$ ⁠, the smallest is 1. Therefore
$\text{absolute maximum value} = \frac{5}{4}\ \text{ at } x = \frac{\pi}{6},\ \frac{5\pi}{6};$ $\text{absolute minimum value} = 1\ \text{ at } x = 0,\ \frac{\pi}{2},\ \pi.$

$y = \cos^2 x + \sin x$ on $[0, \pi]$ (horizontal and vertical scales differ). The graph never goes above $\frac{5}{4}$ or below 1; each value is reached at the points marked.

Why is the table enough? $f$ is continuous on the closed interval $[0, \pi]$ ⁠, so by Theorem 5 it has an absolute maximum and minimum there. If one of them occurred at an interior point, Theorem 6 says $f' = 0$ there, so that point is in our list; otherwise it occurs at an end, also in our list. Either way the extreme values are among the five values computed.

Theorems 5 and 6 and the Working Rule, Section 6.4.1. An absolute extreme value may be attained at several points. Lesson: absolute extrema

Result

On $[0, \pi]$ ⁠, the absolute maximum value of $f$ is $\frac{5}{4}$ , attained at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ ⁠; the absolute minimum value is 1, attained at $x = 0$ ⁠, $\frac{\pi}{2}$ and $\pi$ ⁠. (The answer key in this reprint has no entry for this question.)

Check it with numbers

A second route: write $s = \sin x$ ⁠, so $\cos^2 x = 1 - s^2$ and $f = 1 + s - s^2$ ⁠. On $[0, \pi]$ ⁠, $s$ runs over $[0, 1]$ ⁠. Completing the square, $1 + s - s^2 = \frac{5}{4} - \left(s - \frac{1}{2}\right)^2$ ⁠, which is largest ( $\frac54$ ⁠) when $s = \frac12$ ⁠; and $1 + s - s^2 = 1 + s(1-s) \ge 1$ ⁠, with equality only at $s = 0$ or $s = 1$ ✓. Sample values: $f(0.3) \approx 0.9127 + 0.2955$ $= 1.2082$ ⁠, $f(1) \approx 1.1334$ ⁠, $f(2.5) \approx 1.2403$ ⁠, all between 1 and 1.25 ✓.

Common mistakes

Missing $x = \frac{5\pi}{6}$ ⁠. $\sin x = \frac12$ has two solutions in $[0, \pi]$ ⁠.
Dividing by $\cos x$ in $f'(x) = 0$ instead of factoring. That throws away $x = \frac{\pi}{2}$ ⁠.
Forgetting the end points 0 and $\pi$ ⁠. Here they tie for the minimum, so leaving them out would hide two of the three points where it is attained.
Using the second derivative test to classify points and stopping there. Local maxima and minima are not the question; the comparison table is.

What to take away

Closed interval and continuous function: list critical points inside, add the ends, tabulate, compare. No sign charts needed.
Factor $f'$ rather than divide, so that no root is lost.
Substituting $s = \sin x$ turns this function into a quadratic, which gives an independent check. The same Working Rule handles Exercise 6.3, Question 8 and Question 12, also trigonometric functions on a closed interval.

Every fact used, and where it comes from

Fact	Used in	Source
Derivatives of $\sin x$ ⁠, $\cos x$ ⁠; chain rule	Step 1	Chapter 5; chain rule Theorem 4, Section 5.3.1 (Basics: the chain rule)
Solutions of $\cos x = 0$ and $\sin x = \frac12$ in $[0, \pi]$ ⁠; values at $\frac{\pi}{6}$ ⁠, $\frac{5\pi}{6}$	Steps 2, 3	Trigonometry (Basics: signs of sin and cos)
Definition of critical point	Step 2	Note after Theorem 2, Section 6.4 (Lesson: maxima and minima)
Existence of absolute extrema; $f' = 0$ at an interior one	Step 4	Theorems 5 and 6, Section 6.4.1
Working Rule for absolute maximum and minimum	Steps 2 to 4	Section 6.4.1 (Lesson: absolute extrema)
All values	Step 3, Check	Recomputed exactly (sympy) and by a search over 200 001 points of $[0, \pi]$ in the site's verification script

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Miscellaneous · Q11: Absolute extrema of cos⁡2x+sin⁡x\cos^2 x+\sin xcos2x+sinx on [0,π][0,\pi][0,π]

Basics you need

Solution

Differentiate and factor

Working Rule, Step 1: critical points in [0,π][0, \pi][0,π]

Working Rule, Steps 2 and 3: values at all candidates

Working Rule, Step 4: pick the largest and smallest

Every fact used, and where it comes from

Miscellaneous · Q11: Absolute extrema of $\cos^2 x+\sin x$ on $[0,\pi]$

Working Rule, Step 1: critical points in $[0, \pi]$