differentiation jan 21, 2014
DESCRIPTION
Stationary Points, To determine Maximum / Minimum / Inflection Points First Derivative Test Second Derivative TestTRANSCRIPT
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Objective: How to find Objective: How to find Stationary Points Stationary Points
&&determine their nature determine their nature (maximum/minimum)(maximum/minimum)
riazidan
Differentiation Chap 9Differentiation Chap 9
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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.
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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.
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ExercisesFind 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.
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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
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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
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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.
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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
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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.
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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.
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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
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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
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
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SUMMARY To find stationary points, solve the equation
0dx
dy
0
maximum
0
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
dx
ydmax 0
2
2
dx
yd
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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
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The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.
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0dx
dy
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.
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e.g.1 Find the coordinates of the stationary points on the curve xxxy 93 23
0dx
dy
Solution:
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 xx
xxxy 93 23
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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
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dx
dy
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 xxdx
dy
1093 23 xxxy
e.g. Consider
Another method for determining the nature of a stationary point.
y
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The notation for the gradient of the gradient is
“d 2 y by d x squared”2
2
dx
yd
At the min of
1093 23 xxxy
dx
dy
The gradient function is given by
963 2 xxdx
dy
1093 23 xxxy
the gradient of the gradient is positive.
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The gradient of the gradient is called the
2nd derivative and is written as
2
2
dx
yd
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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
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.
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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
dx
yd
1266 At , 1x 2
2
dx
yd
10931 y
1093 23 xxxy
To distinguish between max and min we use the 2nd derivative,
662
2
xdx
yd
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SUMMARY
To find stationary points, solve the equation
0dx
dy
0
maximum
0
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
dx
ydmax 0
2
2
dx
yd