“Teach A Level Maths”
Vol. 1: AS Core Modules
14: Stationary Points –
another notation and
applications
© Christine Crisp
Stationary Points
Module C1
Module C2
AQA
Edexcel
MEI/OCR
OCR
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Stationary Points
An Alternative Notation
The notation f ( x ) for a function of x can be used
instead of y.
d2y
When f ( x ) is used, instead of using
for the
2
dx
2nd derivative, we write
f
//
( x)
Stationary Points
e.g.
(a) Find the coordinates of the stationary
points on the curve y  f ( x ) where
f ( x)  3 x 2  x 3
(b) Determine the nature of the stationary points
and sketch the curve
Solution: (a) f / ( x )  6 x  3 x 2  f // ( x )  6  6 x
Stationary f / ( x )  0 
6x  3x 2  0
points:


3 x( 2  x )  0
x  0 or x  2
We now need to find the y-coordinates of the st. pts.
We have
and
Stationary Points
f ( x )  3 x 2  x 3 , f / ( x )  6 x  3 x 2 , f // ( x )  6  6 x
f / ( x)  0
when
x  0 and x  2
Now, f (0)  0  (0, 0) is a st. pt.
f ( 2)  3( 2) 2  ( 2) 3  4  ( 2, 4) is a st. pt.
f // ( x )  6  6 x  f // (0)  6  0  min at (0, 0)
f // ( 2)   6  0  max at (2, 4)
(b) We use the stationary
points to sketch the
curve.
Tip: When curve sketching,
always find the point on the
y-axis. ( In this example, it
is a stationary point so we
already know it. )
( 2, 4)
x
x
( 0, 0 )
y  3x2  x3
Stationary Points
SUMMARY
 To find stationary points, solve the equation
dy
0
or
f / ( x)  0
dx
 Determine the nature of the stationary points
• either by finding the gradients on the left
and right of the stationary points

•





maximum
0
or by finding the value of the 2nd derivative
at the stationary points
d2y
2
or
minimum
0
 0  max
d2y
2
 0  min
dx
dx
f // ( x )  0  max f // ( x )  0  min
Stationary Points
Exercises
Find the coordinates of the stationary points of the
following functions, determine the nature of each
and sketch the functions.
3
2
f ( x )  x  6 x  30
Ans. ( 0, 30 ) is a max.
( 4 ,  2 ) is a min.
1.
2.
3
2
f ( x)   x 3  3 x 2  9 x  1
Ans. (  3,  28 ) is a min.
(1, 4 )
is a max.
y  x  6 x  30
y   x3  3x2  9x  1
Stationary Points
Applications of Stationary Points
3
2
y

x

6
x
 9x  2
e.g.
Suppose that x represents a length in cm.
Suppose also that the y-variable is replaced by A
where A represents an area measured in cm2. Then,
A  x  6x  9x  2
3
2
Parts of the graph
have no meaning . . .
maximum
x
x
minimum
A  x 3  6x 2  9x  2
Stationary Points
Applications of Stationary Points
3
2
y

x

6
x
 9x  2
e.g.
Suppose that x represents a length in cm.
Suppose also that the y-variable is replaced by A
where A represents an area measured in cm2. Then,
A  x  6x  9x  2
3
2
Parts of the graph
have no meaning . . .
maximum
x
x
minimum
since length and area
3
2
A

x

6
x
 9x  2
cannot be negative.
For values of x between 0 and 4,
x  1 gives the largest area, 6cm2
x  3 gives the smallest area, 2cm2
Stationary Points
maximum
x
x
minimum
A  x 3  6x 2  9x  2
The stationary points give the greatest and least
values of the area, A, and the values of x at
which they occur.
If we can find a function that describes a situation
involving greatest and/or least values of a variable,
we can use stationary points to find these values.
An equation describing a situation is called a model.
Stationary Points
e.g.1 The number in a population of foxes in a
certain country area increases and decreases over
the years. A model for the number in the
population, y, is given by the equation
y  2 x 3  19x 2  40x  100,
where x is the number of years since 1990.
(a) How many foxes were in the population in 1990?
(b) Use Calculus to find the year in which the
population was least and give its size.
Solution: (a) In 1990, x = 0, so y = 100
There were 100 foxes in 1990.
Stationary Points
(b) Use Calculus to find the year in which the
population was least and give its’ size.
We want to find the minimum value of y.
y  2 x 3  19x 2  40x  100,
dy

 6 x 2  38 x  40
dx
dy
 0  6 x 2  38 x  40  0
dx
 2(3 x 2  19x  20)  0

2( x  5)(3 x  4)  0
 x  5 or x 
4
3
Since y is a cubic function, one of these values
gives the minimum and one gives the maximum.
Stationary Points
The quickest way to see which one gives the
minimum is to use the 2nd derivative.
2
dy
2
d
y
 6 x  38 x  40 
 12x  38
2
dx
dx
x5 
d2y
dx
2
 12(5)  38  42
Since 42 > 0, x = 5 gives the min.
The minimum value is given by y:
y  2 x 3  19x 2  40x  100,
x  5  y  2(5) 3  19(5) 2  40(5)  100
y  75

The smallest number of foxes was 75 in 1995.
Stationary Points
e.g.2 I want to make a fruit cage from 2 pieces of
netting. The smaller piece will make the top and I’ll
use the longer piece, which is 10m long, for the
sides. The cage is to be rectangular and to enclose
the largest possible area. One side won’t need to
be netted as there is a wall on that side. How long
should the 3 sides be?
Solution: We draw a diagram and choose letters
for the unknown lengths and area.
wall
The lengths are x
x
x
metres and y metres.
A
The area is A m2.
y
Stationary Points
We want to find the
maximum value of A, so we
need an expression for A
that we can differentiate.
wall
x
A
x
y
The area of a rectangle  length  breadth
 A  xy
      (1)
We can’t differentiate with 3 variables so we need
to substitute for one of them.
Since the length of netting is 10m, we know that
Rearranging,
2 x  y  10
y  10  2 x       (2)
Stationary Points
wall
A  xy
y  10  2 x
x
      (1)
      (2)
Substituting for y in (1),
 A  x(10  2 x )  A  10x  2 x 2
We can now find the stationary points.
dA
 10  4 x
dx
dA
 0  10  4 x  0
dx

10  4 x
x  25

Substitute x  2  5 in ( 2)  y  5
Substitute in (1),
A  xy  A  12 5
A
y
x
Stationary Points
wall
We have
x
x  2  5, y  5 and A  12  5
A
x
y
We know that we have found the maximum since
A  10x  2 x 2
and we recognise
the shape of this
quadratic.
A
 10 x 
 2 x22
Stationary Points
Stationary Points
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.
Stationary Points
An Alternative Notation
The notation f ( x ) for a function of x can be used
instead of y.
d2y
When f ( x ) is used, instead of using
for the
2
dx
2nd derivative, we write
//
f ( x)
Stationary Points
e.g.
(a) Find the coordinates of the stationary
points on the curve y  f ( x ) where
f ( x)  3 x 2  x 3
(b) Determine the nature of the stationary points
and sketch the curve
Solution: (a) f / ( x )  6 x  3 x 2  f // ( x )  6  6 x
Stationary f / ( x )  0 
6x  3x 2  0
points:


3 x( 2  x )  0
x  0 or x  2
We now need to find the y-coordinates of the st. pts.
We have
and
Stationary Points
f ( x )  3 x 2  x 3 , f / ( x )  6 x  3 x 2 , f // ( x )  6  6 x
f / ( x)  0
when
x  0 and x  2
Now, f (0)  0  (0, 0) is a st. pt.
f ( 2)  3( 2) 2  ( 2) 3  4  ( 2, 4) is a st. pt.
f // ( x )  6  6 x  f // (0)  6  0  min at (0, 0)
f // ( 2)   6  0  max at (2, 4)
(b) We use the stationary
points to sketch the
curve.
Tip: When curve sketching,
always find the point on the
y-axis. ( In this example, it
is a stationary point so we
already know it. )
( 2, 4)
x
x
( 0, 0 )
y  3x2  x3
Stationary Points
SUMMARY
 To find stationary points, solve the equation
dy
0
or
f / ( x)  0
dx
 Determine the nature of the stationary points
• either by finding the gradients on the left
and right of the stationary points

•





maximum
0
or by finding the value of the 2nd derivative
at the stationary points
d2y
or
minimum
0
2
 0  max
d2y
2
 0  min
dx
dx
f // ( x )  0  max f // ( x )  0  min
Stationary Points
Applications of Stationary Points
If we can find a function that describes a situation
involving greatest and/or least values of a variable,
we can use stationary points to find these values.
An equation describing a situation is called a model.
Stationary Points
e.g.1 The number in a population of foxes in a
certain country area increases and decreases over
the years. A model for the number in the
population, y, is given by the equation
y  2 x 3  19x 2  40x  100,
where x is the number of years since 1990.
(a) How many foxes were in the population in 1990?
(b) Use Calculus to find the year in which the
population was least and give its’ size.
Solution: (a) In 1990, x = 0, so y = 100
There were 100 foxes in 1990.
Stationary Points
(b) Use Calculus to find the year in which the
population was least and give its size.
We want to find the minimum value of y.
y  2 x 3  19x 2  40x  100,
dy

 6 x 2  38 x  40
dx
dy
 0  6 x 2  38 x  40  0
dx
 2(3 x 2  19x  20)  0

2( x  5)(3 x  4)  0
 x  5 or x 
4
3
Since y is a cubic function, one of these values
gives the minimum and one gives the maximum.
Stationary Points
The quickest way to see which one gives the
minimum is to use the 2nd derivative.
2
dy
2
d
y
 6 x  38 x  40 
 12x  38
2
dx
dx
x5 
d2y
dx
2
 12(5)  38  42
Since 42 > 0, x = 5 gives the min.
The minimum value is given by y:
y  2 x 3  19x 2  40x  100,
x  5  y  2(5) 3  19(5) 2  40(5)  100
y  75

The smallest number of foxes was 75 in 1995.
Stationary Points
e.g.2 I want to make a fruit cage from 2 pieces of
netting. The smaller piece will make the top and I’ll
use the longer piece, which is 10m long, for the
sides. The cage is to be rectangular and to enclose
the largest possible area. One side won’t need to
be netted as there is a wall on that side. How long
should the 3 sides be?
Solution: We draw a diagram and choose letters
for the unknown lengths and area.
wall
The lengths are x
x
x
metres and y metres.
A
The area is A m2.
y
Stationary Points
We want to find the
maximum value of A, so we
need an expression for A
that we can differentiate.
wall
x
A
x
y
The area of a rectangle  length  breadth
 A  xy
      (1)
We can’t differentiate with 3 variables so we need
to substitute for one of them.
Since the length of netting is 10m, we know that
Rearranging,
2 x  y  10
y  10  2 x       (2)
Stationary Points
wall
A  xy
y  10  2 x
x
      (1)
      (2)
Substituting for y in (1),
 A  x(10  2 x )  A  10x  2 x 2
We can now find the stationary points.
dA
 10  4 x
dx
dA
 0  10  4 x  0
dx

10  4 x
x  25

Substitute x  2  5 in ( 2)  y  5
Substitute in (1),
A  xy  A  12 5
A
y
x
Stationary Points
wall
We have
x
x  2  5, y  5 and A  12  5
A
x
y
We know that we have found the maximum since
A  10x  2 x 2
and we recognise
the shape of this
quadratic.
A  10x  2 x 2