Chapter 6 · Application of Derivatives

Example 20: Local extrema of $3x^4+4x^3-12x^2+12$

ExamplesSecond derivative test · Section 6.4

The question

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

f(x) = 3x^4 + 4x^3 - 12x^2 + 12.

Exam answer

\begin{aligned} f'(x) &= 12x^3 + 12x^2 - 24x \\ &= 12x(x - 1)(x + 2) \end{aligned}

$f'(x) = 0 \Rightarrow x = -2, 0, 1$ ⁠.

f''(x) = 36x^2 + 24x - 24

\begin{aligned} f''(-2) &= 72 > 0 \\ f''(0) &= -24 < 0 \\ f''(1) &= 36 > 0 \end{aligned}

So $x = 0$ is a point of local maxima, and $x = -2$ and $x = 1$ are points of local minima.

\begin{aligned} f(0) &= 12 \\ f(-2) &= 48 - 32 - 48 + 12 = -20 \\ f(1) &= 3 + 4 - 12 + 12 = 7 \end{aligned}

Hence the local maximum value is $f(0) = 12$ ⁠, and the local minimum values are $f(-2) = -20$ and $f(1) = 7$ ⁠.

The book's last sentence prints $x = -1$ where it means the point of local minima $x = 1$ ⁠; this is a misprint.

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

The one idea this problem is built on

At a critical point where $f'(c) = 0$ ⁠, the second derivative tells us which way the graph bends. If $f''(c) < 0$ ⁠, the graph bends downwards like a cap, so $c$ is the top of a hill (local maximum); if $f''(c) > 0$ ⁠, it bends upwards like a cup, so $c$ is the bottom of a valley (local minimum). This second derivative test needs only one number at each critical point, instead of signs on both sides.

Given, to find, and the plan

Given: $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$ for all real $x$ ⁠.
To find: The local maximum and local minimum values of $f$ (and the points where they occur).
Plan: Find $f'$ ⁠, factorise it fully and get the critical points; find $f''$ and evaluate it at each critical point; apply Theorem 4; compute the values of $f$ ⁠; cross-check with a sign chart of $f'$ (Theorem 3).

Basics you need

1Second Derivative Test (Theorem 4, Section 6.4)

Let $f$ be twice differentiable at $c$ ⁠. (i) If $f'(c) = 0$ and $f''(c) < 0$ ⁠, then $c$ is a point of local maxima and $f(c)$ the local maximum value. (ii) If $f'(c) = 0$ and $f''(c) > 0$ ⁠, then $c$ is a point of local minima. (iii) If $f'(c) = 0$ and $f''(c) = 0$ ⁠, the test fails. Lesson: first and second derivative tests

2Critical points and Theorem 2

For a function differentiable everywhere, the only candidates for local extrema are the zeros of $f'$ ⁠. Lesson: maxima and minima

3Factorising

Take out the common factor $12x$ first, then factorise the remaining quadratic: $x^2 + x - 2$ needs two numbers with product $-2$ and sum $1$ ⁠, namely $2$ and $-1$ ⁠. Basics: factorising and the discriminant

4Sign of a product

Used for the cross-check: the sign of $12x(x-1)(x+2)$ on each interval is the product of the factor signs. Basics: sign of a product

Solution

First derivative and critical points

By the power, constant-multiple and sum rules,
$\begin{aligned} f'(x) &= 12x^3 + 12x^2 - 24x \\ &= 12x(x^2 + x - 2) \\ &= 12x(x-1)(x+2). \end{aligned}$
A product is zero only when a factor is zero, so
$f'(x) = 0 \iff x = 0, \ x = 1 \ \text{ or } \ x = -2.$
$f$ is a polynomial, differentiable everywhere, so these three are all the critical points.

Taking out $12x$ (every term has at least one $x$ and a factor 12) turns a cubic into "linear times quadratic", and the quadratic factorises by Basic 3. Check: $(x-1)(x+2)$ $= x^2 + 2x - x - 2$ $= x^2 + x - 2$ ⁠.

Basics: factorising
Second derivative

Differentiate $f'(x) = 12x^3 + 12x^2 - 24x$ once more:
$\begin{aligned} f''(x) &= 36x^2 + 24x - 24 \\ &= 12(3x^2 + 2x - 2). \end{aligned}$
Differentiate the expanded form of $f'$ ⁠, not the factorised one: it is simpler and avoids the product rule. $f$ is a polynomial, so it is twice differentiable everywhere and Theorem 4 may be used at every critical point.
Apply the Second Derivative Test at each critical point
$\begin{aligned} f''(0) &= 12(0 + 0 - 2) = -24 < 0, \\ f''(1) &= 12(3 + 2 - 2) = 36 > 0, \\ f''(-2) &= 12(12 - 4 - 2) = 72 > 0. \end{aligned}$
By Theorem 4:
- $x = 0$ is a point of local maxima ( $f'(0) = 0$ ⁠, $f''(0) < 0$ ⁠);
- $x = 1$ and $x = -2$ are points of local minima ( $f' = 0$ ⁠, $f'' > 0$ ⁠).
None of the three second derivatives is 0, so the test gives a definite answer at every critical point; there is no need to fall back on the first derivative test (contrast Example 21, where $f''(1) = 0$ ⁠).

Theorem 4(i), (ii), Section 6.4. Lesson: first and second derivative tests
The local maximum and minimum values
$\begin{aligned} f(0) &= 0 + 0 - 0 + 12 = 12, \\ f(1) &= 3 + 4 - 12 + 12 = 7, \\ f(-2) &= 48 - 32 - 48 + 12 = -20. \end{aligned}$
(For $f(-2)$ ⁠: $(-2)^4 = 16$ ⁠, $(-2)^3 = -8$ ⁠, $(-2)^2 = 4$ ⁠, so $3\cdot16 + 4\cdot(-8) - 12\cdot4 + 12$ ⁠.)

So the local maximum value is $f(0) = 12$ , and the local minimum values are $f(1) = 7$ and $f(-2) = -20$ .

The question asks for values, so the last step is to put each point back into $f$ (not into $f'$ or $f''$ ⁠).

A misprint to watch for: the book's last sentence says the local minimum values "at $x = -1$ and $-2$ are $f(1) = 7$ and $f(-2) = -20$ ". The " $-1$ " should be $1$ ⁠: the critical point is $x = 1$ (indeed $f'(-1) = 12(-1)(-2)(1)$ $= 24 \ne 0$ ⁠, so $-1$ is not a critical point, and $f(-1) = -1$ ⁠). The values 7 and $-20$ are correct.
Cross-check with the first derivative test

Sign chart of $f'(x) = 12x(x-1)(x+2)$ (not to scale; the positive factor 12 is left out). $f'$ goes $-$ ⁠, $+$ ⁠, $-$ ⁠, $+$ ⁠.

At $-2$ ⁠, $f'$ changes from $-$ to $+$ ⁠: local minima. At 0, from $+$ to $-$ ⁠: local maxima. At 1, from $-$ to $+$ ⁠: local minima. This is exactly what Theorem 4 gave.

The graph of $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$ ⁠: two valleys, at $(-2, -20)$ and $(1, 7)$ ⁠, and a hill top at $(0, 12)$ ⁠. Green where rising, rose where falling. Vertical scale much smaller than horizontal.

The two tests are independent routes to the same answer: the sign chart looks at slopes beside each point, the second derivative at the bending at the point.

Basics: sign of a product

Result

$f'(x) = 12x(x-1)(x+2)$ ⁠, $f''(x) = 12(3x^2 + 2x - 2)$ ⁠. Local maximum value $f(0) = 12$ (at $x = 0$ ⁠, since $f''(0) = -24 < 0$ ⁠); local minimum values $f(1) = 7$ and $f(-2) = -20$ (at $x = 1$ and $x = -2$ ⁠, since $f''(1) = 36 > 0$ and $f''(-2) = 72 > 0$ ⁠). This agrees with the book's solution (whose " $x = -1$ " in the last sentence is a misprint for $x = 1$ ⁠).

Check it with numbers

Near	Left	At the point	Right
$x = -2$	$f(-2.1) = -19.6197$	$-20$	$f(-1.9) = -19.6597$
$x = 0$	$f(-0.1) = 11.8763$	12	$f(0.1) = 11.8843$
$x = 1$	$f(0.9) = 7.1643$	7	$f(1.1) = 7.1963$

At $-2$ and 1 the middle value is the smallest of the three; at 0 it is the largest. Note also $f(2) = 44 > 12$ ⁠: the local maximum value is not the largest value of $f$ ⁠.

Common mistakes

Losing the root $x = 0$ ⁠. Dividing $12x^3 + 12x^2 - 24x = 0$ by $x$ throws away $x = 0$ ⁠; factor out $12x$ instead.
Reversing the test. $f''(c) < 0$ means a local maximum (cap shape), $f''(c) > 0$ a local minimum (cup shape).
Reporting $f''(c)$ as the value. The local extreme values are $f(c)$ ⁠: 12, 7 and $-20$ ⁠, not $-24$ ⁠, 36 and 72.
Copying the book's " $x = -1$ ". The minimum value 7 occurs at $x = 1$ ⁠.

What to take away

Second derivative test: solve $f'(x) = 0$ ⁠, then the sign of $f''$ at each root decides (negative: maximum, positive: minimum, zero: no conclusion).
A quartic with positive leading coefficient can have two valleys and one hill between them.
"Local" values need not be the overall largest or smallest values.

Every fact used, and where it comes from

Fact	Used in	Source
Derivatives of polynomials; $x^2 + x - 2 = (x-1)(x+2)$	Steps 1, 2	Derivative rules; Basics: factorising
Local extrema only at critical points	Step 1	Theorem 2, Section 6.4 (Lesson)
Second Derivative Test	Step 3	Theorem 4, Section 6.4 (Lesson)
First Derivative Test	Step 5	Theorem 3, Section 6.4
Values of $f$ ⁠, $f'$ ⁠, $f''$ ⁠; decimals in the check table	Steps 3–5; Check	Recomputed in the site's verification script

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Example 20: Local extrema of 3x4+4x3−12x2+123x^4+4x^3-12x^2+123x4+4x3−12x2+12

Basics you need

Solution

First derivative and critical points

Second derivative

Apply the Second Derivative Test at each critical point

The local maximum and minimum values

Cross-check with the first derivative test

Every fact used, and where it comes from

Example 20: Local extrema of $3x^4+4x^3-12x^2+12$