28: harder stationary points © christine crisp “teach a level maths” vol. 1: as core modules
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28: Harder Stationary 28: Harder Stationary PointsPoints
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 1: AS Core Vol. 1: AS Core ModulesModules
Harder Stationary Points
Module C1
AQA
Edexcel
OCR
MEI/OCR
Module C2
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Harder Stationary Points
The stationary points of a curve are the points where the gradient is zero
We may be able to determine the nature of a stationary point just by knowing the shape of a curve.
e.g.1 We know the curve has a minimum because the sign of the term ( positive ) tells us that the graph has the following shape
9123 2 xxy2x
Harder Stationary Points
x 10,3
2714
31 ,
x
Plotting the stationary points and using our knowledge that a cubic is a continuous function ( we can draw it with a single stroke ) means we must get the following:
e.g.1 The cubic curve has 2 stationary points.
32531 xxxy
They are and 2714
31 , 10,3
1,0 x
The y-intercept is also useful
Harder Stationary Points
2714
31 ,
x
x 10,3
1,0 x
Plotting the stationary points and using our knowledge that a cubic is a continuous function ( we can draw it with a single stroke ) means we must get the following:
e.g.1 The cubic curve has 2 stationary points.
32531 xxxy
They are and 2714
31 , 10,3
Harder Stationary Points
Using the 2nd derivative is usually the easiest method.
We may not know the shape of some functions, so we need to determine the nature of the stationary points by another method.
Harder Stationary Points
Distinguish between the max and the min.
xxxf
1)( Solutio
n:2/ 1)( xxf
1)( xxxf
2/ 1
1)(x
xf
e.g.2 Calculate the coordinates of the stationary points on the graph of where
)(xfy x
xxf1
)(
Multiply by :
2x 012 x
must be written in the form before we can differentiate
x1
1x
0)(/ xfFor st. pts. 01
12
x
12 x1xthis quadratic equation has
no linear term so there is no need to factorize
Harder Stationary Points
2/ 1)( xxf 3// 2)( xxf
N.B. The maximum has a smaller y -value than the minimum !
To distinguish between the stationary points we need the 2nd derivative
The stationary points are ( 1, 2 ) and (1, 2)
3
2
x
2)1(// f
3
//
)1(
2)1(f 2
0
0
)2,1( is a min
is a max)2,1(
It’s interesting to see what the graph looks like.
,211)1( f 211)1( f
Calculate y-values at x = 1 and 1:
xxxf
1)(
Harder Stationary Points
x
xxf1
)(0
10)0( f
is infinite, so x = 0 (the y-axis) is an asymptote0
1
So, we now have
)2,1(x
)2,1( x(max
)
(min)
x = 0
Harder Stationary Points
“ approaches x “)(xf
xxxf
1)(
Also, as
,x
“ x approaches infinity “
01
x
xxf )(so
“ approaches zero “x
1
)2,1(x
)2,1( x(max
)
(min)
x = 0
is also an asymptote
xy
y = x
Harder Stationary Points
xxy
1
)2,1(x
)2,1( x(max)
(min)
Asymptote, x = 0
We can now complete the curve.
Harder Stationary Points
xxy
910
Exercise
xxxf
910)(
1. Find the stationary points on the curve
where
)(xfy
Determine the nature of the stationary points.
10xy and
The asymptotes are
0x
Ans: )16,3()4,3( is a
maximum
is a minimum
The question didn’t ask for the graph but it looks like this:
Harder Stationary Points
Harder Stationary Points
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Harder Stationary Points
Distinguish between the max and the min.
xxxf
1)( Solutio
n:2/ 1)( xxf
1)( xxxf
2/ 1
1)(x
xf
e.g.2 Calculate the coordinates of the stationary points on the graph of where
)(xfy x
xxf1
)(
Multiply by :
2x 012 x
must be written in the form before we can differentiate
x1
1x
0)(/ xfFor st. pts. 01
12
x
12 x1xthis quadratic equation has
no linear term so there is no need to factorize
Harder Stationary Points
2/ 1)( xxf 3// 2)( xxf
N.B. The maximum has a smaller y -value than the minimum !
To distinguish between the stationary points we need the 2nd derivative
The stationary points are ( 1, 2 ) and (1, 2)
3
2
x
2)1(// f
3
//
)1(
2)1(f 2
0
0
)2,1( is a min
is a max)2,1(
211)1( f 211)1( f
Calculate y-values at x = 1 and 1: x
xxf1
)(