Identifying Stationary Points

Differentiation 3.6

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.

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 or3x 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

Tip: Watch out for common factors when finding stationary points.

Exercises

Find 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.

2. 11232 23 xxxy

21 xx or 61 yx

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

1266 2 xxdxdySolution:

0)2(60 2 xxdxdy

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

0)2)(1(6 xx

Point of Inflection

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

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.

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

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.

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 of1093 23 xxxy

the gradient of the gradient is positive.

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: 0dxdy

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

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

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

SUMMARY

To find stationary points, solve the equation

0dxdy

0

maximum0

minimum

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

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

exercise 16.01 and 16.02

second derivative exercise 16.03

stationary points and points of inflection 16.04