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Identifying Stationary Points. Differentiation 3.6. The stationary points of a curve are the points where the gradient is zero. e.g. A local maximum. x. x. A local minimum. The word local is usually omitted and the points called maximum and minimum points. - PowerPoint PPT Presentation
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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
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