Increasing and Decreasing Functions

Use differentiation to determine where functions are increasing or decreasing

SPEC 7.3 • DIFFERENTIATION

What Does Increasing Mean?

Increasing

A function is increasing when as x increases, y also increases.

The graph goes uphill (from left to right).

Decreasing

A function is decreasing when as x increases, y decreases.

The graph goes downhill (from left to right).

Key Insight

The gradient tells us if a function is increasing or decreasing at any point!

Formal Definitions

Increasing Function

A function f(x) is increasing on an interval if:

x_{1} < x_{2} \Rightarrow f (x_{1}) < f (x_{2})

For any two points in the interval, a larger x gives a larger y.

Decreasing Function

A function f(x) is decreasing on an interval if:

x_{1} < x_{2} \Rightarrow f (x_{1}) > f (x_{2})

For any two points in the interval, a larger x gives a smaller y.

The Derivative Test

The derivative tells us the gradient at any point. We can use this to determine behaviour:

ConditionMeaningFunction Behaviour

\frac{d y}{d x} > 0

Positive gradientINCREASING

\frac{d y}{d x} < 0

Negative gradientDECREASING

\frac{d y}{d x} = 0

Zero gradientStationary point (neither)

Method

To find intervals: (1) Find dy/dx, (2) Solve dy/dx = 0, (3) Test sign in each interval

Worked Example01

Find the intervals where $y = x^{2} - 4 x$ is increasing and decreasing.

1

Find the derivative:

\frac{d y}{d x} = 2 x - 4

2

Find critical points:

2 x - 4 = 0 \Rightarrow x = 2

3

Test intervals:

•

When

x < 2

: e.g. x = 0,

\frac{d y}{d x} = - 4 < 0

→ DECREASING

•

When

x > 2

: e.g. x = 3,

\frac{d y}{d x} = 2 > 0

→ INCREASING

AnswerDecreasing for x < 2, Increasing for x > 2

Worked Example02

Find the intervals where $y = x^{3} - 3 x$ is increasing and decreasing.

1

Find the derivative:

\frac{d y}{d x} = 3 x^{2} - 3 = 3 (x^{2} - 1) = 3 (x - 1) (x + 1)

2

Critical points:

x = - 1

and

x = 1

3

Test three intervals:

•

x < -1: e.g. x = -2,

\frac{d y}{d x} = 3 (4 - 1) = 9 > 0

→ INCREASING

•

-1 < x < 1: e.g. x = 0,

\frac{d y}{d x} = - 3 < 0

→ DECREASING

•

x > 1: e.g. x = 2,

\frac{d y}{d x} = 9 > 0

→ INCREASING

AnswerIncreasing for x < -1 and x > 1; Decreasing for -1 < x < 1

Worked Example03

Find the intervals where $y = 2 x^{3} - 9 x^{2} + 12 x - 5$ is increasing.

1

Find derivative:

\frac{d y}{d x} = 6 x^{2} - 18 x + 12 = 6 (x^{2} - 3 x + 2) = 6 (x - 1) (x - 2)

2

Critical points:

x = 1

and

x = 2

3

Test intervals:

•

x < 1: e.g. x = 0,

\frac{d y}{d x} = 6 (- 1) (- 2) = 12 > 0

→ INCREASING

•

1 < x < 2: e.g. x = 1.5,

\frac{d y}{d x} = 6 (0.5) (- 0.5) = - 1.5 < 0

→ DECREASING

•

x > 2: e.g. x = 3,

\frac{d y}{d x} = 6 (2) (1) = 12 > 0

→ INCREASING

AnswerIncreasing for x < 1 and x > 2

Sign Analysis Method

A systematic approach to determine increasing/decreasing intervals:

Step-by-Step Method

Find dy/dx and factorise if possible

Solve dy/dx = 0 to find critical values

Draw a sign table with critical values marked

Test a value in each interval

State intervals where function is increasing/decreasing

Tip

For factored form like

(x - a) (x - b)

: The sign changes at each root. If you know one sign, you can determine the others!

Worked Example04

Find the intervals where $y = x + \frac{1}{x}$ for $x > 0$ is increasing and decreasing.

1

Rewrite:

y = x + x^{- 1}

2

Differentiate:

\frac{d y}{d x} = 1 - x^{- 2} = 1 - \frac{1}{x ^{2}} = \frac{x ^{2} - 1}{x ^{2}}

3

Critical point (for x > 0):

x^{2} - 1 = 0 \Rightarrow x = 1

4

Test intervals (x > 0 only):

•

0 < x < 1: e.g. x = 0.5,

\frac{d y}{d x} = \frac{0.25 - 1}{0.25} = - 3 < 0

→ DECREASING

•

x > 1: e.g. x = 2,

\frac{d y}{d x} = \frac{4 - 1}{4} = 0.75 > 0

→ INCREASING

AnswerDecreasing for 0 < x < 1; Increasing for x > 1

Proving Always Increasing/Decreasing

Some functions are always increasing (or decreasing) for all real values of x.

Method

To prove f(x) is always increasing:

Show that

f^{'} (x) > 0

for all x (or

f^{'} (x) \geq 0

with equality only at isolated points)

Common Techniques

Show dy/dx is a perfect square:

(a x + b)^{2} \geq 0

Complete the square to show always positive

Show discriminant < 0 for quadratic (no real roots, always same sign)

Worked Example05

Prove that $f (x) = x^{3} + 3 x - 2$ is an increasing function for all real values of x.

1

Find derivative:

f^{'} (x) = 3 x^{2} + 3

2

Since

x^{2} \geq 0

for all real x...

3

3 x^{2} \geq 0

for all real x

4

Therefore

f^{'} (x) = 3 x^{2} + 3 \geq 3 > 0

for all real x

5

Since

f^{'} (x) > 0

for all x, f(x) is always increasing. ∎

Worked Example06

Find the range of values of k for which $f (x) = x^{3} + k x^{2} + 5 x$ is an increasing function for all real values of x.

1

Find derivative:

f^{'} (x) = 3 x^{2} + 2 k x + 5

2

For f(x) to be always increasing, we need

f^{'} (x) > 0

for all x

3

This is a quadratic in x. Since coefficient of x² is positive (3 > 0)...

4

The quadratic is always positive when it has NO real roots

5

Discriminant:

b^{2} - 4 a c = (2 k)^{2} - 4 (3) (5) = 4 k^{2} - 60

6

Need:

4 k^{2} - 60 < 0 \Rightarrow k^{2} < 15

Answer

- 15 < k < 15

Practice

Question 1: Find the intervals where

y = x^{3} - 12 x

is increasing and decreasing.

\frac{d y}{d x} = 3 x^{2} - 12 = 3 (x^{2} - 4) = 3 (x - 2) (x + 2)

Critical points: x = -2 and x = 2

Testing intervals:

x < -2: dy/dx > 0 → INCREASING

-2 < x < 2: dy/dx < 0 → DECREASING

x > 2: dy/dx > 0 → INCREASING

AnswerIncreasing for x < -2 and x > 2; Decreasing for -2 < x < 2

Practice

Question 2: Prove that

f (x) = 2 x^{3} + 6 x + 1

is an increasing function for all real values of x.

f^{'} (x) = 6 x^{2} + 6

Since

x^{2} \geq 0

for all real x

6 x^{2} \geq 0

for all real x

Therefore

f^{'} (x) = 6 x^{2} + 6 \geq 6 > 0

for all real x

Since f'(x) > 0 for all x, f(x) is always increasing. ∎

Practice

Question 3: Find the range of values of k for which

f (x) = x^{3} - 3 x^{2} + k x

is an increasing function for all real values of x.

f^{'} (x) = 3 x^{2} - 6 x + k

For always increasing:

f^{'} (x) > 0

for all x

Coefficient of x² is 3 > 0, so need no real roots

Discriminant:

b^{2} - 4 a c = 36 - 12 k < 0

36 < 12 k \Rightarrow k > 3

Answerk > 3

Summary

Key Concepts

Derivative SignFunction Behaviour

\frac{d y}{d x} > 0

INCREASING

\frac{d y}{d x} < 0

DECREASING

\frac{d y}{d x} = 0

Stationary point

Exam Skills

Find intervals: Solve dy/dx = 0, then test sign in each region

Prove always increasing: Show dy/dx > 0 for all x (use discriminant or perfect square)

Find parameter ranges: Set up condition for no real roots