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

Revision Note

Paul

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 equals straight f left parenthesis x right parenthesisstationary points can be found using the following process

  • STEP 1
    Find the derivative,space fraction numerator straight d y over denominator straight d x end fraction equals straight f apostrophe left parenthesis x right parenthesis

  • STEP 2
    Solve the equation space straight f to the power of apostrophe left parenthesis x right parenthesis equals 0 

    • The solution(s) are the bold italic x-coordinate(s) of any stationary points

      • Remember, 'stationary points' includes turning points

  • STEP 3
    Find the corresponding bold italic y-coordinates (if necessary)

    • Substitute eachspace x-coordinate(s) intospace straight f left parenthesis x right parenthesis

    • 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. straight f apostrophe left parenthesis x right parenthesis equals 0

    • But not all points with a zero derivative are maximum or minimum points

  • A local minimum point,space left parenthesis x comma space straight f left parenthesis x right parenthesis right parenthesis 

    • will have the lowest value of straight f left parenthesis x right parenthesis in the local vicinity of the x value

    • But straight f open parentheses x close parentheses may reach a lower value further away

  • A local maximum point, left parenthesis x comma space straight f left parenthesis x right parenthesis right parenthesis 

    • will have the highest value of straight f left parenthesis x right parenthesis in the local vicinity of the x value

    • But straight f open parentheses x close parentheses may reach a higher value further away

  • A global maximum (or global minimum) point

    • has the highest (or lowest) value ofspace straight f left parenthesis x right parenthesis for all values ofspace x in the domain of straight 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 plus-or-minusinfinity as x tends to plus-or-minusinfinity

      • 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 functionspace straight f left parenthesis x right parenthesis

  • STEP 1
    Find straight f to the power of apostrophe left parenthesis x right parenthesis and solve straight f to the power of apostrophe left parenthesis x right parenthesis equals 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 straight f to the power of apostrophe left parenthesis x minus h right parenthesis and straight f to the power of apostrophe left parenthesis x plus h right parenthesis for smallspace 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

    • straight f to the power of apostrophe left parenthesis x minus h right parenthesis less than 0 comma space space straight f to the power of apostrophe left parenthesis x right parenthesis equals 0 comma space space straight f to the power of apostrophe left parenthesis x plus h right parenthesis greater than 0

  • At a local maximum point 

    • the derivative changes from positive to negative

    • straight f to the power of apostrophe left parenthesis x minus h right parenthesis greater than 0 comma space space straight f to the power of apostrophe left parenthesis x right parenthesis equals 0 comma space space straight f to the power of apostrophe left parenthesis x plus h right parenthesis less than 0

      Stationary points of a cubic graph
  • 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 straight f to the power of apostrophe open parentheses x close parentheses equals 0 there

What is the second derivative?

  • It is possible to differentiate a function more than once

    • If you differentiate y equals straight f open parentheses x close parentheses, you find its (first) derivative  fraction numerator straight d y over denominator straight d x end fraction equals straight f to the power of apostrophe open parentheses x close parentheses

      • e.g. the (first) derivative of 2 x cubed is 6 x squared

    • Then you can differentiate the derivative again to find the second derivative  fraction numerator straight d squared y over denominator straight d x squared end fraction equals straight f to the power of apostrophe apostrophe end exponent open parentheses x close parentheses

      • e.g. the derivative of 6 x squared is 12 x

      • so the second derivative of 2 x cubed 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 straight f to the power of apostrophe left parenthesis x right parenthesis and solve straight f to the power of apostrophe left parenthesis x right parenthesis equals 0

    • The solutions are the the x-coordinates of any stationary points

  • STEP 2
    Find the second derivative straight f to the power of apostrophe apostrophe end exponent open parentheses x close parentheses

    • Do this by differentiating straight f to the power of apostrophe open parentheses x close parentheses from Step 1

  • STEP 3
    Find the value of straight f to the power of apostrophe apostrophe end exponent open parentheses x close parentheses at each of the stationary points

    • i.e., by substituting the x-coordinate of each point into straight f to the power of apostrophe apostrophe end exponent open parentheses x close parentheses  

  • If  bold f to the power of bold apostrophe bold apostrophe end exponent stretchy left parenthesis x stretchy right parenthesis bold greater than bold 0

    • then the stationary point is a local minimum

  • If  bold f to the power of bold apostrophe bold apostrophe end exponent stretchy left parenthesis x stretchy right parenthesis bold less than bold 0 

    • then the stationary point is a local maximum

  • If  bold f to the power of bold apostrophe bold apostrophe end exponent begin bold style stretchy left parenthesis x stretchy right parenthesis end style bold equals bold 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 equals straight f left parenthesis x right parenthesis, where straight f is the function defined by  straight f left parenthesis x right parenthesis equals 2 x cubed minus 3 x squared minus 36 x plus 25.

Start by finding straight f to the power of apostrophe open parentheses x close parentheses

straight f to the power of apostrophe open parentheses x close parentheses equals 6 x squared minus 6 x minus 36


Solve straight f to the power of apostrophe open parentheses x close parentheses equals 0 to find the x-coordinates of any stationary points

table row cell 6 x squared minus 6 x minus 36 end cell equals 0 row cell 6 open parentheses x squared minus x minus 6 close parentheses end cell equals 0 row cell x squared minus x minus 6 end cell equals 0 row cell open parentheses x minus 3 close parentheses open parentheses x plus 2 close parentheses end cell equals 0 end table

x equals negative 2 space space space or space space space x equals 3


Substitute into straight f open parentheses x close parentheses to find the corresponding y-coordinates

table row cell straight f open parentheses negative 2 close parentheses end cell equals cell 2 open parentheses negative 2 close parentheses cubed minus 3 open parentheses negative 2 close parentheses squared minus 36 open parentheses negative 2 close parentheses plus 25 end cell row blank equals cell negative 16 minus 12 plus 72 plus 25 end cell row blank equals 69 end table

table row cell straight f open parentheses 3 close parentheses end cell equals cell 2 open parentheses 3 close parentheses cubed minus 3 open parentheses 3 close parentheses squared minus 36 open parentheses 3 close parentheses plus 25 end cell row blank equals cell 54 minus 27 minus 108 plus 25 end cell row blank equals cell negative 56 end cell end table


So the stationary points are open parentheses negative 2. space 69 close parentheses and open parentheses 3 comma space minus 56 close parentheses

Method 1: using the second derivative to test the points

Differentiate straight f to the power of apostrophe open parentheses x close parentheses to find the second derivative straight f to the power of apostrophe apostrophe end exponent open parentheses x close parentheses

table row cell straight f to the power of apostrophe apostrophe end exponent open parentheses x close parentheses end cell equals cell fraction numerator straight d over denominator straight d x end fraction open parentheses 6 x squared minus 6 x minus 36 close parentheses end cell row blank equals cell 12 x minus 6 end cell end table

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

straight f to the power of apostrophe apostrophe end exponent open parentheses negative 2 close parentheses equals 12 open parentheses negative 2 close parentheses minus 6 equals negative 30 less than 0

That is less than zero, so open parentheses negative 2 comma space 69 close parentheses is a local maximum point

straight f to the power of apostrophe apostrophe end exponent open parentheses 3 close parentheses equals 12 open parentheses 3 close parentheses minus 6 equals 30 greater than 0


That is greater than zero, so open parentheses 3 comma space minus 56 close parentheses is a local minimum point


Method 2: using the first derivative to test the points

Test the values of straight f to the power of apostrophe open parentheses x close parentheses at either side of the stationary points
(Note that, where applicable, x equals 0 is a very convenient test value!)

table row cell straight f to the power of apostrophe open parentheses negative 3 close parentheses end cell equals cell 6 open parentheses negative 3 close parentheses squared minus 6 open parentheses negative 3 close parentheses minus 36 end cell row blank equals cell 54 plus 18 minus 36 end cell row blank equals 36 row blank greater than 0 end table


table row cell straight f to the power of apostrophe open parentheses 0 close parentheses end cell equals cell 6 open parentheses 0 close parentheses squared minus 6 open parentheses 0 close parentheses minus 36 end cell row blank equals cell 0 plus 0 minus 36 end cell row blank equals cell negative 36 end cell row blank less than 0 end table


table attributes columnalign right center left columnspacing 0px end attributes row cell straight f to the power of apostrophe open parentheses 4 close parentheses end cell equals cell 6 open parentheses 4 close parentheses squared minus 6 open parentheses 4 close parentheses minus 36 end cell row blank equals cell 96 minus 24 minus 36 end cell row blank equals 36 row blank greater than 0 end table

At  open parentheses negative 2 comma space 69 close parentheses  the derivative changes from positive to negative, so that is a local maximum point

At  open parentheses 3 comma space minus 56 close parentheses  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


stretchy left parenthesis negative 2 comma space 69 stretchy right parenthesis  is a local maximum point

stretchy left parenthesis 3 comma space minus 56 stretchy right parenthesis  is a local minimum point 

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Paul

Author: Paul

Expertise: Maths Content Creator (Previous)

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams – one of the many reasons he is excited to be a member of the SME team.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.