Chapter 6 · Application of Derivatives

Example 11: Where $4x^3-6x^2-72x+30$ increases and decreases

ExamplesIncreasing and decreasing functions · Section 6.3

The question

Find the intervals in which the function $f$ given by $f(x) = 4x^3 - 6x^2 - 72x + 30$ is (a) increasing (b) decreasing.

Exam answer

\begin{aligned} f'(x) &= 12x^2 - 12x - 72 \\ &= 12(x - 3)(x + 2) \end{aligned}

$f'(x) = 0 \Rightarrow x = -2, 3$ ⁠.

In $(-\infty, -2)$ ⁠, $f'(x) > 0$ ⁠.

In $(-2, 3)$ ⁠, $f'(x) < 0$ ⁠.

In $(3, \infty)$ ⁠, $f'(x) > 0$ ⁠.

(a) $f$ is increasing in $(-\infty, -2)$ and in $(3, \infty)$ ⁠.

(b) $f$ is decreasing in $(-2, 3)$ ⁠.

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

The one idea this problem is built on

The sign of the derivative tells us the direction of the graph: where $f'(x) > 0$ the graph climbs, where $f'(x) < 0$ it falls. So the whole question reduces to one task: find where $f'(x)$ is positive and where it is negative. Here $f'$ is a quadratic that factorises as $12(x-3)(x+2)$ ⁠; its two roots cut the number line into three intervals, and a product of two factors has a fixed sign on each of them.

Given, to find, and the plan

Given: $f(x) = 4x^3 - 6x^2 - 72x + 30$ for every real $x$ (a polynomial, so defined, continuous and differentiable on all of $\R$ ⁠).
To find: (a) the intervals on which $f$ is increasing; (b) the intervals on which it is decreasing.
Plan: Differentiate; factorise $f'(x)$ ⁠; find where $f'(x) = 0$ ⁠; make a sign chart of $f'$ on the intervals between those points; read off the answer with Theorem 1 of Section 6.3; check with values of $f$ ⁠.

Basics you need

1Increasing and strictly increasing (Definition 1)

On an interval I, $f$ is increasing if $x_1 < x_2$ in I gives $f(x_1) \le f(x_2)$ ⁠, and strictly increasing if it gives $f(x_1) < f(x_2)$ ⁠. Decreasing and strictly decreasing are the same with $\ge$ and $>$ ⁠. A strictly increasing function is in particular increasing, so proving "strictly" answers the question with room to spare. Lesson: increasing and decreasing functions

2The derivative test (Theorem 1, Section 6.3)

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ ⁠, then $f$ is increasing on $[a,b]$ when $f'(x) \ge 0$ on $(a,b)$ ⁠, and decreasing when $f'(x) \le 0$ there. The book proves it with the Mean Value Theorem: $f(x_2) - f(x_1)$ $= f'(c)(x_2 - x_1)$ for some $c$ strictly between $x_1$ and $x_2$ ⁠. If $f' > 0$ (strictly), that difference is strictly positive.

3Factorising a quadratic

To factorise $x^2 - x - 6$ ⁠, look for two numbers with product $-6$ and sum $-1$ ⁠: they are $-3$ and $2$ ⁠. So $x^2 - x - 6 = (x-3)(x+2)$ ⁠. Basics: factorising and the discriminant

4Sign of a product

A product is positive when its factors have an even number of minus signs, negative when the number is odd, and zero only when a factor is zero. A factor $x - a$ is negative left of $a$ and positive right of $a$ ⁠. Basics: sign of a product

5Interval notation

$(-\infty, -2)$ means all $x < -2$ ⁠; $[-2, 3]$ means $-2 \le x \le 3$ ⁠. A round bracket leaves the end out, a square bracket puts it in. Basics: interval notation

Solution

Differentiate

Use the power rule $\frac{d}{dx}x^n = nx^{n-1}$ on each term, the constant-multiple rule, and the sum rule (the derivative of a sum is the sum of the derivatives). The constant 30 has derivative 0.
$\begin{aligned} f'(x) &= 4\cdot 3x^2 - 6\cdot 2x - 72\cdot 1 + 0 \\ &= 12x^2 - 12x - 72. \end{aligned}$
The derivative measures the slope of the graph at each point. Positive slope means the graph is going up as we move right, which is exactly what "increasing" describes. So to locate the rising and falling stretches we need $f'$ ⁠, not $f$ ⁠.

A polynomial is differentiable (and therefore continuous) at every real number, so the conditions of Theorem 1 hold on every interval we shall use.
Factorise $f'(x)$ and find where it is zero

Every coefficient is a multiple of 12, so take 12 out, then factorise the quadratic (Basic 3):
$\begin{aligned} f'(x) &= 12(x^2 - x - 6) \\ &= 12(x-3)(x+2). \end{aligned}$
Check by expanding: $(x-3)(x+2)$ $= x^2 + 2x - 3x - 6$ $= x^2 - x - 6$ ⁠. A product is zero only when one of its factors is zero, so
$f'(x) = 0 \iff x = 3 \ \text{ or } \ x = -2.$
The derivative is a polynomial, so it is continuous: it cannot jump from positive to negative without passing through 0. Hence $f'$ can change sign only at $x = -2$ and $x = 3$ ⁠. These two points cut the real line into three intervals, $(-\infty,-2)$ ⁠, $(-2,3)$ and $(3,\infty)$ ⁠, and on each one $f'$ keeps a single sign.

Taking out the common factor 12 first keeps the numbers small; 12 is positive, so it never changes the sign of $f'$ ⁠. Basics: factorising

Sign chart of $f'(x) = 12(x+2)(x-3)$

On each interval, find the sign of each factor (Basic 4) and multiply:

Interval	$x+2$	$x-3$	$f'(x)$	Test value
$(-\infty,-2)$	$-$	$-$	$+$	$f'(-3) = 12(-1)(-6) = 72$
$(-2,3)$	$+$	$-$	$-$	$f'(0) = 12(2)(-3) = -72$
$(3,\infty)$	$+$	$+$	$+$	$f'(4) = 12(6)(1) = 72$

Sign chart (not to scale). The dashed lines are at

x=-2

and

x=3

⁠, where

f'(x) = 0

⁠. Two minus signs make a plus, one minus sign makes a minus.

Left of

-2

⁠, both

x+2

and

x-3

are negative (for example at

x = -3

they are

-1

and

-6

⁠), and negative times negative is positive. Between

-2

and

3

only

x - 3

is negative, so the product is negative. Right of 3 both are positive. The test values confirm each row with an actual number.

One test value per interval is enough only because

f'

cannot change sign inside an interval (Step 2). Basics: sign of a product

Read off where $f$ increases and decreases

On $(-\infty,-2]$ ⁠. Take any $x_1 < x_2$ in this interval. By the Mean Value Theorem (as in the book's proof of Theorem 1) there is a $c$ with $x_1 < c < x_2$ and
$f(x_2) - f(x_1) = f'(c)\,(x_2 - x_1).$
Since $c < x_2 \le -2$ ⁠, $c$ lies in $(-\infty,-2)$ ⁠, where $f'(c) > 0$ ⁠. Also $x_2 - x_1 > 0$ ⁠. So $f(x_2) - f(x_1) > 0$ ⁠: $f$ is strictly increasing on $(-\infty,-2]$ , hence increasing there.

On $[-2,3]$ ⁠. The same argument, now with $-2 \le x_1 < c < x_2 \le 3$ ⁠, puts $c$ in $(-2,3)$ ⁠, where $f'(c) < 0$ ⁠. So $f(x_2) < f(x_1)$ ⁠: $f$ is strictly decreasing on $[-2,3]$ .

On $[3,\infty)$ ⁠. Here $3 \le x_1 < c$ ⁠, so $f'(c) > 0$ and $f$ is strictly increasing on $[3,\infty)$ .

The book states its answer with the open intervals $(-\infty,-2)$ ⁠, $(-2,3)$ ⁠, $(3,\infty)$ ⁠, where $f'$ is strictly positive or negative. Because $c$ always lies strictly between $x_1$ and $x_2$ ⁠, the end points $-2$ and $3$ can be included at no cost: $f$ is continuous there, and the book does exactly this in Example 12. Both forms are correct; the closed form says a little more.

This is Theorem 1 of Section 6.3 (with the Remark after it for the unbounded intervals $(-\infty,-2]$ and $[3,\infty)$ ⁠, where the same Mean Value Theorem argument applies). Lesson: increasing and decreasing functions
What the graph looks like, and what the answer does not say

Putting the pieces together: the graph rises to the point $(-2, f(-2))$ ⁠, falls to $(3, f(3))$ ⁠, then rises for ever. The two turning values are
$\begin{aligned} f(-2) &= -32 - 24 + 144 + 30 = 118, \\ f(3) &= 108 - 54 - 216 + 30 = -132. \end{aligned}$

The graph of $f(x) = 4x^3 - 6x^2 - 72x + 30$ ⁠: green where $f'(x) > 0$ (rising), rose where $f'(x) < 0$ (falling). Vertical scale: 1 unit of $y$ is much smaller than 1 unit of $x$ ⁠.

Two things the answer does not say:
- $f$ is neither increasing nor decreasing on $\R$ ⁠. $-3 < -2$ and $f(-3) = 84 < 118 = f(-2)$ ⁠, so $f$ is not decreasing on $\R$ ⁠; $-2 < 3$ and $f(-2) = 118 > -132 = f(3)$ ⁠, so $f$ is not increasing on $\R$ ⁠. (The book makes this remark too.)
- $f$ is not increasing on the union $(-\infty,-2) \cup (3,\infty)$ ⁠. $-2.5 < 3.5$ ⁠, both in the union, yet $f(-2.5) = 110 > f(3.5) = -124$ ⁠. The answer means "increasing on each of the two intervals separately".
"Increasing on I" compares every pair of points of I, so one pair in the wrong order disproves it. A union of two separate intervals allows pairs from different pieces, and the valley between them breaks the order.

Result

$f'(x) = 12(x-3)(x+2)$ ⁠. (a) $f$ is increasing on $(-\infty,-2)$ and on $(3,\infty)$ ; (b) $f$ is decreasing on $(-2,3)$ . In fact $f$ is strictly increasing on $(-\infty,-2]$ and on $[3,\infty)$ ⁠, and strictly decreasing on $[-2,3]$ ⁠; it is neither increasing nor decreasing on $\R$ ⁠. This agrees with the book's solution.

Check it with numbers

$x$	$-3$	$-2.5$	$-2$	$0$	$3$	$3.5$	$4$
$f(x)$	84	110	118	30	$-132$	$-124$	$-98$

Up from 84 to 118 on $[-3,-2]$ ⁠, down from 118 through 30 to $-132$ on $[-2,3]$ ⁠, up again from $-132$ to $-98$ on $[3,4]$ ⁠: exactly the pattern of the sign chart.

Common mistakes

Solving $f'(x) = 0$ and stopping. The roots $-2$ and $3$ are only the boundaries; the answer needs the sign of $f'$ on each interval between them.
Wrong factor signs. $x + 2$ is negative to the left of $-2$ ⁠, not to the left of 2. Test each interval with a number if unsure.
Writing "increasing on $(-\infty,-2)\cup(3,\infty)$ ". As Step 5 shows, $f$ is not increasing on that union; say "on $(-\infty,-2)$ and on $(3,\infty)$ ".
Thinking the end points must be excluded. Including $-2$ and $3$ is also correct, because $f$ is continuous there (Step 4).

What to take away

Recipe: differentiate, factorise $f'$ ⁠, mark the zeros, fill a sign chart, translate $+$ into "increasing" and $-$ into "decreasing".
$f' > 0$ on an interval gives strictly increasing, by the Mean Value Theorem; the end points can be added when $f$ is continuous there.
Answer interval by interval; never merge separate intervals with $\cup$ in an "increasing on" statement.

Every fact used, and where it comes from

Fact	Used in	Source
Power, constant-multiple and sum rules	Step 1	Derivatives of polynomials (Class XI; Chapter 5)
$x^2 - x - 6 = (x-3)(x+2)$ ⁠; a product is 0 only if a factor is 0	Step 2	Basics: factorising
Signs of products of linear factors	Step 3	Basics: sign of a product
Increasing, strictly increasing, decreasing	Steps 4, 5	Definition 1, Section 6.3 (Lesson)
Sign of $f'$ decides increasing or decreasing; proof by the Mean Value Theorem	Step 4	Theorem 1 and its proof, and the Remark after it, Section 6.3
Mean Value Theorem: $f(x_2) - f(x_1)$ $= f'(c)(x_2 - x_1)$ for some $c$ between $x_1$ and $x_2$	Step 4	Used without proof in the book's proof of Theorem 1, which cites it as "Theorem 8 in Chapter 5"; this reprint's Chapter 5 does not contain it (a stale reference)
Values $f(-3), \dots, f(4)$ and $f'(-3), f'(0), f'(4)$	Steps 3, 5; Check	Arithmetic, recomputed in the site's verification script

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Example 11: Where 4x3−6x2−72x+304x^3-6x^2-72x+304x3−6x2−72x+30 increases and decreases

Basics you need

Solution

Differentiate

Factorise f′(x)f'(x)f′(x) and find where it is zero

Sign chart of f′(x)=12(x+2)(x−3)f'(x) = 12(x+2)(x-3)f′(x)=12(x+2)(x−3)

Read off where fff increases and decreases

What the graph looks like, and what the answer does not say

Every fact used, and where it comes from

Example 11: Where $4x^3-6x^2-72x+30$ increases and decreases

Factorise $f'(x)$ and find where it is zero

Sign chart of $f'(x) = 12(x+2)(x-3)$

Read off where $f$ increases and decreases