Stationary Points & Turning Points (Edexcel IGCSE Further Maths)

Spec point — 9D

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,$\frac{dy}{dx}=f'(x)$

• 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

Spec point — 9D

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 $\mathrm{f} ( x )$ 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 $\mathrm{f} ( x )$ 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 $\mathrm{f}^{'} ( x ) = 0$ there

What is the second derivative?

• It is possible to differentiate a function more than once

  • If you differentiate $y = \mathrm{f} ( x )$, you find its (first) derivative $\frac{ \mathrm{d} y}{ \mathrm{d} x} = \mathrm{f}^{'} ( x )$

    • e.g. the (first) derivative of $2x^{3}$ is $6x^{2}$

  • Then you can differentiate the derivative again to find the second derivative$\frac{ \mathrm{d}^{2} y}{ \mathrm{d} x^{2}} = \mathrm{f}^{''} ( x )$

    • e.g. the derivative of $6x^{2}$ is $12x$
    • so the second derivative of $2x^{3}$ is $12x$

• 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 $\mathrm{f}^{''} ( x )$

  • Do this by differentiating $\mathrm{f}^{'} ( x )$ from Step 1

• STEP 3 Find the value of $\mathrm{f}^{''} ( x )$ at each of the stationary points

  • i.e., by substituting the $x$-coordinate of each point into $\mathrm{f}^{''} ( x )$

• 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

Exam Hint

• 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 $f(x)=2x^{3}-3x^{2}-36x+25$.

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

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

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

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

$x=-2\mathrm{or}x=3$

Substitute into $\mathrm{f} ( x )$ to find the corresponding $y$-coordinates

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

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

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

Method 1: using the second derivative to test the points

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

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

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

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

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

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

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

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

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

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

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

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

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