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Differentiation I

Stationary points

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Summary

Stationary points

In a nutshell

Some curves have stationary points, which are points where the gradient is equal to zero. The nature of a stationary point (whether it is a local maximum, local minimum or a point of inflection) can be determined by looking to the left and right of the stationary point. Alternatively, the second derivative can be used.

Finding a stationary point

Since stationary points are points on a curve where the gradient is zero, it follows that you can differentiate the equation of a curve, then set this derivative to zero before solving.

Example 1

Find the stationary points of the curve

$y=2x^3+3x^2-72x+1$ 

As a cubic, you know it can have up to two stationary points. Finding them begins with differentiating:

$\dfrac{\text dy}{\text dx}=6x^2+6x-72$ 

Since you are looking for points where the gradient is zero, set the derivative to zero and solve:

$0=6x^2+6x-72\\\space\\0=x^2+x-12\\\space\\0=(x+4)(x-3)$

$\begin{aligned} x&=3 \\ x&=4 \end{aligned}$ 

Inserting each of these back into the equation of the curve will find the corresponding $y$ -coordinates. Thus the stationary points are:

$\underline{(3,-134)}$ and $\underline{(-4,209)}$ 

The nature of a stationary point

Stationary points can be classified as either (local) maxima, (local) minima, or points of inflection. The "local" is to indicate that they may not be the highest or lowest point on the whole graph, but that in that local area, they are. Examples are given below:

Maths; Differentiation I; KS5 Year 12; Stationary points

A feature of a maximum is that the gradient of the curve just to the left of the stationary point is positive, while the gradient just to the right is negative. Conversely, just to the left of a minimum, the gradient is negative and just to the right, the gradient is positive.

A point of inflection is the third classification of stationary point. If just to the left of this stationary point the gradient of the curve is negative, then it will also be negative just to the right. Similarly, if the gradient of the curve is positive just to the left of the point, then it will also be positive just to the right. An example is given below at the point $(0,1)$ :

How to classify a stationary point

There are two methods which can be used to find the nature of a stationary point.

Substitution

Once you have the coordinates of a stationary point (in particular the $x$ -coordinate) you can check the sign of the gradient just to the left and just to the right by substituting suitable $x$ values.

Example 2

Classify the stationary point $(3,-134)$ of the curve $y=f(x)$ where

$f(x)=2x^3+3x^2-72+1$ 

From above you have that the derivative is

$f'(x)=6x^2+6x-72$ 

Checking the gradient just to the left ( $x=2.5$ ) and just to the right ( $x=3.5$ ):

$x$	$2.5$	$3$	$3.5$
$f'(x)$	$-19.5$	$0$	$22.5$

Hence to the left of the point the gradient is negative and to the right it is positive.

The stationary point is a minimum.

Second order derivatives

Another method to classify a stationary point is to use the second derivative. Insert the $x$ -coordinate of the stationary point into the equation for the second derivative and take note of the sign of the output. If the second derivative is positive, the stationary point is a minimum. If it is negative, it is a maximum. If it is equal to zero, then this is inconclusive. You may have any one of the three types of stationary point, and looking to the left and right of it will determine which.

Example 3

Classify the stationary point $(-4,209)$ of the curve $y=f(x)$ where

$f(x)=2x^3+3x^2-72+1$

From before you have found the first derivative:

$f'(x)=6x^2+6x-72$ 

Now find the second derivative:

$f''(x)=12x+6$ 

Insert the $x$ -coordinate into this equation:

$f''(-4)=12(-4)+6=-42$ 

Since this is negative, it follows that this stationary point is a maximum.

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