Chapter 6 · Application of Derivatives

Example 33: Where $f(x)=\frac{3}{10}x^4-\frac{4}{5}x^3-3x^2+\frac{36}{5}x+11$ increases and decreases

ExamplesIncreasing and decreasing functions · Section 6.3 · Miscellaneous Examples

The question

Find intervals in which the function given by

f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11

is (a) increasing (b) decreasing.

Exam answer

\begin{aligned} f'(x) &= \frac65x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5} \\ &= \frac65(x^3 - 2x^2 - 5x + 6) \\ &= \frac65(x - 1)(x + 2)(x - 3) \end{aligned}

$f'(x) = 0$ gives $x = -2$ ⁠, $x = 1$ or $x = 3$ ⁠.

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

In $(-2, 1)$ ⁠, $f'(x) > 0$ ⁠, so $f$ is increasing.

In $(1, 3)$ ⁠, $f'(x) < 0$ ⁠, so $f$ is decreasing.

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

(a) $f$ is increasing in $(-2, 1)$ and $(3, \infty)$ ⁠.

(b) $f$ is decreasing in $(-\infty, -2)$ and $(1, 3)$ ⁠.

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 function goes up, where $f'(x) < 0$ it goes down (Theorem 1, Section 6.3). So we factorise $f'$ completely, mark its zeros on the number line, and read off the sign of each factor on each piece. A sign chart does the bookkeeping.

Given, to find, and the plan

Given: $f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11$ ⁠, defined for all real $x$ ⁠.
To find: The intervals where $f$ is (a) increasing and (b) decreasing.
Plan: Differentiate term by term, take out the common factor $\frac{6}{5}$ ⁠, factorise the cubic using the factor theorem, find the zeros of $f'$ ⁠, make a sign chart on the four intervals they create, and conclude with Theorem 1. Then decide whether the end points may be included.

Basics you need

1Increasing and decreasing (Definition 1, Section 6.3)

On an interval I, $f$ is increasing if $x_1 < x_2$ in I implies $f(x_1) \le f(x_2)$ ⁠, and strictly increasing if it implies $f(x_1) < f(x_2)$ ⁠; decreasing and strictly decreasing are defined the same way with the inequalities reversed. Lesson: increasing and decreasing

2Theorem 1 (Section 6.3)

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ ⁠, then $f$ is increasing on $[a, b]$ if $f'(x) \ge 0$ for each $x \in (a, b)$ ⁠, and decreasing if $f'(x) \le 0$ ⁠. When the inequality is strict, the book's proof (Mean Value Theorem) gives $f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$ with $f'(c) \ne 0$ ⁠, so the function is then strictly increasing or strictly decreasing.

3Factor theorem

If a polynomial $p$ has $p(a) = 0$ ⁠, then $x - a$ is a factor of $p(x)$ ⁠. For a cubic with integer coefficients and leading coefficient 1, try the divisors of the constant term first. Basics: factorising

4Sign of a product

A product is negative when an odd number of its factors are negative, and positive when an even number are. Basics: sign of a product

5Interval notation

$(a, b)$ leaves out both ends, $[a, b]$ includes them, and $(-\infty, a)$ means all $x < a$ ⁠. Basics: interval notation

Solution

Domain and smoothness

$f$ is a polynomial, so it is defined, continuous and differentiable at every real $x$ ⁠.

Theorem 1 needs continuity on the closed interval and differentiability inside it. For a polynomial both hold everywhere, so the theorem can be used on any interval we like, and the only places where the sign of $f'$ can change are its zeros.

Polynomials are continuous and differentiable on all of $\R$ (Chapter 5).
Differentiate term by term
$\begin{aligned} f'(x) &= \frac{3}{10}(4x^3) - \frac{4}{5}(3x^2) \\ &\quad - 3(2x) + \frac{36}{5} \\ &= \frac{6}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5}. \end{aligned}$
The derivative of a sum is the sum of the derivatives, and each term is a constant times a power of $x$ ⁠. The constant 11 has derivative 0.

Power rule $\frac{d}{dx}x^n = nx^{n-1}$ ⁠; $\frac{3}{10}\cdot 4 = \frac{12}{10} = \frac{6}{5}$ ⁠, $\frac{4}{5}\cdot 3 = \frac{12}{5}$ ⁠.
Take out the common factor $\frac{6}{5}$
$f'(x) = \frac{6}{5}\left(x^3 - 2x^2 - 5x + 6\right).$
Check each term: $\frac{6}{5}\cdot 2 = \frac{12}{5}$ ⁠, $\frac{6}{5}\cdot 5 = 6$ ⁠, $\frac{6}{5}\cdot 6 = \frac{36}{5}$ ⁠.

The positive constant $\frac{6}{5}$ does not change the sign of $f'$ ⁠, so all the sign information sits in the cubic, which now has integer coefficients and leading coefficient 1, ready for the factor theorem.

Factoring out a common factor is the distributive law used backwards.
Factorise the cubic

Let $p(x) = x^3 - 2x^2 - 5x + 6$ ⁠. Try divisors of 6: $p(1) = 1 - 2 - 5 + 6 = 0$ ⁠. So $x - 1$ is a factor. Dividing,
$\begin{aligned} p(x) &= (x - 1)(x^2 - x - 6) \\ &= (x - 1)(x + 2)(x - 3). \end{aligned}$
Hence
$f'(x) = \frac{6}{5}(x - 1)(x + 2)(x - 3),$
which is the textbook's "on simplification" form.

To know the sign of $f'$ we need it as a product of simple factors whose signs are obvious. Expanding back confirms the division: $\begin{aligned} (x - 1)(x^2 - x - 6) &= x^3 - x^2 - 6x \\ &\quad - x^2 + x + 6 \\ &= x^3 - 2x^2 - 5x + 6. \end{aligned}$ The quadratic factorises because $-3 \cdot 2 = -6$ and $-3 + 2 = -1$ ⁠.

Factor theorem; a quadratic $x^2 + bx + c$ factorises as $(x - \alpha)(x - \beta)$ when $\alpha\beta = c$ and $\alpha + \beta = -b$ ⁠. Basics: factorising
Zeros of $f'$ ⁠: the points that cut the line
$f'(x) = 0 \iff x = -2,\ 1 \ \text{or}\ 3.$
These three points divide the real line into four disjoint intervals:
$(-\infty, -2),\quad (-2, 1),\quad (1, 3),\quad (3, \infty).$
Each factor is zero at only one point and has a constant sign on either side of it. So $f'$ can change sign only at $-2$ ⁠, 1 and 3, and on each of the four intervals its sign is constant.

Zero-product rule: a product is zero exactly when one of its factors is zero.

Sign chart

Sign of each factor on each interval, and their product. The positive constant

\frac{6}{5}

does not affect the sign.

Interval	$x + 2$	$x - 1$	$x - 3$	$f'(x)$	Test point
$(-\infty, -2)$	−	−	−	−	$f'(-3) = -28.8$
$(-2, 1)$	+	−	−	+	$f'(0) = 7.2$
$(1, 3)$	+	+	−	−	$f'(2) = -4.8$
$(3, \infty)$	+	+	+	+	$f'(4) = 21.6$

(-\infty, -2)

all three factors are negative (odd number), so the product is negative. Moving right past each zero flips exactly one factor, so the sign of

f'

alternates. The test points (the textbook uses

x = -3

and

x = 0

⁠) confirm each row with actual numbers.

An odd number of negative factors gives a negative product. Basics: sign of a product

Conclude with Theorem 1

(a) $f'(x) > 0$ on $(-2, 1)$ and on $(3, \infty)$ ⁠, so $f$ is increasing in $(-2, 1)$ and in $(3, \infty)$ .

(b) $f'(x) < 0$ on $(-\infty, -2)$ and on $(1, 3)$ ⁠, so $f$ is decreasing in $(-\infty, -2)$ and in $(1, 3)$ .

Because the inequalities are strict, $f$ is in fact strictly increasing and strictly decreasing on these intervals (Basic 2).

For an unbounded interval such as $(3, \infty)$ ⁠, take any $x_1 < x_2$ in it; they lie in a closed interval $[x_1, x_2]$ where Theorem 1 applies, so $f(x_1) < f(x_2)$ ⁠. That is exactly Definition 1 for the whole interval.

Theorem 1 (Section 6.3), applied on each interval. Lesson: increasing and decreasing
May the end points be included?

Yes. $f$ is continuous at $-2$ ⁠, 1 and 3, and Theorem 1 asks only for $f' > 0$ (or $< 0$ ⁠) inside the interval. So $f$ is also strictly increasing on $[-2, 1]$ and $[3, \infty)$ ⁠, and strictly decreasing on $(-\infty, -2]$ and $[1, 3]$ ⁠. Each critical point belongs to an increasing piece and a decreasing piece at the same time; the textbook simply states the open intervals.

What is not true is that $f$ is increasing on the union $(-2, 1) \cup (3, \infty)$ ⁠: for example $0.9 < 3.1$ ⁠, both points lie in the union, but $f(0.9) \approx 14.6636$ is larger than $f(3.1) \approx 8.3628$ ⁠.

The answer must be stated interval by interval. Between the two increasing pieces lies a decreasing piece that undoes part of the rise, so the function is not increasing across the gap.

Theorem 1 needs continuity on the closed interval and the sign condition only on the open interval. Basics: interval notation
Picture the result

The graph of $f$ ⁠: falling (rose) on $(-\infty, -2]$ and $[1, 3]$ ⁠, rising (blue) on $[-2, 1]$ and $[3, \infty)$ ⁠. The turning points are at $x = -2$ ⁠, 1 and 3.

The graph turns exactly where $f'$ changes sign, which is a useful check on the sign chart. (These turning points are local minima at $-2$ and 3 and a local maximum at 1, by the first derivative test of Section 6.4.)

$f(-2) = -4.2$ ⁠, $f(1) = 14.7$ ⁠, $f(3) = 8.3$ ⁠, computed from the formula for $f$ ⁠.

Result

$f'(x) = \frac{6}{5}(x - 1)(x + 2)(x - 3)$ ⁠.

(a) $f$ is increasing in $(-2, 1)$ and in $(3, \infty)$ . (b) $f$ is decreasing in $(-\infty, -2)$ and in $(1, 3)$ .

Both are strict, and the end points may be included: increasing on $[-2, 1]$ and $[3, \infty)$ ⁠, decreasing on $(-\infty, -2]$ and $[1, 3]$ ⁠. (Textbook's own solution: the same four open intervals.)

Check it with numbers

Values of $f$ at whole numbers: $f(-3) = 8.3$ ⁠, $f(-2) = -4.2$ ⁠, $f(0) = 11$ ⁠, $f(1) = 14.7$ ⁠, $f(2) = 11.8$ ⁠, $f(3) = 8.3$ ⁠, $f(4) = 17.4$ ⁠. They fall from $-3$ to $-2$ ⁠, rise from $-2$ to 1, fall from 1 to 3, and rise from 3 to 4, exactly as the sign chart says.

Common mistakes

Losing the sign while factoring. Taking out $\frac{6}{5}$ from $-6x$ leaves $-5x$ ⁠, not $-6x$ ⁠. Expand back to check.
Writing the answer as one union. " $f$ is increasing on $(-2, 1) \cup (3, \infty)$ " is false (Step 8). List the intervals separately.
Deciding signs from one test point only. Use the factor signs on every interval; test points are a check, not the argument.
Forgetting that $f$ is defined everywhere. The outer intervals are unbounded: $(-\infty, -2)$ and $(3, \infty)$ ⁠.

What to take away

Recipe: differentiate, factorise $f'$ completely, mark its zeros, sign chart, conclude with Theorem 1.
Factor out positive constants first; they never change the sign.
With simple zeros (each factor to the first power), the sign of $f'$ alternates from one interval to the next.

Every fact used, and where it comes from

Fact	Used in	Source
Polynomials are continuous and differentiable everywhere	Steps 1, 8	Chapter 5
Power rule, sum rule	Step 2	Chapter 5
Factor theorem; factorising a quadratic	Step 4	Algebra
Sign of a product	Step 6	Algebra
Increasing, strictly increasing (and decreasing)	Steps 7, 8	Definition 1, Section 6.3 (Lesson: increasing and decreasing)
Sign of $f'$ decides increasing or decreasing	Steps 7, 8	Theorem 1 and its proof by the Mean Value Theorem, Section 6.3
Turning points as local extrema	Step 9	Theorem 3 (First Derivative Test), Section 6.4 (Lesson: derivative tests)
Textbook working for this problem	Result	Example 33 and Fig 6.22, Miscellaneous Examples, Chapter 6

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Example 33: Where f(x)=310x4−45x3−3x2+365x+11f(x)=\frac{3}{10}x^4-\frac{4}{5}x^3-3x^2+\frac{36}{5}x+11f(x)=103​x4−54​x3−3x2+536​x+11 increases and decreases

Basics you need

Solution

Domain and smoothness

Differentiate term by term

Take out the common factor 65\frac{6}{5}56​

Factorise the cubic

Zeros of f′f'f′⁠: the points that cut the line

Sign chart

Conclude with Theorem 1

May the end points be included?

Picture the result

Every fact used, and where it comes from

Example 33: Where $f(x)=\frac{3}{10}x^4-\frac{4}{5}x^3-3x^2+\frac{36}{5}x+11$ increases and decreases

Take out the common factor $\frac{6}{5}$

Zeros of $f'$ ⁠: the points that cut the line