Open Math›Year 11›Mathematics Advanced›Introduction to differentiation›Applications of the derivative

Applications of the derivative

Determine intervals where a function is increasing or decreasing using the sign of f'(x), find and classify stationary points of polynomials using sign tests, and interpret the derivative in context including position and velocity.

Worked examples

Finding stationary points and increasing/decreasing intervals

Straightforward

Problem

For

f(x) = x^2 - 4x + 7

, find the stationary point and determine the intervals on which

f

is increasing and decreasing.

Differentiate

f(x)

f'(x) = 2x - 4

Solve

f'(x) = 0

to find the

x

-coordinate of the stationary point.

2x - 4 = 0 \implies x = 2

Find the

y

-coordinate of the stationary point.

f(2) = (2)^2 - 4(2) + 7 = 4 - 8 + 7 = 3

Stationary point:

(2,\, 3)

Determine where

f

is increasing and decreasing using the sign of

f'(x)

For

x < 2

: test

x = 0 \Rightarrow f'(0) = -4 < 0

—

f

is **decreasing**.

For

x > 2

: test

x = 3 \Rightarrow f'(3) = 2 > 0

—

f

is **increasing**.

Answer

Stationary point at

(2, 3)

f

is decreasing for

x < 2

and increasing for

x > 2

Classifying stationary points using a sign test

Moderate

Problem

Find and classify all stationary points of

f(x) = x^3 - 3x

Differentiate and solve

f'(x) = 0

f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1)

Setting

f'(x) = 0

x = -1

x = 1

Apply a sign test to classify

x = -1

Test

x = -2

f'(-2) = 3(4) - 3 = 9 > 0

— increasing.

Test

x = 0

f'(0) = -3 < 0

— decreasing.

Sign changes from

+

-

, so

x = -1

is a **local maximum**.

Apply a sign test to classify

x = 1

Test

x = 0

f'(0) = -3 < 0

— decreasing.

Test

x = 2

f'(2) = 3(4) - 3 = 9 > 0

— increasing.

Sign changes from

-

+

, so

x = 1

is a **local minimum**.

Find the

y

-coordinates of each stationary point.

f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2 \quad \Rightarrow \quad

local max at

(-1,\, 2)

f(1) = (1)^3 - 3(1) = 1 - 3 = -2 \quad \Rightarrow \quad

local min at

(1,\, -2)

Answer

Local maximum at

(-1,\, 2)

; local minimum at

(1,\, -2)

Interpreting velocity and rest from a position function

Challenging

Problem

The displacement of a particle is

x(t) = t^3 - 6t^2 + 9t

metres at time

t \geq 0

seconds. Find when the particle is at rest, and determine the direction of motion on each interval between rest times.

Differentiate to find the velocity function.

v(t) = x'(t) = 3t^2 - 12t + 9

Solve

v(t) = 0

to find when the particle is at rest.

3t^2 - 12t + 9 = 0

3(t^2 - 4t + 3) = 0

3(t - 1)(t - 3) = 0

t = 1

t = 3

Determine the sign of

v(t)

on each interval to find the direction of motion.

For

0 < t < 1

: test

t = 0.5 \Rightarrow v(0.5) = 3(0.25) - 6 + 9 = 3.75 > 0

— moving in the **positive direction**.

For

1 < t < 3

: test

t = 2 \Rightarrow v(2) = 12 - 24 + 9 = -3 < 0

— moving in the **negative direction**.

For

t > 3

: test

t = 4 \Rightarrow v(4) = 48 - 48 + 9 = 9 > 0

— moving in the **positive direction**.

State the displacements at the rest times for a complete picture.

x(1) = 1 - 6 + 9 = 4

x(3) = 27 - 54 + 27 = 0

m

The particle starts at

x(0) = 0

, moves right to

x = 4

m, turns around, moves left back to

x = 0

m at

t = 3

, then continues moving right.

Answer

The particle is at rest at

t = 1

s and

t = 3

s. It moves in the positive direction for

0 < t < 1

, in the negative direction for

1 < t < 3

, and in the positive direction for

t > 3

Practise

Q1·Straightforward

Find the

x

-coordinate of the stationary point of

f(x) = x^2 - 6x + 5

by differentiating and solving

f'(x) = 0

Explanation

Differentiating:

f'(x) = 2x - 6

.

Setting

f'(x) = 0

2x - 6 = 0 \implies x = 3

The stationary point occurs at

x = 3

Q2·Straightforward

The function

f(x) = x^3 - 3x^2 + 1

has derivative

f'(x) = 3x^2 - 6x

. Determine whether

f

is increasing or decreasing at

x = 4

Increasing, because $f'(4) > 0$Decreasing, because $f'(4) > 0$Decreasing, because $f'(4) < 0$Increasing, because $f'(4) < 0$

Explanation

Substituting

x = 4

f'(4) = 3(4)^2 - 6(4) = 48 - 24 = 24

Since

f'(4) = 24 > 0

, the function is **increasing** at

x = 4

Q3·Straightforward

Find the

x

-coordinate of the stationary point of

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

by solving

f'(x) = 0

Explanation

Differentiating:

f'(x) = -2x + 4

.

Setting

f'(x) = 0

-2x + 4 = 0 \implies x = 2

The stationary point occurs at

x = 2

Q4·Straightforward

The function

f(x) = x^2 - 8x + 15

has a stationary point at

x = 4

. On which interval is

f

decreasing?

$x < 4$$x > 4$$x < 0$$x > 0$

Explanation

Differentiating:

f'(x) = 2x - 8

.

The function is decreasing where

f'(x) < 0

2x - 8 < 0 \implies x < 4

f

is decreasing on

x < 4

and increasing on

x > 4

Q5·Moderate

For

f(x) = x^3 - 3x^2 - 9x + 5

, the derivative is

f'(x) = 3x^2 - 6x - 9 = 3(x + 1)(x - 3)

. Use a sign test on

f'(x)

to classify the stationary point at

x = -1

Local maximumLocal minimumStationary point of inflectionGlobal maximum

Explanation

Testing the sign of

f'(x) = 3(x + 1)(x - 3)

near

x = -1

f'(-2) = 3(-2 + 1)(-2 - 3) = 3(-1)(-5) = 15 > 0

f'(0) = 3(0 + 1)(0 - 3) = 3(1)(-3) = -9 < 0

Since

f'

changes sign from positive to negative at

x = -1

, the function changes from increasing to decreasing. This means

x = -1

is a **local maximum**.

Q6·Moderate

For

f(x) = x^3 - 3x^2 - 9x + 5

with

f'(x) = 3(x + 1)(x - 3)

, apply a sign test to classify the stationary point at

x = 3

Local minimumLocal maximumStationary point of inflectionGlobal minimum

Explanation

Testing the sign of

f'(x) = 3(x + 1)(x - 3)

near

x = 3

f'(2) = 3(2 + 1)(2 - 3) = 3(3)(-1) = -9 < 0

f'(4) = 3(4 + 1)(4 - 3) = 3(5)(1) = 15 > 0

Since

f'

changes sign from negative to positive at

x = 3

, the function changes from decreasing to increasing. This means

x = 3

is a **local minimum**.

Q7·Moderate

Find the

y

-coordinate of the local maximum of

f(x) = 2x^3 - 9x^2 + 12x - 4

Explanation

Differentiating:

f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2)

.

Stationary points at

x = 1

and

x = 2

.

Sign test:

f'(0) = 6(0 - 1)(0 - 2) = 12 > 0

f'(1.5) = 6(1.5 - 1)(1.5 - 2) = 6(0.5)(-0.5) = -1.5 < 0

f'(3) = 6(3 - 1)(3 - 2) = 12 > 0

f'

changes from positive to negative at

x = 1

, so

x = 1

is the local maximum.

Substituting

x = 1

f(1) = 2(1)^3 - 9(1)^2 + 12(1) - 4 = 2 - 9 + 12 - 4 = 1

Q8·Moderate

For

f(x) = x^3 - 6x^2 + 12x - 8

, the derivative simplifies to

f'(x) = 3(x - 2)^2

. Apply a sign test to classify the stationary point at

x = 2

Stationary point of inflection, because $f'(x)$ does not change sign at $x = 2$Local minimum, because $f'(x) \geq 0$ on both sides of $x = 2$Local maximum, because the curve is concave down at $x = 2$Local minimum, because $f''(x) > 0$ at $x = 2$

Explanation

Testing the sign of

f'(x) = 3(x - 2)^2

near

x = 2

f'(1) = 3(1 - 2)^2 = 3(1) = 3 > 0

f'(3) = 3(3 - 2)^2 = 3(1) = 3 > 0

Since

f'(x) \geq 0

on both sides of

x = 2

and the sign does **not** change, the function continues increasing through

x = 2

. This means

x = 2

is a **stationary point of inflection**, not a local maximum or minimum.

Q9·Challenging

The displacement of a particle is given by

x(t) = t^3 - 6t^2 + 9t

metres at time

t

seconds, where

t \geq 0

. The velocity is

v(t) = x'(t)

. Find the velocity (in m/s) at

t = 2

Explanation

Differentiating:

v(t) = x'(t) = 3t^2 - 12t + 9

.

Substituting

t = 2

v(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3 \text{ m/s}

The negative velocity means the particle is moving in the negative direction at

t = 2

Q10·Challenging

For the particle with displacement

x(t) = t^3 - 6t^2 + 9t

metres, the particle is momentarily at rest (that is,

v(t) = 0

) at

t = 1

. Find the next time (in seconds) after

t = 1

when the particle is again momentarily at rest.

Explanation

The velocity is

v(t) = 3t^2 - 12t + 9

.

Setting

v(t) = 0

3t^2 - 12t + 9 = 0 \implies 3(t^2 - 4t + 3) = 0 \implies 3(t - 1)(t - 3) = 0

t = 1

t = 3

.

The next time after

t = 1

when the particle is at rest is

t = 3

seconds.

Q11·Challenging

A function

f(x)

is increasing for

x < -2

, has a local maximum at

x = -2

, is decreasing for

-2 < x < 3

, has a local minimum at

x = 3

, and is increasing for

x > 3

. Which of the following best describes the graph of

y = f'(x)

$f'(x)$ has zeros at $x = -2$ and $x = 3$, is positive for $x < -2$, negative for $-2 < x < 3$, and positive for $x > 3$$f'(x)$ has zeros at $x = -2$ and $x = 3$, is negative for $x < -2$, positive for $-2 < x < 3$, and negative for $x > 3$$f'(x)$ has local maxima at $x = -2$ and $x = 3$$f'(x)$ is always positive except at $x = -2$ and $x = 3$

Explanation

The key connections between

y = f(x)

and

y = f'(x)

are:

- Stationary points of

f

(local max and local min) occur where

f'(x) = 0

. So

f'(-2) = 0

and

f'(3) = 0

.
- Where

f

is increasing,

f'(x) > 0

: this gives

f'(x) > 0

for

x < -2

and for

x > 3

.
- Where

f

is decreasing,

f'(x) < 0

: this gives

f'(x) < 0

for

-2 < x < 3

.

The correct description is:

f'(x)

has zeros at

x = -2

and

x = 3

, is positive for

x < -2

, negative for

-2 < x < 3

, and positive for

x > 3

Q12·Challenging

A particle moves so that its position is

x(t) = -2t^2 + 12t - 10

metres at time

t \geq 0

seconds. The particle reaches its maximum displacement when

v(t) = 0

. Find the time (in seconds) at which this occurs.

Explanation

Differentiating:

v(t) = x'(t) = -4t + 12

.

Setting

v(t) = 0

-4t + 12 = 0 \implies t = 3

Since

v(t) = -4t + 12 > 0

for

t < 3

and

v(t) < 0

for

t > 3

, the particle is moving in the positive direction before

t = 3

and in the negative direction after

t = 3

. So the maximum displacement occurs at

t = 3

seconds.

(The maximum displacement is

x(3) = -2(9) + 12(3) - 10 = -18 + 36 - 10 = 8

metres.)