Classifying Stationary Points (Cambridge International IGCSE Mathematics)

Classifying stationary points

What are the different types of stationary points?

• You need to know two different types of stationary points:

  • Maximum points (this is where the graph reaches a “peak”)

  • Minimum points (this is where the graph reaches a “trough”)

    • These are also called turning points
    • Deciding which is which is called classifying them (or finding the nature of them)

• If a graph has multiple turning points, the one you are interested in can be described as a local maximum or minimum point

  • Other parts of the graph may still reach higher/lower values elsewhere

How do I classify turning points using graphs?

• You can use the shape of common graphs to classify different turning points

• To classify the turning point on a quadratic curve (parabola), remember that:

  • A positive quadratic (positive x² term) must have a minimum point
  • A negative quadratic (negative x² term) must have a maximum point

• To classify two different turning points on a cubic curve, remember that:

  • A positive cubic has a maximum point on the left and a minimum on the right
  • A negative cubic has a minimum point on the left and a maximum on the right

How do I classify turning points using the gradient function (derivative)?

• An alternative method if no graph is available is to use the gradient function (derivative) $\frac{dy}{dx}$

  • By looking at the gradient just before and just after a turning point, you can determine if it is a maximum or a minimum
  • Find the gradient just before and just after a turning point by substituting values near to the value of x into the gradient function
  • It helps to imagine what the tangents to the curve would look like

How do I classify turning points using the second derivative?

• Another method is to use the second derivative, $\frac{d^{2}y}{dx^{2}}$, which is the derivative-of-the-derivative

  • $\frac{d^{2}y}{dx^{2}}$ is pronounced "d-two-y by dx-squared"

• Differentiate the equation once to get $\frac{dy}{dx}$ then again to get $\frac{d^{2}y}{dx^{2}}$

  • This is the same as differentiating the original equation for $y$ twice

• If the stationary point is at $x=a$, substitute $x=a$ into the expression for $\frac{d^{2}y}{dx^{2}}$ to get a numerical value

  • If the result is negative, $\frac{d^{2}y}{dx^{2}}<0$, then the stationary point is a maximum point

    • This is often the opposite of what people expect!

  • If the result is positive, $\frac{d^{2}y}{dx^{2}}>0$, then the stationary point is a minimum point

  • If the value is zero, $\frac{d^{2}y}{dx^{2}}=0$, then the test has failed

    • You will need to try another method to classify the turning point

Worked Example The turning points on the curve with equation $y=2x^{3}+3x^{2}-12x+1$ are (-2, 21) and (1, -6).

Determine the nature of these turning points.

Answer:

Method 1: Using the graph

Make a quick sketch of the curve It is a positive cubic so you know that it will start at the bottom left, have two turning points and go towards the top right You know the coordinates of the turning points and the y-intercept: (0, 1) (the sketch does not have to be perfect!)

Final answer: **From the graph, which is a positive cubic, ** (-2, 21) is to the left of (1, -6)

Final answer: Therefore, (-2, 21) is a maximum point and (1, -6) is a minimum point

Method 2: Differentiation (1^st** derivative)**

Differentiate the original equation to find the 1^st derivative, the gradient function, $\frac{dy}{dx}$

$\begin{gathered} 2 x^{3} & \rightarrow 6 x^{2} \\ 3 x^{2} & \rightarrow 6 x \\ - 12 x & \rightarrow - 12 \\ 1 \rightarrow 0 \end{gathered}$

$\frac{dy}{dx}=6x^{2}+6x-12$

Find the gradient just before and just after each turning point by substituting a suitable x-value into the gradient function

Turning point	Gradient before	Gradient after
(-2, 21)	sub x *= -3 *6(-3)2* + 6(-3) - 12 = 24*	sub x = -1 6(-1)2 + 6(-1) - 12 = -12
(1, -6)	sub x *= 0 *6(0)2* + 6(0) - 12 = -12*	sub x *= 2 *6(2)2* + 6(2) - 12 = 24*

Final answer: **The gradient just before (-2, 21) is positive and the gradient just after is negative, therefore the turning point here is a maximum point **

Final answer: The gradient just before (1, -6) is negative and the gradient just after is positive, therefore the turning point here is a minimum point

Method 3: Differentiation (2^nd** derivative)**

Differentiate the original equation to find the 1^st derivative

$\begin{gathered} 2 x^{3} & \rightarrow 6 x^{2} \\ 3 x^{2} & \rightarrow 6 x \\ - 12 x & \rightarrow - 12 \\ 1 \rightarrow 0 \end{gathered}$

$\frac{dy}{dx}=6x^{2}+6x-12$

Differentiate again to find the 2^nd derivative

$\begin{gathered} 6 x^{2} & \rightarrow 12 x \\ 6 x & \rightarrow 6 \\ - 12 & \rightarrow 0 \end{gathered}$

$\frac{d^{2}y}{dx^{2}}=12x+6$

Find the nature of each turning point by substituting the value of its x-coordinate into the 2^nd derivative A negative result indicates a maximum point (or positive for a minimum point)

For (-2, 21): $\frac{ \mathrm{d}^{2} y}{ \mathrm{d} x^{2}} = 12 ( - 2 ) + 6 = - 18$

Final answer: The second derivative is negative, therefore the turning point at (-2, 21) is a maximum point

For (1, -6): $\frac{ \mathrm{d}^{2} y}{ \mathrm{d} x^{2}} = 12 ( 1 ) + 6 = 18$

Final answer: The second derivative is positive, therefore the turning point at (1, -6) is a minimum point