Extreme Points, Maximum and Minimum Turning Points

Understanding Extreme Points

In everyday life, we often look for the "best" value, like the biggest profit, the smallest cost, or the shortest distance. In mathematics, these optimal values are known as extreme values, which consist of maximum values and minimum values. The points where these values occur are called extreme points.

These extreme points are closely related to stationary points, which are points where the gradient of the curve is zero ( $f'(x)=0$ ). There are two main ways to test and determine the type of a stationary point: the First Derivative Test and the Second Derivative Test.

First Derivative Test

This method focuses on the change in the function's behavior, i.e., whether it's increasing or decreasing, around a stationary point. By observing the sign of $f'(x)$ , we can identify the type of stationary point.

Maximum Turning Point: Occurs if the function changes from increasing to decreasing. This means the sign of $f'(x)$ changes from positive (+) to negative (-). Think of it as the peak of a hill.
Minimum Turning Point: Occurs if the function changes from decreasing to increasing. The sign of $f'(x)$ changes from negative (-) to positive (+). This is like the bottom of a valley.
Horizontal Inflection Point: Occurs if the function does not change direction (it continues to increase or decrease). The sign of $f'(x)$ does not change (positive to positive or negative to negative).

Take a look at the visualization below to understand the concept of turning points.

First Derivative Test Visualization

A visualization of the turning point concept. The horizontal helper lines show a zero gradient at the maximum and minimum points.

Let's apply this to an example. Determine the stationary values of the function $f(x) = x(x-3)^2$ .

Solution:

Step $1$ : Find the first derivative

First, let's expand the function: $f(x) = x(x^2 - 6x + 9) = x^3 - 6x^2 + 9x$ .

Its derivative is:

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

Step $2$ : Find the stationary points

Set $f'(x) = 0$ .

3x^2 - 12x + 9 = 0

The stationary points occur at $x = 1$ and $x = 3$ .

Step $3$ : Test the sign around the stationary points

Around $x = 1$ : For $x < 1$ (e.g., $x=0$ ), $f'(0) = 9$ (positive). For $x > 1$ (e.g., $x=2$ ), $f'(2) = -3$ (negative). Since the sign changes from (+) to (-), $x=1$ is a maximum turning point.
Around $x = 3$ : For $x < 3$ (e.g., $x=2$ ), $f'(2) = -3$ (negative). For $x > 3$ (e.g., $x=4$ ), $f'(4) = 9$ (positive). Since the sign changes from (-) to (+), $x=3$ is a minimum turning point.

Step $4$ : Determine the stationary values

To get the turning point values ( $y$ -values), substitute the stationary point $x$ -coordinates ( $x = 1$ and $x = 3$ ) back into the original function $f(x)$ .

Maximum value: $f(1) = 1(1-3)^2 = 4$ .
Minimum value: $f(3) = 3(3-3)^2 = 0$ .

So, the maximum turning point value of the function is $4$ , which occurs at the point $(1,4)$ , and its minimum turning point value is $0$ , which occurs at the point $(3,0)$ .

Extreme Points Visualization

A visualization of the curve

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

with horizontal tangent lines at its maximum and minimum points.

Second Derivative Test

This method is often faster because it doesn't require testing intervals. It uses the second derivative, $f''(x)$ , to determine the concavity of the curve at a stationary point.

Suppose $f'(c) = 0$ .

If $f''(c) < 0$ , then $f(c)$ is a maximum turning point value. This means the curve is concave down at that point, like an upside-down bowl.
If $f''(c) > 0$ , then $f(c)$ is a minimum turning point value. This means the curve is concave up at that point, like an open bowl.
If $f''(c) = 0$ , this test fails, and we must go back to using the First Derivative Test. This point is likely an inflection point.

Determine the extreme values of $f(x) = \frac{1}{3}x^3 - x^2 - 3x + 2$ using the second derivative test.

Solution:

Step $1$ : Find the first and second derivatives

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

Step $2$ : Find the stationary points

x^2 - 2x - 3 = 0

The stationary points occur at $x = 3$ and $x = -1$ .

Step $3$ : Test the stationary points with the second derivative

For $x = 3$ : $f''(3) = 2(3) - 2 = 4$ . Since $4 > 0$ , this is a minimum turning point value.
For $x = -1$ : $f''(-1) = 2(-1) - 2 = -4$ . Since $-4 < 0$ , this is a maximum turning point value.

Step $4$ : Calculate the extreme values

Substitute $x=3$ and $x=-1$ into the original function to get their extreme values.

Minimum value: $f(3) = \frac{1}{3}(3)^3 - (3)^2 - 3(3) + 2 = 9 - 9 - 9 + 2 = -7$ .
Maximum value: $f(-1) = \frac{1}{3}(-1)^3 - (-1)^2 - 3(-1) + 2 = -\frac{1}{3} - 1 + 3 + 2 = \frac{11}{3}$ .

If we look at the visualization below, the curve is concave down at the point $x=-1$ and concave up at the point $x=3$ .

Second Derivative Test Visualization

A graph of

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

. The helper lines mark the maximum point, where the curve is concave down, and the minimum point, where the curve is concave up. The curve is concave down at the point

x=-1

and concave up at the point

x=3

Exercises

Determine the stationary point for the function $y = x^2 + 2x - 3$ .

Answer Key

Solution:

Step $1$ : Find the first derivative

$f'(x) = 2x + 2$

Step $2$ : Find the stationary point

Set $f'(x) = 0$ .

$2x + 2 = 0 \implies x = -1$

Step $3$ : Find the y-value

Substitute $x = -1$ into the original function.

$y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$

So, the stationary point is $(-1, -4)$ . With the Second Derivative Test ( $f''(x)=2 > 0$ ), this point is a minimum.

Understanding Extreme Points

First Derivative Test

Maximum Turning Point: Occurs if the function changes from increasing to decreasing. This means the sign of $f'(x)$ changes from positive (+) to negative (-). Think of it as the peak of a hill.
Minimum Turning Point: Occurs if the function changes from decreasing to increasing. The sign of $f'(x)$ changes from negative (-) to positive (+). This is like the bottom of a valley.
Horizontal Inflection Point: Occurs if the function does not change direction (it continues to increase or decrease). The sign of $f'(x)$ does not change (positive to positive or negative to negative).

Take a look at the visualization below to understand the concept of turning points.

First Derivative Test Visualization

A visualization of the turning point concept. The horizontal helper lines show a zero gradient at the maximum and minimum points.

Let's apply this to an example. Determine the stationary values of the function $f(x) = x(x-3)^2$ .

Solution:

Step $1$ : Find the first derivative

First, let's expand the function: $f(x) = x(x^2 - 6x + 9) = x^3 - 6x^2 + 9x$ .

Its derivative is:

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

Step $2$ : Find the stationary points

Set $f'(x) = 0$ .

3x^2 - 12x + 9 = 0

The stationary points occur at $x = 1$ and $x = 3$ .

Step $3$ : Test the sign around the stationary points

Around $x = 1$ : For $x < 1$ (e.g., $x=0$ ), $f'(0) = 9$ (positive). For $x > 1$ (e.g., $x=2$ ), $f'(2) = -3$ (negative). Since the sign changes from (+) to (-), $x=1$ is a maximum turning point.
Around $x = 3$ : For $x < 3$ (e.g., $x=2$ ), $f'(2) = -3$ (negative). For $x > 3$ (e.g., $x=4$ ), $f'(4) = 9$ (positive). Since the sign changes from (-) to (+), $x=3$ is a minimum turning point.

Step $4$ : Determine the stationary values

To get the turning point values ( $y$ -values), substitute the stationary point $x$ -coordinates ( $x = 1$ and $x = 3$ ) back into the original function $f(x)$ .

Maximum value: $f(1) = 1(1-3)^2 = 4$ .
Minimum value: $f(3) = 3(3-3)^2 = 0$ .

So, the maximum turning point value of the function is $4$ , which occurs at the point $(1,4)$ , and its minimum turning point value is $0$ , which occurs at the point $(3,0)$ .

Extreme Points Visualization

A visualization of the curve

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

with horizontal tangent lines at its maximum and minimum points.

Second Derivative Test

This method is often faster because it doesn't require testing intervals. It uses the second derivative, $f''(x)$ , to determine the concavity of the curve at a stationary point.

Suppose $f'(c) = 0$ .

If $f''(c) < 0$ , then $f(c)$ is a maximum turning point value. This means the curve is concave down at that point, like an upside-down bowl.
If $f''(c) > 0$ , then $f(c)$ is a minimum turning point value. This means the curve is concave up at that point, like an open bowl.
If $f''(c) = 0$ , this test fails, and we must go back to using the First Derivative Test. This point is likely an inflection point.

Determine the extreme values of $f(x) = \frac{1}{3}x^3 - x^2 - 3x + 2$ using the second derivative test.

Solution:

Step $1$ : Find the first and second derivatives

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

Step $2$ : Find the stationary points

x^2 - 2x - 3 = 0

The stationary points occur at $x = 3$ and $x = -1$ .

Step $3$ : Test the stationary points with the second derivative

For $x = 3$ : $f''(3) = 2(3) - 2 = 4$ . Since $4 > 0$ , this is a minimum turning point value.
For $x = -1$ : $f''(-1) = 2(-1) - 2 = -4$ . Since $-4 < 0$ , this is a maximum turning point value.

Step $4$ : Calculate the extreme values

Substitute $x=3$ and $x=-1$ into the original function to get their extreme values.

Minimum value: $f(3) = \frac{1}{3}(3)^3 - (3)^2 - 3(3) + 2 = 9 - 9 - 9 + 2 = -7$ .
Maximum value: $f(-1) = \frac{1}{3}(-1)^3 - (-1)^2 - 3(-1) + 2 = -\frac{1}{3} - 1 + 3 + 2 = \frac{11}{3}$ .

If we look at the visualization below, the curve is concave down at the point $x=-1$ and concave up at the point $x=3$ .

Second Derivative Test Visualization

A graph of

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

. The helper lines mark the maximum point, where the curve is concave down, and the minimum point, where the curve is concave up. The curve is concave down at the point

x=-1

and concave up at the point

x=3

Exercises

Determine the stationary point for the function $y = x^2 + 2x - 3$ .

Answer Key

Solution:

Step $1$ : Find the first derivative

$f'(x) = 2x + 2$

Step $2$ : Find the stationary point

Set $f'(x) = 0$ .

$2x + 2 = 0 \implies x = -1$

Step $3$ : Find the y-value

Substitute $x = -1$ into the original function.

$y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$

So, the stationary point is $(-1, -4)$ . With the Second Derivative Test ( $f''(x)=2 > 0$ ), this point is a minimum.