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

Module C1AQA Edexce

lOCR MEI/OCR

Module C2

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Stationary Points

xxxy 93 23

0dxdy

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

0dxdy

Solution: xxxy 93 23

dxdy 963 2 xx

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

r3x 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 xdxdy

2 x15)2(4)2(2 2 yx

42 xdxdy1.

Stationary Points

2. 11232 23 xxxy

21 xx or 61 yx

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

1266 2 xxdxdySolutio

n:

0)2(60 2 xxdxdy

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 PointsSo, 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 xdxdy

0420 xdxdy

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 dxdy

On the right of x = 2 e.g. at x = 3, 24)3(2

dxdy 0

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

dxdy

but the gradient of the gradient is negative.

The gradient function is given by

963 2 xxdxdy

1093 23 xxxye.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

dxyd

dxdy

Another method for determining the nature of a stationary point.

The gradient function is given by

963 2 xxdxdy

1093 23 xxxye.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

dxyd

Solution: 1093 23 xxxy

Stationary points: 0

dxdy

0963 2 xx

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

1x3x or

dxdy

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

is called the

2nd derivative2

2

dxyd

Stationary Points

3x 10)3(9)3(3)3( 23 y 37

1x 5

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

0 min at )5,1(

3xAt , 2

2

dxyd

1266 1xAt , 2

2

dxyd

10931 y

1093 23 xxxy

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

662

2 x

dxyd

Stationary PointsSUMMAR

Y To find stationary points, solve the equation

0dxdy

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

dxydmax 02

2

dxyd

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

Stationary Points

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Stationary Points

0dxdy

xxxy 93 23

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

0dxdy

Solution:

dxdy 963 2 xx

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

r3x 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 xx

xxxy 93 23

Stationary Points

For a max we have

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.

Determining the nature of a Stationary Point

Stationary Points

dxdy

At the max of the gradient is 0, but the gradient of the gradient is negative.

1093 23 xxxy

The gradient function is given by

963 2 xxdxdy

1093 23 xxxye.g. Consider

Another method for determining the nature of a stationary point.

y

Stationary Points

The notation for the gradient of the gradient is

“d 2 y by d x squared”2

2

dxyd

At the min of 1093 23 xxxy

dxdy

The gradient function is given by

963 2 xxdxdy

1093 23 xxxy

the gradient of the gradient is positive.

Stationary Points

The gradient of the gradient is called the 2nd derivative and is written as

2

2

dxyd

Stationary Points

66 x963 2 xx

e.g. Find the stationary points on the curve

and distinguish between

the max and the min.

1093 23 xxxy

2

2

dxyd

Solution: 1093 23 xxxy

Stationary points: 0

dxdy

0963 2 xx

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

1x3x or

dxdy

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

Stationary Points

3x 10)3(9)3(3)3( 23 y 37

1x 5

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

0 min at )5,1(

At , 3x 2

2

dxyd

1266 At , 1x 2

2

dxyd

10931 y

1093 23 xxxy

To distinguish between max and min we use the 2nd derivative,

662

2 x

dxyd

Stationary PointsSUMMAR

Y To find stationary points, solve the equation

0dxdy

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

dxydmax 02

2

dxyd