13: stationary points © christine crisp “teach a level maths” vol. 1: as core modules
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13: Stationary Points13: Stationary Points
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 1: AS Core Vol. 1: AS Core ModulesModules
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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.
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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.
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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.
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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
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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
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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.
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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
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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.
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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.
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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
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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
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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
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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
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Stationary Points