Stationary Points, To determine Maximum / Minimum / Inflection Points First Derivative Test Second Derivative Test

1. Differentiation Chap 9
Objective: How to find
Stationary Points
&
determine their nature
(maximum/minimum)
riazidan

2. The stationary points of a curve are the points where
the gradient is zero
e.g.
y = x3 − 3x2 − 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.

3. e.g.1 Find the coordinates of the stationary points
on the curve y = x 3 − 3 x 2 − 9 x
y = x3 − 3x2 − 9x
Solution:
dy
⇒
= 3x2 − 6x − 9
dx
dy
⇒ 3 x 2 − 6 x − 9 = 0 ⇒ 3( x 2 − 2 x − 3) = 0
=0
dx
3( x − out + 1 = 0 ⇒ x = 3
Tip: Watch 3)( xfor )common factors or x = −1
x = 3 when finding )stationary points.
⇒ y = ( 3 3 − 3( 3) 2 − 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)

4. Exercises
Find the coordinates of the stationary points of the
following functions
2
1. y = x − 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)

5. y = 2 x 3 + 3 x 2 − 12 x + 1
2.
Solution:
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 )

6. 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
−

7. So, for a max the gradients are
0 At the max
On the left of
On the right of
the max
the max
−
+
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.

8. e.g.2 Find the coordinates of the stationary point of the
2
curve y = x − 4 x + 1 . Is the point a max or min?
− − − − − − (1)
y = x2 − 4x + 1
Solution:
dy
⇒
= 2x − 4
dx
dy
=0
⇒
2x − 4 = 0 ⇒ x = 2
dx
y = ( 2) 2 − 4( 2) + 1
⇒ y = −3
Substitute in (1):
dy
= 2(1) − 4 = − 2 < 0
On the left of x = 2 e.g. at x = 1,
dx
dy
On the right of x = 2 e.g. at x = 3,
= 2( 3) − 4 = 2 > 0
dx
+
−
⇒ ( 2, − 3) is a min
We have
0

9. 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
3
2
At the max of y = x + 3 x − 9 x + 10 the gradient is 0
but the gradient of the gradient is negative.

10. 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.
d2y
The notation for the gradient of the gradient is
dx 2
“d 2 y by d x squared”

11. 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.
y = x 3 + 3 x 2 − 9 x + 10
Solution:
dy
d2y
2
⇒
= 3x + 6x − 9 ⇒
= 6x + 6
2
dx
dx
2
dy
2 d y
Stationary points:
= 0 ⇒ 3 x + 6 x −is called the
9=0
dx
dx 2 nd
2 derivative
⇒ 3( x 2 + 2 x − 3) = 0
⇒ 3( x + 3)( x − 1) = 0
⇒
x = −3 or x = 1
We now need to find the y-coordinates of the st. pts.

12. 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
= 6( −3) + 6 = −12 < 0 ⇒ max at (−3, 37 )
At x = −3 ,
2
dx
2
At x = 1 ,
d2y
dx
2
= 6 + 6 = 12 > 0 ⇒ min at (1, 5)

13. 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
0
+
−
⇒
maximum
or by finding the value of the 2nd derivative
at the stationary points
d2y
dx
2
< 0 ⇒ max
d2y
dx
2
> 0 ⇒ min

14. Exercises
Find the coordinates of the stationary points of the
following functions, determine the nature of each
and sketch the functions.
3
2
3
2
y = x + 3x − 2
Ans. (0, − 2) is a min.
1.
(−2 , 2)
2.
y = x + 3x − 2
is a max.
y = 2 + 3x − x3
Ans. (−1, 0)
(1 , 4)
is a min.
is a max.
y = 2 + 3x − x3

16. 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.

17. The stationary points of a curve are the points where
the gradient is zero
e.g.
y = x3 − 3x2 − 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.

18. e.g.1 Find the coordinates of the stationary points
y = x3 − 3x2 − 9x
on the curve
Solution:
⇒
dy
=0
dx
⇒
y = x3 − 3x2 − 9x
dy
= 3x2 − 6x − 9
dx
3x2 − 6x − 9 = 0 ⇒
3( x 2 − 2 x − 3) = 0
3( x − 3)( x + 1) = 0 ⇒ x = 3 or x = −1
x = 3 ⇒ y = ( 3) 3 − 3( 3) 2 − 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)

19. Determining the nature of a Stationary Point
For a max we have
On the left of
the max
+
0
At the max
−
On the right of
the max
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.

20. Another method for determining the nature of a
stationary point.
e.g. Consider
y
y = x 3 + 3 x 2 − 9 x + 10
The gradient function
is given by
dy
= 3x2 + 6x − 9
dx
dy
dx
3
2
At the max of y = x + 3 x − 9 x + 10 the gradient is
0, but the gradient of the gradient is negative.

21. 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.
d2y
The notation for the gradient of the gradient is
dx 2
“d 2 y by d x squared”

22. The gradient of the gradient is called the 2nd
derivative and is written as
d2y
dx 2

23. e.g. Find the stationary points on the curve
3
y = xand 3distinguish between the max
+ x 2 − 9 x + 10
and the=min.+ 3 x 2 − 9 x + 10
y x3
Solution:
dy
d2y
2
⇒
= 3x + 6x − 9 ⇒
= 6x + 6
2
dx
dx
dy
Stationary points:
= 0 ⇒ 3x2 + 6x − 9 = 0
dx
⇒ 3( x 2 + 2 x − 3) = 0
⇒ 3( x + 3)( x − 1) = 0
⇒
x = −3 or x = 1
We now need to find the y-coordinates of the st. pts.

24. 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,
d2y
2
= 6x + 6
dx
d2y
At x = −3 , 2 = 6( −3) + 6 = −12 < 0 ⇒ max at (−3, 37 )
dx
At
x =1 ,
d2y
dx
2
= 6 + 6 = 12 > 0 ⇒ min at (1, 5)

25. 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
0
−
+
−
⇒ maximum
⇒ minimum +
0
• or by finding the value of the 2nd derivative
at the stationary points
d2y
dx
2
< 0 ⇒ max
d2y
dx
2
> 0 ⇒ min