Chapter 6 · Application of Derivatives

Exercise 6.3 · Q7: Maximum and minimum of $3x^4-8x^3+12x^2-48x+25$ on $[0,3]$

Exercise 6.3Absolute extrema on a closed interval · Section 6.4.1

The question

7. Find both the maximum value and the minimum value of

3x^4-8x^3+12x^2-48x+25

on the interval $[0,3]$ ⁠.

Exam answer

Let

\begin{aligned} f(x) &= 3x^4 - 8x^3 + 12x^2 \\ &\quad - 48x + 25 \end{aligned}

Then

\begin{aligned} f'(x) &= 12x^3 - 24x^2 + 24x - 48 \\ &= 12[x^2(x - 2) + 2(x - 2)] \\ &= 12(x - 2)(x^2 + 2) \end{aligned}

As $x^2 + 2 > 0$ ⁠, $f'(x) = 0$ gives only $x = 2$ ⁠, which lies in $[0, 3]$ ⁠.

\begin{aligned} f(0) &= 25 \\ f(2) &= -39 \\ f(3) &= 16 \end{aligned}

Maximum value $f(0) = 25$ ⁠; minimum value $f(2) = -39$ ⁠.

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

The one idea this problem is built on

On a closed interval, the greatest and least values of a polynomial are found among the end points and the points inside where the derivative is zero. Here the derivative factorises as $12(x-2)(x^2+2)$ ⁠, and $x^2+2$ is never zero, so there is just one interior candidate, $x=2$ ⁠. Three values to compare, and we are done.

Given, to find, and the plan

Given: $f(x)=3x^4-8x^3+12x^2-48x+25$ on the closed interval $[0,3]$ ⁠.
To find: The maximum value and the minimum value of $f$ on $[0,3]$ (absolute, Section 6.4.1).
Plan: Working Rule: (1) critical points in $[0,3]$ by factorising $f'$ ⁠; (2) end points $0$ ⁠, $3$ ⁠; (3) evaluate $f$ at each; (4) compare.

Basics you need

1Theorems 5, 6 and the Working Rule

A continuous function on $[a,b]$ has an absolute maximum and minimum value (Theorem 5); at an interior one, $f'=0$ (Theorem 6). So compare $f$ at the critical points inside and at both ends. Lesson: absolute extrema on a closed interval

2Factorising by grouping

$x^3-2x^2+2x-4$ $=x^2(x-2)+2(x-2)$ $=(x-2)(x^2+2)$ ⁠. Basics: factorising and the discriminant

3A sum of a square and a positive number is positive

$x^2+2\ge2\gt0$ for every real $x$ ⁠, so it contributes no zeros.

Solution

Differentiate and factorise
$\begin{aligned} f'(x)&=12x^3-24x^2+24x-48\\ &=12(x^3-2x^2+2x-4)\\ &=12\left[x^2(x-2)+2(x-2)\right]\\ &=12(x-2)(x^2+2). \end{aligned}$
Power rule term by term, then take out the common factor $12$ ⁠. The cubic factorises by grouping the first two terms and the last two terms; both groups contain $(x-2)$ ⁠. Expanding back: $(x-2)(x^2+2)=x^3+2x-2x^2-4$ ⁠. ✓

Factorising by grouping. Basics: factorising
Critical points in $[0,3]$

$f'(x)=0$ needs $x-2=0$ or $x^2+2=0$ ⁠. The second is impossible for real $x$ ( $x^2+2\ge2$ ⁠). So the only critical point is $x=2$ ⁠, and it lies in $[0,3]$ ⁠.

$f$ is a polynomial, differentiable everywhere, so there are no points of non-differentiability to add.

Working Rule, step 1. Lesson: absolute extrema
Evaluate at the critical point and the end points
$\begin{aligned} f(0)&=25,\\ f(2)&=48-64+48-96+25=-39,\\ f(3)&=243-216+108-144+25=16. \end{aligned}$
At $x=2$ ⁠: $3\cdot16=48$ ⁠, $8\cdot8=64$ ⁠, $12\cdot4=48$ ⁠, $48\cdot2=96$ ⁠. At $x=3$ ⁠: $3\cdot81=243$ ⁠, $8\cdot27=216$ ⁠, $12\cdot9=108$ ⁠, $48\cdot3=144$ ⁠. Writing each term separately keeps the arithmetic checkable.

Working Rule, steps 2 and 3.
Compare

$x$ $0$ (end) $2$ (critical) $3$ (end)

$f(x)$ $25$ $-39$ $16$

The largest is $25$ and the smallest is $-39$ ⁠. So the maximum value is $25$ ⁠, at $x=0$ ⁠, and the minimum value is $-39$ ⁠, at $x=2$ .

The sign of $f'=12(x-2)(x^2+2)$ agrees: negative on $(0,2)$ ( $f$ falls from $25$ to $-39$ ⁠), positive on $(2,3)$ ( $f$ rises to $16$ ⁠). Since $16\lt25$ ⁠, the climb back does not overtake the starting value.

Working Rule, step 4; Theorem 1, Section 6.3.

$x$	$0$ (end)	$2$ (critical)	$3$ (end)
$f(x)$	$25$	$-39$	$16$

f(x)=3x^4-8x^3+12x^2-48x+25

[0,3]

⁠: it starts at its highest value

25

⁠, falls to its lowest value

-39

x=2

⁠, then climbs only to

16

x=3

⁠.

Result

On $[0,3]$ ⁠: maximum value $25$ (at $x=0$ ⁠) and minimum value $-39$ (at $x=2$ ⁠).

(Textbook answer: Minima at x = 2, minimum value = −39, Maxima at x = 0, maximum value = 25.)

Check it with numbers

$f(1)=3-8+12-48+25=-16$ ⁠, between the values at $0$ and $2$ ⁠. ✓

$f(1.9)=-38.6557$ and $f(2.1)=-38.6237$ ⁠, both just above $-39$ ⁠. ✓

$f(2.9)=3.7923$ ⁠, $f(3)=16$ ⁠: still climbing at the right end, as $f'(3)=12\cdot1\cdot11=132\gt0$ says. ✓

Common mistakes

Missing the end point $x=0$ ⁠. The maximum is there; $f'(0)=-48\ne0$ ⁠, so it never shows up as a critical point.
Trying to solve $x^2+2=0$ ⁠. It has no real roots; say so and move on.
Arithmetic slips in $f(2)$ and $f(3)$ ⁠. Write out each term separately.

What to take away

For absolute extremes on a closed interval, end points are as important as critical points.
Factorise the derivative completely; factors like $x^2+2$ contribute no critical points.
A sign check of $f'$ between the candidates confirms the comparison table.

Every fact used, and where it comes from

Fact	Used in	Source
Absolute extremes exist on $[a,b]$ ⁠; $f'=0$ at an interior one	Steps 2–4	Theorems 5 and 6, Section 6.4.1 (Lesson: absolute extrema)
Working Rule	Steps 2–4	Working Rule, Section 6.4.1; compare Example 27
Power rule	Step 1	Chapter 5 (Continuity and Differentiability)
Factorising by grouping; $x^2+2\gt0$	Steps 1, 2	Algebra (Basics)
Sign of $f'$ gives rising/falling	Step 4	Theorem 1, Section 6.3 (Lesson: increasing and decreasing)

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Exercise 6.3 · Q7: Maximum and minimum of 3x4−8x3+12x2−48x+253x^4-8x^3+12x^2-48x+253x4−8x3+12x2−48x+25 on [0,3][0,3][0,3]

Basics you need

Solution

Differentiate and factorise

Critical points in [0,3][0,3][0,3]

Evaluate at the critical point and the end points

Compare

Every fact used, and where it comes from

Exercise 6.3 · Q7: Maximum and minimum of $3x^4-8x^3+12x^2-48x+25$ on $[0,3]$

Critical points in $[0,3]$