CHAPTER 3
SECTION 3.4
CONCAVITY AND
THE SECOND DERIVATIVE
TEST
Definition of Concavity and
Figure 3.24
Sketch 4 graphs a)1 decreasing and concave up
b)1 increasing and concave up, c)1 decreasing and concave down,
d)1 increasing and concave down
y
y
y
x
y
x
x
x
b
a
y
y
y
x
c
x
y
x
d
x
Concave upward
y
y
y
x
x
y
x
x
• Look at these two graphs. Each is concave upward, but
one is decreasing and the other is increasing. We need
to be able to determine concavity from the function and
not just from the graph. For each of the graphs above
sketch the tangent lines to the graph at a number of
different points.
Concave upward
y
y
x
x
• As we move from left
to right the slopes of
the tangent lines are
getting less negative.
That is they are
increasing.
Concave upward
y
x
• As we move from left
to right the slopes of
the tangent lines are
getting larger. That is
they are increasing.
When a graph is concave
upward
The slope of the tangent lines are
increasing.
Concave downward
y
y
y
x
x
y
x
x
• Look at these two graphs. Each is concave downward,
but one is decreasing and the other is increasing. We
need to be able to determine concavity from the function
and not just from the graph. For each of the graphs
above sketch the tangent lines to the graph at a number
of different points.
Concave downward
y
x
• As we move from left
to right, the slopes of
the tangent lines are
getting more
negative.
• They are decreasing.
Concave downward
y
x
• As we move from left
to right the slopes of
the tangent lines are
getting smaller. That
is they are
decreasing.
When a graph is concave
downward
The slopes of the tangent lines
are decreasing.
Putting it all together
• For a function f that is differentiable on an
interval I, the graph of f is
• (i) Concave up on I, if the slope of the tangent
line is increasing on I or
• (ii) Concave down on I, if the slope of the
tangent line is decreasing on I
Linking knowledge
• (i) Concave up on I, if the slope of the tangent line is
increasing on I.
• If the slope of the tangent line is increasing and the
slope of the tangent line is represented by the first
derivative and to determine when something is
increasing we had to take the derivative, then to find
where the slope of the tangent line (f ‘(x)) is
increasing we will need to take the derivative of f ‘(x)
or find the second derivative f “(x)
I know, this is a very large run on sentence.
Linking knowledge
• (ii) Concave down on I, if the slope of the tangent
line is decreasing on I
• If the slope of the tangent line is decreasing and the
slope of the tangent line is represented by the first
derivative and to determine when something is
decreasing we had to take the derivative, then to find
where the slope of the tangent line (f ‘(x)) is
decreasing we will need to take the derivative of
• f ‘(x) or find the second derivative f “(x)
Definition of concavity
• For a function f that is differentiable on an
interval I, the graph of f is
• (i) Concave up on I, if f’ is increasing on I or
• (ii) Concave down on I, if f’ is decreasing on I
Theorem 3.7
Test for concavity
Let f be a function w hose second derivative exists
on an open interval I.
1. If f ( x)  0 for all x in I, then
th e graph of f is concave upward in I.
2. If f ( x)  0 for all x in I, then
th e graph of f is concave downward in I.
Putting it all together
•
•
•
•
•
Given the function f(x)
f(x) = 0
f(x) undefined
f(x)>0
f(x)<0
x-intercepts
vertical asymptote
Q-1 or Q-2
Q-3 or Q-4
Putting it all together
Find f ( x)
f ( x)  0
f ( x) undefined
critical number
critical number(cor ner pt or
ver tical asymptote)
f ( x)  0
f ( x)  0
f is increasing
f is decreasing
Putting it all together
Find f ( x)
f ( x)  0
f ( x) undefined
f ( x)  0
f ( x)  0
" hypercriti cal" number
???????
f  is increasing ;
f is concave upward
f  is decreasing
f is concave downward
Determining concavity
• Determine the open
intervals on which the
graph is concave
upward or concave
downward.
6
f ( x)  2
x 3
• Concavity
find
second derivative.
• Find hypercritical
numbers.
• Set up a chart
• Find concavity
f ( x)  6x 2  3
1
f ( x)  6( x 2  3)  2 (2 x)
 12 x
f ( x) 
2
2
x  3
f ( x) 
f ( x) 
 12x  3  2( x 2  3)( 2 x)( 12 x)
2
2
36x 2  1
x
2
x
2
 3
4
 3
3
c = 1; c = -1 and f”
is defined on the
entire # line
f ( x) 


36 x 2  1
x
2

3
3
Setting up the chart
interval
Test
points
Sign of f”
f‘
concave
(-∞, -1)
-2
+
inc
upward
(-1,1)
0
-
dec
downward
(1,∞)
2
+
inc
upward
Points of inflection
• A point of inflection for
the graph of f is that
point where the
concavity changes.
Theorem 3.7 Test for Concavity
Definition of Point of Inflection and
Figure 3.28
Theorem 3.8 Points of Inflection
Theorem 3.9 Second Derivative
Test and Figure 3.31
• Example 1: Graph the function f given
by
f (x)  x 3  3x 2  9x  13,
• and find the relative extrema.
• 1st find graph the function.
• Example 1 (continued):
• 2nd solve f (x) = 0.
3x 2  6x  9  0
x 2  2x  3  0
(x  3)(x  1)  0
x3  0
x 1  0
or
x
 3
x
 1
• Thus, x = –3 and x = 1 are critical values.
• Example 1 (continued):
• 3rd use the Second Derivative Test with –
3 and 1.
f (3)  6(3)  6  18  6  12  0 : Relative maximum
f (1)  6(1)  6  6  6  12  0 : Relative minimum
• Lastly, find the values of f (x) at –3 and 1.
f (3)  (3)3  3(3)2  9(3)  13  14
f (1)  (1)3  3(1)2  9(1)  13  18
• So, (–3, 14) is a relative maximum and (1, –18) is a
• relative minimum.
Second Derivative Test
If c is a critical number of f’(x) and…
a.
If f’’(c) > 0 then ________________________
b.
If f’’(c) < 0 then ________________________
c.
If f’’(c) = 0 or undefined then __________________________________
Second Derivative Test
If c is a critical number of f’(x) and…
a.
(c, f(c)) is a relative min
If f’’(c) > 0 then ________________________
b.
(c, f(c)) is a relative max
If f’’(c) < 0 then ________________________
c.
the test fails (use 1st Derivative test)
If f’’(c) = 0 or undefined then __________________________________
Concave downward
Concave upward
Inflection Points
If f’(x) > 0 ______________ If f’(x) < 0 ______________ If f’(x) = 0 ______________
If f’’(x) > 0 ______________ If f’’(x) < 0 ______________ If f’’(x) = 0 ______________
______________
______________
________________
The second derivative gives the same information about the first derivative that
the first derivative gives about the original function.
For f(x) to increase, _____________
For f’(x) to increase, _____________
For f(x) to decrease, _____________
For f’(x) to decrease, _____________
Concave downward
Concave upward
Slopes increase
Slopes decrease
Inflection Points Where concavity changes
Occur at critical numbers of f”(x)
f(x) is constant
f(x) decreases
f(x) increases
If f’(x) = 0 ______________
If f’(x) < 0 ______________
If f’(x) > 0 ______________
f’(x) is constant
f’(x) increases If f’’(x) < 0 ______________
f’(x)
If f’’(x) = 0 ______________
If f’’(x) > 0 ______________
decreases
f(x)
is
conc
up
f(x) is a straight line
f(x)
is conc down
______________
________________
______________
The second derivative gives the same information about the first derivative that
the first derivative gives about the original function.
f’(x) > 0
For f(x) to increase, _____________
f’’(x) > 0
For f’(x) to increase, _____________
f’(x) < 0
For f(x) to decrease, _____________
f’’(x) < 0
For f’(x) to decrease, _____________

Sketch f  x   6 x  3
2

1
Include extrema, inflection points, and intervals of concavity.

Sketch f  x   6 x  3
2

1
No VA’s
Include extrema, inflection points, and intervals of concavity.

f '  x   6 x  3
12 x

2
2
x 3

x


2

2
 2x 
Critical numbers:

 3   12    12 x  2  x
f ''  x 
 x  3
12  x  3   x  3    4 x  

 x  3
2
2
2
2
2
2
2


12x  0
x 0
x2  3  0
None
smooth

 3  2 x  
4
2
4
12  3 x 2  3 
x
2
3

36  x 2  1
x
2
3

3
3
Critical numbers:
x2  1  0
x  1
x2  3  0
None
1. Find the extrema of f  x   2sin x  cos2x on 0,2 
1. Find the extrema of f  x   2sin x  cos2x on 0,2 
f '  x   2cos x  2sin2x
f ''  x   2sin x  4cos2x
2nd Derivative Test
 6 , 32 


,1

rel
min
at
f ''  2   2
2 
5 3
f ''  56   3 rel max at  6 , 2 
f ''  32   6 rel min at  32 , 3 
f ''  6   3
rel max at
0  2cos x  2sin2x
0  2cos x  2  2sin x cos x 
0  2cos x 1 2sin x 
0  2cos x 0  1  2sin x
1
0  cos x
 sin x
2
 3
 5
,

x
, x
2 2
6 6
Crit numbers:
f  6   32
f  2   1
f  56   32
f  32   3
2. Sketch
f  x   2sin  2x  on 0,4 
f  x   2sin  2x  on 0,4 
2. Sketch
f '  x   cos  2x 
Crit numbers:
2nd Derivative Test
f ''     21
f '' 3   21
 ,2
3 , 2
rel max at
rel min at
f ''  x    21 sin  2x 
Intervals:
0,   2 ,4 
f ’’(test pt)
3

f(x)
down
up
Inf pt
x  2
 2 ,0
x   ,3 ,5 ,7
f  3   2
Crit numbers:


Test values:
f    2
0  cos  2x 
5 7
 3
x
,2

,
2
2
2 2
f  2   0
0  sin  2x 
x
  ,2 ,3 ,4
2
x  2 ,4 ,6 ,8
f  x   sin  2x 
2
2
4

-2
Find a Function
Describe the function at the point x=3
based on the following:
f (3)  4
(3, 4)
f (3)  0
f (3)  6
3
Find a Function
Describe the function at the point x=5
based on the following:
f (5)  0
f (5)  0
f (5)  0
5
Find a Function
Given the function is continuous at the
point x=2, sketch a graph based on the
following:
f ( 2)  3
f (2) DNE
f ( x)  0 for x  2
f ( x)  0 for x  2
f ( x)  0 for all x  2
(2,3)
2
WHY? BECAUSE f’(x) is POSITVE!!!!!!!!!!!!!!!