Using Differentiation
Stationary Points
© Christine Crisp
Stationary Points
The stationary points of a curve are the points where
the gradient is zero
e.g.
y  x 3  3x 2  9x
A local maximum
x
dy
0
dx
x
A local minimum
The word local is usually omitted and the points called
maximum and minimum points.
Stationary Points
e.g.1 Find the coordinates of the stationary points
on the curve y  x 3  3 x 2  9 x
Solution:
y  x3  3x2  9x
dy

 3x 2  6x  9
dx
dy
2
2

3
x

6
x

9

0
3
(
x
 2 x  3)  0

0
dx
3( x  3out
)( xfor
 1)common
 0 factors
Tip: Watch
x  3 or x  1
3
x  3 when
 finding
y  ( 3)stationary
 3( 3) 2  points.
9( 3)
 27  27  27   27
x  1  y  ( 1) 3  3( 1) 2  9( 1)
 1  3  9  5
The stationary points are (3, -27) and ( -1, 5)
Exercises
Stationary Points
Find the coordinates of the stationary points of the
following functions
1. y  x 2  4 x  5
2.
y  2 x 3  3 x 2  12 x  1
Solutions:
dy
1.
 2x  4
dx
dy
 0  2x  4  0
dx
 x2
x  2  y  (2) 2  4(2)  5  1
Ans: St. pt. is ( 2, 1)
Stationary Points
2.
Solution:
y  2 x 3  3 x 2  12 x  1
dy
 6 x 2  6 x  12
dx
dy
 0  6( x 2  x  2)  0  6( x  1)( x  2)  0
dx
 x  1 or x  2
x  1  y  6
x  2  y  2(2) 3  3(2) 2  12(2)  1  21
Ans: St. pts. are ( 1, 6) and ( 2, 21 )
Stationary Points
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 left of
a maximum,
the gradient is
positive

On the right of
a maximum,
the gradient is
negative

Stationary Points
So, for a max the gradients are
At the max
On the left of
On the right of
the max
the max

0

The opposite is true for a minimum

0

Calculating the gradients on the left and right of a
stationary point tells us whether the point is a max or a
min.
Stationary Points
e.g.2 Find the coordinates of the stationary point of the
curve y  x 2  4 x  1 . Is the point a max or min?
Solution:
dy
0
dx
y  x  4 x  1       (1)
dy

 2x  4
dx

2x  4  0  x  2
Substitute in (1):
2
 y  3
y  ( 2) 2  4( 2)  1
dy
On the left of x = 2 e.g. at x = 1,
 2(1)  4  2  0
dx
dy
On the right of x = 2 e.g. at x = 3,
 2( 3)  4  2  0
dx
We have

  ( 2,  3) is a min
0
Stationary Points
Another method for determining the nature of a
stationary point.
e.g.3 Consider
y  x 3  3 x 2  9 x  10
The gradient function
is given by
dy
 3x2  6x  9
dx
dy
dx
At the max of y  x 3  3 x 2  9 x  10 the gradient is 0
but the gradient of the gradient is negative.
Stationary Points
Another method for determining the nature of a
stationary point.
e.g.3 Consider
y  x 3  3 x 2  9 x  10
The gradient function
is given by
dy
 3x2  6x  9
dx
dy
dx
At the min of
y  x 3  3 x 2  9 x  10
the gradient of the
gradient is positive.
2
d
y
The notation for the gradient of the gradient is
2
dx
“d 2 y by d x squared”
Stationary Points
e.g.3 ( continued ) Find the stationary points on the
curve y  x 3  3 x 2  9 x  10 and distinguish between
the max and the min.
3
2
Solution: y  x  3 x  9 x  10
2
dy
d
y
2

 3x  6x  9 
 6x  6
2
dx
dx
2
dy
2
d
Stationary points:
 0  3 x  6 xy is9 called
 0 the
dx
dx 2
 3( x  2 x  3)  0
 3( x  3)( x  1)  0
 x  3 or x  1
2
2nd derivative
We now need to find the y-coordinates of the st. pts.
Stationary Points
y  x 3  3 x 2  9 x  10
x  3  y  ( 3) 3  3( 3) 2  9( 3)  10  37
x  1  y  1  3  9  10  5
To distinguish between max and min we use the 2nd
derivative, at the stationary points.
d2y
2
 6x  6
dx
d y
At x  3 ,
 6( 3)  6  12  0  max at (3, 37)
2
dx
2
At x  1 ,
d2y
dx
2
 6  6  12  0  min at (1, 5)
Stationary Points
SUMMARY
 To find stationary points, solve the equation
dy
0
dx
 Determine the nature of the stationary points
• either by finding the gradients on the left
and right of the stationary points

•


minimum
0



maximum
0
or by finding the value of the 2nd derivative
at the stationary points
d2y
dx
2
 0  max
d2y
dx
2
 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
y  x  3x  2
Ans. (0,  2) is a min.
(2 , 2) is a max.
1.
2.
3
2
y  x  3x  2
y  2  3x  x3
Ans. (1, 0)
(1 , 4)
is a min.
is a max.
y  2  3x  x3
Exercise



Consider y=x2-4
What ar the coordinates of the stationary
point
Find the 2nd differential and comment on
whether this point is a max or min??