Stationary Points & Turning Points (Edexcel IGCSE Further Pure Maths)

Revision Note

Written by: Paul

Reviewed by: Dan Finlay

Stationary Points & Turning Points

What is the difference between a stationary point and a turning point?

A stationary point is a point at which the derivative of a function is equal to zero
- The tangent to the curve of the function is horizontal (gradient = 0)
A turning point is a point at which
- the derivative of a function is equal to zero
- AND the derivative changes sign (from negative to positive, or positive to negative)
  - i.e. the curve changes from ‘going upwards’ to ‘going downwards’ (or vice versa)
- Turning points will either be (local) minimum or (local) maximum points
All turning points are also stationary points
- But not all stationary points are turning points

How do I find stationary points and turning points?

For the function $y = f (x)$ , stationary points can be found using the following process
STEP 1
Find the derivative, $\frac{d y}{d x} = f' (x)$
STEP 2
Solve the equation $f^{'} (x) = 0$
- The solution(s) are the $x$ -coordinate(s) of any stationary points
  - Remember, 'stationary points' includes turning points
STEP 3
Find the corresponding $y$ -coordinates (if necessary)
- Substitute each $x$ -coordinate(s) into $f (x)$
- If the question only asks for the $x$ -coordinates, you can skip this step!
More work is needed to find if a stationary point is a turning point
- i.e. if it is a (local) maximum or (local) minimum
- See the following note

Testing for Local Minimum & Maximum Points

What are local minimum and maximum points?

Local minimum and maximum points are two types of stationary point
- The derivative at stationary points equals zero
  - i.e. $f' (x) = 0$
- But not all points with a zero derivative are maximum or minimum points
A local minimum point, $(x, f (x))$
- will have the lowest value of $f (x)$ in the local vicinity of the $x$ value
- But $f (x)$ may reach a lower value further away
A local maximum point, $(x, f (x))$
- will have the highest value of $f (x)$ in the local vicinity of the $x$ value
- But $f (x)$ may reach a higher value further away
A global maximum (or global minimum) point
- has the highest (or lowest) value of $f (x)$ for all values of $x$ in the domain of $f$
  - A global maximum (or minimum) may not be a local maximum (or minimum)
  - and vice versa
- Not all functions have a global maximum (or minimum)
  - e.g. some functions tend to $\pm$ infinity as $x$ tends to $\pm$ infinity
  - or as $x$ approaches some other value (vertical asymptotes)

How can I use the derivative to identify local minimum & maximum points?

The nature of a stationary point can be determined using the derivative
For the function $f (x)$ …
STEP 1
Find $f^{'} (x)$ and solve $f^{'} (x) = 0$
- The solutions are the the $x$ -coordinates of any stationary points
STEP 2
Find the sign of the derivative just either side of each stationary point
- i.e. evaluate $f^{'} (x - h)$ and $f^{'} (x + h)$ for small $h$
  - Choose a convenient value for $h$
  - Just make sure you don't 'jump over' any other stationary points!

At a local minimum point
- the derivative changes from negative to positive
- $f^{'} (x - h) < 0, f^{'} (x) = 0, f^{'} (x + h) > 0$
At a local maximum point
- the derivative changes from positive to negative
- $f^{'} (x - h) > 0, f^{'} (x) = 0, f^{'} (x + h) < 0$
If the derivative does not change sign at the point
- (i.e. if it is positive on both sides or negative on both sides)
- then the point is not a local minimum or a local maximum
- But it is still a stationary point if $f^{'} (x) = 0$ there

What is the second derivative?

It is possible to differentiate a function more than once
- If you differentiate $y = f (x)$ , you find its (first) derivative $\frac{d y}{d x} = f^{'} (x)$
  - e.g. the (first) derivative of $2 x^{3}$ is $6 x^{2}$
- Then you can differentiate the derivative again to find the second derivative $\frac{d^{2} y}{d x^{2}} = f^{''} (x)$
  - e.g. the derivative of $6 x^{2}$ is $12 x$
  - so the second derivative of $2 x^{3}$ is $12 x$
The second derivative can be used to test the nature of a stationary point

How can I use the second derivative to identify local minimum & maximum points?

STEP 1
Find $f^{'} (x)$ and solve $f^{'} (x) = 0$
- The solutions are the the $x$ -coordinates of any stationary points
STEP 2
Find the second derivative $f^{''} (x)$
- Do this by differentiating $f^{'} (x)$ from Step 1
STEP 3
Find the value of $f^{''} (x)$ at each of the stationary points
- i.e., by substituting the $x$ -coordinate of each point into $f^{''} (x)$

If $f^{''} (x) > 0$
- then the stationary point is a local minimum
If $f^{''} (x) < 0$
- then the stationary point is a local maximum
If $f^{''} (x) = 0$
- then the test does not tell you anything
- the stationary point may be a local maximum, a local minimum, or neither
- In this case, use the first derivative test instead

Examiner Tips and Tricks

Don't just assume that a zero derivative corresponds to a maximum or minimum point
- Especially if a question asks you to justify that a point is a maximum or minimum
The second derivative test is usually much quicker for identifying maximum and minimum points
- But it is good to understand how to use the first derivative test as well
Your calculator may be able to show graphs of functions
- You can use this to check your work

Worked Example

Find the coordinates and determine the nature of any stationary points on the graph of $y = f (x)$ , where $f$ is the function defined by $f (x) = 2 x^{3} - 3 x^{2} - 36 x + 25$ .

Start by finding $f^{'} (x)$

$f^{'} (x) = 6 x^{2} - 6 x - 36$

Solve $f^{'} (x) = 0$ to find the $x$ -coordinates of any stationary points

$\begin{array}{rcl} 6 x^{2} - 6 x - 36 & = & 0 \\ 6 (x^{2} - x - 6) & = & 0 \\ x^{2} - x - 6 & = & 0 \\ (x - 3) (x + 2) & = & 0 \end{array}$

$x = - 2 or x = 3$

Substitute into $f (x)$ to find the corresponding $y$ -coordinates

$\begin{array}{rcl} f (- 2) & = & 2 {(- 2)}^{3} - 3 {(- 2)}^{2} - 36 (- 2) + 25 \\ = & - 16 - 12 + 72 + 25 \\ = & 69 \end{array}$

$\begin{array}{rcl} f (3) & = & 2 {(3)}^{3} - 3 {(3)}^{2} - 36 (3) + 25 \\ = & 54 - 27 - 108 + 25 \\ = & - 56 \end{array}$

So the stationary points are $(- 2 . 69)$ and $(3, - 56)$

Method 1: using the second derivative to test the points

Differentiate $f^{'} (x)$ to find the second derivative $f^{''} (x)$

$\begin{array}{rcl} f^{''} (x) & = & \frac{d}{d x} (6 x^{2} - 6 x - 36) \\ = & 12 x - 6 \end{array}$

Now substitute in the $x$ -coordinates of the two stationary points
This will give the value of the second derivative at those points

$f^{''} (- 2) = 12 (- 2) - 6 = - 30 < 0$

That is less than zero, so $(- 2, 69)$ is a local maximum point

$f^{''} (3) = 12 (3) - 6 = 30 > 0$

That is greater than zero, so $(3, - 56)$ is a local minimum point

Method 2: using the first derivative to test the points

Test the values of $f^{'} (x)$ at either side of the stationary points
(Note that, where applicable, $x = 0$ is a very convenient test value!)

$\begin{array}{rcl} f^{'} (- 3) & = & 6 {(- 3)}^{2} - 6 (- 3) - 36 \\ = & 54 + 18 - 36 \\ = & 36 \\ > & 0 \end{array}$

$\begin{array}{rcl} f^{'} (0) & = & 6 {(0)}^{2} - 6 (0) - 36 \\ = & 0 + 0 - 36 \\ = & - 36 \\ < & 0 \end{array}$

$\begin{array}{rcl} f^{'} (4) & = & 6 {(4)}^{2} - 6 (4) - 36 \\ = & 96 - 24 - 36 \\ = & 36 \\ > & 0 \end{array}$

At $(- 2, 69)$ the derivative changes from positive to negative, so that is a local maximum point

At $(3, - 56)$ the derivative changes from negative to positive, so that is a local minimum point

Note that for this function both stationary points are also turning points

$(- 2, 69)$ is a local maximum point

$(3, - 56)$ is a local minimum point

Last updated: 16 October 2024

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