Chapter 6 · Application of Derivatives

Exercise 6.2 · Q5: Where $2x^3-3x^2-36x+7$ increases and decreases

Exercise 6.2Increasing and decreasing functions · Section 6.3

The question

Find the intervals in which the function $f$ given by $f(x) = 2x^3 - 3x^2 - 36x + 7$ is

(a) increasing (b) decreasing

Exam answer

\begin{aligned} f'(x) &= 6x^2 - 6x - 36 \\ &= 6(x - 3)(x + 2) \end{aligned}

$f'(x) = 0$ gives $x = -2$ or $x = 3$ ⁠.

In $(-\infty, -2)$ ⁠, $f'(x) > 0$ ⁠, so $f$ is increasing.

In $(-2, 3)$ ⁠, $f'(x) < 0$ ⁠, so $f$ is decreasing.

In $(3, \infty)$ ⁠, $f'(x) > 0$ ⁠, so $f$ is increasing.

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

(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 derivative of a cubic is a quadratic, and a factorised quadratic shows its sign at a glance. Here $f'(x) = 6(x - 3)(x + 2)$ ⁠. Its two roots, $-2$ and $3$ ⁠, cut the line into three intervals, and the sign of the product on each interval decides whether $f$ rises or falls there. This is the book's Example 11 pattern.

Given, to find, and the plan

Given: $f(x) = 2x^3 - 3x^2 - 36x + 7$ ⁠, $x$ real.
To find: The intervals where $f$ is (a) increasing, (b) decreasing.
Plan: (1) Differentiate and factorise $f'$ ⁠. (2) Find its zeros. (3) Sign chart: sign of each factor on each interval. (4) Conclude with Theorem 1.

Basics you need

1Increasing and decreasing (Definition 1)

On an interval I, increasing means $x_1 < x_2 \Rightarrow f(x_1) \le f(x_2)$ ⁠; decreasing means $x_1 < x_2 \Rightarrow f(x_1) \ge f(x_2)$ ⁠; "strictly" uses $<$ and $>$ ⁠. Lesson: increasing and decreasing functions

2The derivative test (Theorem 1)

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ ⁠: $f'(x) \ge 0$ on $(a, b)$ makes $f$ increasing on $[a, b]$ ⁠, and $f'(x) \le 0$ makes it decreasing (the theorem as printed). The Mean Value step of its proof shows that $f' > 0$ gives strictly increasing and $f' < 0$ strictly decreasing.

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 it has an even number of negative factors and negative when it has an odd number. Each linear factor $x - r$ is negative for $x < r$ and positive for $x > r$ ⁠. Basics: sign of a product

Solution

Differentiate and factorise
$\begin{aligned} f'(x) &= 6x^2 - 6x - 36\\ &= 6(x^2 - x - 6)\\ &= 6(x - 3)(x + 2). \end{aligned}$
Check the factorisation by expanding: $(x - 3)(x + 2) = x^2 + 2x - 3x - 6 = x^2 - x - 6$ ⁠. ✓

First line: power rule term by term, $\frac{d}{dx}(2x^3) = 6x^2$ ⁠, $\frac{d}{dx}(3x^2) = 6x$ ⁠, $\frac{d}{dx}(36x) = 36$ ⁠, and the constant 7 has derivative 0. Taking out the common factor 6 leaves a simpler quadratic, and the factorised form is what lets us read signs.

Product $-6$ ⁠, sum $-1$ ⁠: the pair $-3, 2$ ⁠. Basics: factorising and the discriminant
Zeros of $f'$ and the three intervals
$f'(x) = 0 \iff x - 3 = 0 \text{ or } x + 2 = 0 \iff x = -2 \text{ or } x = 3.$
The points $-2$ and $3$ divide the real line into $(-\infty, -2)$ ⁠, $(-2, 3)$ and $(3, \infty)$ ⁠.

A product is zero exactly when one factor is zero (6 is never zero). Between consecutive zeros no factor changes sign, so $f'$ has one fixed sign on each of the three intervals.
Sign chart
- $x < -2$ ⁠: both $x + 2 < 0$ and $x - 3 < 0$ ⁠; two negatives, so $f'(x) > 0$ ⁠. (Test: $f'(-3) = 6(-6)(-1) = 36$ ⁠.)
- $-2 < x < 3$ ⁠: $x + 2 > 0$ but $x - 3 < 0$ ⁠; one negative, so $f'(x) < 0$ ⁠. (Test: $f'(0) = -36$ ⁠.)
- $x > 3$ ⁠: both factors positive, so $f'(x) > 0$ ⁠. (Test: $f'(4) = 6(1)(6) = 36$ ⁠.)
Sign chart (not to scale) for $f'(x) = 6(x + 2)(x - 3)$ ⁠. The constant 6 is positive, so the sign of $f'$ is the sign of $(x + 2)(x - 3)$ ⁠.

Reading each factor separately is safer than guessing from one test point: it shows why the sign is what it is on the whole interval. The test points then confirm it.

Even number of negative factors: positive; odd number: negative. Basics: sign of a product
Conclude with Theorem 1

$f$ is a polynomial, so continuous and differentiable everywhere. For any $x_1 < x_2$ inside one of the intervals, Theorem 1 on $[x_1, x_2]$ applies:
- $f' > 0$ on $(-\infty, -2)$ and on $(3, \infty)$ ⁠: $f$ is increasing on each of them.
- $f' < 0$ on $(-2, 3)$ ⁠: $f$ is decreasing there.
$y = 2x^3 - 3x^2 - 36x + 7$ ⁠: rising to $(-2, 51)$ ⁠, falling to $(3, -74)$ ⁠, rising again. The vertical scale is compressed.

Note the wording "on each of them". $f$ is not increasing on the union $(-\infty, -2) \cup (3, \infty)$ taken as one set: for example $-3 < 4$ but $f(-3) = 34 > f(4) = -57$ ⁠. This is why the book says "increasing in the intervals $(-\infty, -2)$ and $(3, \infty)$ ".

Values used: $f(-2) = -16 - 12 + 72 + 7 = 51$ and $f(3) = 54 - 27 - 108 + 7 = -74$ ⁠.
The endpoints $-2$ and $3$

Because $f$ is continuous at $-2$ and $3$ ⁠, Theorem 1 also lets us close the intervals: $f$ is increasing on $(-\infty, -2]$ and $[3, \infty)$ ⁠, and decreasing on $[-2, 3]$ ⁠. The answer key, like Example 11, writes the open intervals; both forms are correct.

At $x = -2$ and $x = 3$ the derivative is 0 and the graph turns, but a single point never spoils the inequality $f(x_1) < f(x_2)$ for a continuous function: the Mean Value step of Theorem 1's proof only needs $f' > 0$ strictly inside $[a, b]$ ⁠, not at its ends.

Result

$f'(x) = 6(x - 3)(x + 2)$ ⁠.
(a) $f$ is increasing on $(-\infty, -2)$ and on $(3, \infty)$ ⁠.
(b) $f$ is decreasing on $(-2, 3)$ ⁠.
(All strictly; the endpoints may be included by continuity.) (Textbook answer: (a) $(-\infty, -2)$ and $(3, \infty)$ ⁠, (b) $(-2, 3)$ ⁠.)

Check it with numbers

$f(-4) = -25$ ⁠, $f(-3) = 34$ ⁠, $f(-2) = 51$ ⁠: rising. $f(-1) = 38$ ⁠, $f(0) = 7$ ⁠, $f(2) = -61$ ⁠, $f(3) = -74$ ⁠: falling. $f(4) = -57$ ⁠, $f(5) = 2$ ⁠: rising again. Slope check at $x = 0$ ⁠: the central difference $\frac{f(0.001) - f(-0.001)}{0.002}$ is about $-36.000$ ⁠, matching $f'(0) = -36$ ⁠.

Common mistakes

Writing "increasing on $(-\infty, -2) \cup (3, \infty)$ ". As Step 4 shows, $f$ is increasing on each interval separately, not on their union.
Factorising wrongly, e.g. $(x + 3)(x - 2)$ ⁠. That expands to $x^2 + x - 6$ ⁠. Always expand to check.
Using one test point for the whole line. Each interval needs its own sign.

What to take away

Factorise $f'$ fully; its zeros are the only places the direction can change.
Build the sign chart factor by factor, then read off increasing (+) and decreasing (−).
Report several intervals as "on each of", never as one union.

Every fact used, and where it comes from

Fact	Used in	Source
Increasing and decreasing on an interval	Steps 4, 5	Definition 1, Section 6.3 (Lesson: increasing and decreasing functions)
Derivative of a polynomial	Step 1	Algebra of derivatives, Section 5.3; Table 5.3
$x^2 - x - 6 = (x - 3)(x + 2)$	Step 1	Factorisation, checked by expanding
Sign of a product from the signs of its factors	Step 3	Rules of signs (Basics: sign of a product)
Sign of $f'$ gives the direction; endpoints by continuity	Steps 4, 5	Theorem 1 and its proof, Section 6.3
Same method and table layout	Steps 2 to 4	Example 11, Section 6.3

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Exercise 6.2 · Q5: Where 2x3−3x2−36x+72x^3-3x^2-36x+72x3−3x2−36x+7 increases and decreases

Basics you need

Solution

Differentiate and factorise

Zeros of f′f'f′ and the three intervals

Sign chart

Conclude with Theorem 1

The endpoints −2-2−2 and 333

Every fact used, and where it comes from

Exercise 6.2 · Q5: Where $2x^3-3x^2-36x+7$ increases and decreases

Zeros of $f'$ and the three intervals

The endpoints $-2$ and $3$