13: Stationary Points13: Stationary Points

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 1: AS Core Vol. 1: AS Core ModulesModules

Stationary Points

xxxy 93 23

0dx

dy

The stationary points of a curve are the points where the gradient is zero

A local maximum

A local minimum

x

x

The word local is usually omitted and the points called maximum and minimum points.

e.g.

Stationary Points

e.g.1 Find the coordinates of the stationary points on the curve xxxy 93 23

0dx

dy

Solution:

xxxy 93 23

dx

dy963 2 xx

0)32(3 2 xx

0)1)(3(3 xx or

3x 1x yx 3

272727 yx 1 )1(9)1(3)1( 23

)3(9)3(3)3( 23

The stationary points are (3, -27) and ( -1, 5)

931

27

5

0963 2 xxTip: Watch out for common factors when finding stationary points.

Stationary PointsExercise

sFind the coordinates of the stationary points of the following functions

542 xxy1. 2. 11232 23 xxxy

Ans: St. pt. is ( 2, 1)

Solutions:

0420 xdx

dy

2 x

15)2(4)2(2 2 yx

42 xdx

dy1.

Stationary Points

2. 11232 23 xxxy

21 xx or

61 yx

211)2(12)2(3)2(22 23 yx

1266 2 xxdx

dySolution:

0)2(60 2 xxdx

dy

Ans: St. pts. are ( 1, 6) and ( 2, 21 )

0)2)(1(6 xx

Stationary Points

On the left of a maximum, the gradient is positive

We need to be able to determine the nature of a stationary point ( whether it is a max or a min ). There are several ways of doing this. e.g.

On the right of a maximum, the gradient is negative

Stationary Points

So, for a max the gradients are

0

The opposite is true for a minimum

0

At the max

On the right of the max

On the left of the max

Calculating the gradients on the left and right of a stationary point tells us whether the point is a

max or a min.

Stationary Points

Solution:

42 xdx

dy

0420 xdx

dy

1)2(4)2( 2 y

2 x

142 xxy )1(

On the left of x = 2 e.g. at x = 1,

3 y

24)1(2 dx

dy

On the right of x = 2 e.g. at x = 3,

24)3(2 dx

dy0

0

We have 0

)3,2( is a min

Substitute in (1):

e.g.2 Find the coordinates of the stationary point of the curve . Is the point a max or min?

142 xxy

Stationary Points

At the max of 1093 23 xxxy

dx

dy

but the gradient of the gradient is negative.

The gradient function is given by

963 2 xxdx

dy

1093 23 xxxy

e.g.3 Consider

the gradient is 0

Another method for determining the nature of a stationary point.

Stationary Points

The notation for the gradient of the gradient is

“d 2 y by d x squared”2

2

dx

yd

dx

dy

Another method for determining the nature of a stationary point.

The gradient function is given by

963 2 xxdx

dy

1093 23 xxxy

e.g.3 Consider

At the min of 1093 23 xxxythe gradient of the gradient is positive.

Stationary Points

66 x963 2 xx

e.g.3 ( continued ) Find the stationary points on the curve and distinguish between the max and the min.

1093 23 xxxy

2

2

dx

yd

Solution:

1093 23 xxxy

Stationary points: 0

dx

dy 0963 2 xx

0)32(3 2 xx0)1)(3(3 xx

1x3x or

dx

dy

We now need to find the y-coordinates of the st. pts.

is called the

2nd derivative2

2

dx

yd

Stationary Points

3x 10)3(9)3(3)3( 23 y 371x 5

126)3(6 max at )37,3(0

0 min at )5,1(

3xAt , 2

2

dx

yd

1266 1xAt , 2

2

dx

yd

10931 y

1093 23 xxxy

To distinguish between max and min we use the 2nd derivative, at the stationary points.

662

2

xdx

yd

Stationary PointsSUMMAR

Y To find stationary points, solve the equation

0dx

dy

0

maximum

0 minimu

m

Determine the nature of the stationary points

• either by finding the gradients on the left and right of the stationary points

• or by finding the value of the 2nd derivative at the stationary points

min 02

2

dx

ydmax 0

2

2

dx

yd

Stationary Points

ExercisesFind the coordinates of the stationary points

of the following functions, determine the nature of each and sketch the functions.

23 23 xxy1.

2. 332 xxy

)2,0( is a min.

)2,2( is a max.

Ans.

)0,1( is a min.

)4,1( is a max.

Ans.

23 23 xxy

332 xxy

Stationary Points