Chapter 6 · Application of Derivatives

Exercise 6.3 · Q10: Maximum of $2x^3-24x+107$ on $[1,3]$ and $[-3,-1]$

Exercise 6.3Absolute extrema on a closed interval · Section 6.4.1

The question

10. Find the maximum value of $2x^3-24x+107$ in the interval $[1,3]$ ⁠. Find the maximum value of the same function in $[-3,-1]$ ⁠.

Exam answer

Let $f(x) = 2x^3 - 24x + 107$ ⁠. Then

\begin{aligned} f'(x) &= 6x^2 - 24 \\ &= 6(x - 2)(x + 2) \end{aligned}

$f'(x) = 0$ gives $x = 2$ or $x = -2$ ⁠.

On $[1, 3]$ the only critical point is $x = 2$ ⁠.

\begin{aligned} f(1) &= 85 \\ f(2) &= 75 \\ f(3) &= 89 \end{aligned}

Maximum value on $[1, 3]$ is $f(3) = 89$ ⁠.

On $[-3, -1]$ the only critical point is $x = -2$ ⁠.

\begin{aligned} f(-3) &= 125 \\ f(-2) &= 139 \\ f(-1) &= 129 \end{aligned}

Maximum value on $[-3, -1]$ is $f(-2) = 139$ ⁠.

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

The one idea this problem is built on

The same function can have its maximum in very different places on different intervals. The derivative $6(x-2)(x+2)$ vanishes at $x=2$ and $x=-2$ ⁠, one inside each interval. On $[1,3]$ the point $x=2$ is a valley, so the maximum is at an end point; on $[-3,-1]$ the point $x=-2$ is a hilltop, so the maximum is there. The Working Rule handles both without needing to know which is which in advance.

Given, to find, and the plan

Given: $f(x)=2x^3-24x+107$ ⁠; the closed intervals $[1,3]$ and $[-3,-1]$ ⁠.
To find: The maximum value of $f$ on each interval.
Plan: (1) Find the critical points of $f$ ⁠. (2) For each interval, keep the critical points inside it, add the end points, evaluate $f$ and pick the largest value.

Basics you need

1Theorems 5, 6 and the Working Rule

A continuous function on a closed interval attains an absolute maximum value; if it occurs at an interior point, $f'=0$ there. So the maximum is the largest of the values at the interior critical points and at the end points. Lesson: absolute extrema on a closed interval

2Difference of squares

$x^2-4=(x-2)(x+2)$ ⁠. Basics: factorising

3Cubes of negative numbers

$(-2)^3=-8$ ⁠, $(-3)^3=-27$ ⁠, $(-1)^3=-1$ ⁠: an odd power keeps the minus sign. Basics: signs

Solution

Critical points
$f'(x)=6x^2-24=6(x^2-4)=6(x-2)(x+2).$
So $f'(x)=0$ at $x=2$ and $x=-2$ ⁠. $f$ is a polynomial, differentiable everywhere, so there are no other critical points.

Power rule: $2x^3\to6x^2$ ⁠, $-24x\to-24$ ⁠, $107\to0$ ⁠. Then take out $6$ and use the difference of squares.

Working Rule, step 1. Lesson: absolute extrema
On $[1,3]$

The critical point $x=2$ is inside $[1,3]$ ⁠; $x=-2$ is not. Candidates: $1,2,3$ ⁠.
$\begin{aligned} f(1)&=2-24+107=85,\\ f(2)&=16-48+107=75,\\ f(3)&=54-72+107=89. \end{aligned}$
The largest is $89$ ⁠: the maximum value on $[1,3]$ is $89$ ⁠, at $x=3$ .

The critical point gives the smallest value here ( $f''(2)=24\gt0$ ⁠, a local minimum), so the maximum has to be at an end point; comparing $85$ and $89$ picks the right one.

Working Rule, steps 2–4.
On $[-3,-1]$

The critical point $x=-2$ is inside $[-3,-1]$ ⁠; $x=2$ is not. Candidates: $-3,-2,-1$ ⁠.
$\begin{aligned} f(-3)&=-54+72+107=125,\\ f(-2)&=-16+48+107=139,\\ f(-1)&=-2+24+107=129. \end{aligned}$
The largest is $139$ ⁠: the maximum value on $[-3,-1]$ is $139$ ⁠, at $x=-2$ .

Signs: $2(-3)^3=2(-27)=-54$ and $-24(-3)=+72$ ⁠. Here the critical point is a hilltop ( $f''(-2)=-24\lt0$ ⁠) and it beats both end points.

Working Rule, steps 2–4. Basics: signs

f(x)=2x^3-24x+107

⁠, with the two intervals shaded and the graph drawn solid over them. On

[-3,-1]

the highest point is the hilltop

(-2,139)

⁠; on

[1,3]

the critical point is a valley

(2,75)

and the highest point is the end

(3,89)

⁠.

Result

On $[1,3]$ ⁠: maximum value 89, at $x=3$ ⁠. On $[-3,-1]$ ⁠: maximum value 139, at $x=-2$ ⁠.

(Textbook answer: Maximum at x = 3, maximum value 89; maximum at x = −2, maximum value = 139.)

Check it with numbers

On $[1,3]$ ⁠: $f(2.9)=48.778-69.6+107$ $=86.178\lt89$ and $f(1.5)=6.75-36+107=77.75\lt89$ ⁠. ✓

On $[-3,-1]$ ⁠: $f(-2.1)=-18.522+50.4+107$ $=138.878$ and $f(-1.9)=-13.718+45.6+107$ $=138.882$ ⁠, both just below $139$ ⁠. ✓

Common mistakes

Taking the critical point as the maximum on $[1,3]$ ⁠. $f(2)=75$ is the minimum there; the maximum is at the end point $x=3$ ⁠.
Using $x=-2$ for the interval $[1,3]$ (or $x=2$ for $[-3,-1]$ ⁠). Only critical points inside the interval count.
Sign slips with negative cubes. $2(-2)^3=-16$ ⁠, and $-24(-2)=+48$ ⁠.

What to take away

The absolute maximum depends on the interval; always redo the comparison for each interval.
A critical point inside the interval can be a valley; the Working Rule's comparison of values avoids having to classify it.

Every fact used, and where it comes from

Fact	Used in	Source
Absolute extremes exist on a closed interval; $f'=0$ at an interior one	Steps 2, 3	Theorems 5 and 6, Section 6.4.1 (Lesson: absolute extrema)
Working Rule	Steps 1–3	Working Rule, Section 6.4.1; compare Example 27
Power rule; difference of squares	Step 1	Chapter 5; algebra (Basics)
Second derivative test (used only as a remark)	Steps 2, 3	Theorem 4, Section 6.4 (Lesson: tests)

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Exercise 6.3 · Q10: Maximum of 2x3−24x+1072x^3-24x+1072x3−24x+107 on [1,3][1,3][1,3] and [−3,−1][-3,-1][−3,−1]

Basics you need

Solution

Critical points

On [1,3][1,3][1,3]

On [−3,−1][-3,-1][−3,−1]

Every fact used, and where it comes from

Exercise 6.3 · Q10: Maximum of $2x^3-24x+107$ on $[1,3]$ and $[-3,-1]$

On $[1,3]$

On $[-3,-1]$