Stationary Points & Turning Points (Edexcel IGCSE Further Pure Maths): Revision Note

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 $y=\mathrm{f}\left(x\right)$, stationary points can be found using the following process
  

• STEP 1
  Find the derivative,$\frac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{f}\text{'}\left(x\right)$
  

• STEP 2
  Solve the equation ${\mathrm{f}}^{\text{'}}\left(x\right)=0$ 

  • The solution(s) are the $\mathit{x}$-coordinate(s) of any stationary points

    • Remember, 'stationary points' includes turning points
      

• STEP 3
  Find the corresponding $\mathit{y}$-coordinates (if necessary)

  • Substitute each$x$-coordinate(s) into$\mathrm{f}\left(x\right)$

  • 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. $\mathrm{f}\text{'}\left(x\right)=0$

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

• A local minimum point,$(x,\mathrm{f}(x\left)\right)$ 

  • will have the lowest value of $\mathrm{f}\left(x\right)$ in the local vicinity of the $x$ value

  • But $\mathrm{f}\left(x\right)$ may reach a lower value further away

• A local maximum point, $(x,\mathrm{f}(x\left)\right)$ 

  • will have the highest value of $\mathrm{f}\left(x\right)$ in the local vicinity of the $x$ value

  • But $\mathrm{f}\left(x\right)$ may reach a higher value further away

• A global maximum (or global minimum) point

  • has the highest (or lowest) value of$\mathrm{f}\left(x\right)$ for all values of$x$ in the domain of $\mathrm{f}$

    • 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$\mathrm{f}\left(x\right)$ …
  

• STEP 1
  Find ${\mathrm{f}}^{\text{'}}\left(x\right)$ and solve ${\mathrm{f}}^{\text{'}}\left(x\right)=0$

  • 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 ${\mathrm{f}}^{\text{'}}(x-h)$ and ${\mathrm{f}}^{\text{'}}(x+h)$ 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

  • ${\mathrm{f}}^{\text{'}}(x-h)<0,{\mathrm{f}}^{\text{'}}\left(x\right)=0,{\mathrm{f}}^{\text{'}}(x+h)>0$

• At a local maximum point 

  • the derivative changes from positive to negative

  • ${\mathrm{f}}^{\text{'}}(x-h)>0,{\mathrm{f}}^{\text{'}}\left(x\right)=0,{\mathrm{f}}^{\text{'}}(x+h)<0$

    

• 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 ${\mathrm{f}}^{\text{'}}\left(x\right)=0$ there

What is the second derivative?

• It is possible to differentiate a function more than once

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

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

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

    • 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 ${\mathrm{f}}^{\text{'}}\left(x\right)$ and solve ${\mathrm{f}}^{\text{'}}\left(x\right)=0$

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

• STEP 2
  Find the second derivative ${\mathrm{f}}^{\text{'}\text{'}}\left(x\right)$

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

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

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

• If  ${\mathbf{f}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\left(x\right)\mathbf{>}\mathbf{0}$
  

  • then the stationary point is a local minimum

• If  ${\mathbf{f}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\left(x\right)\mathbf{<}\mathbf{0}$ 

  • then the stationary point is a local maximum

• If  ${\mathbf{f}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\left(x\right)\mathbf{=}\mathbf{0}$

  • 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