Transcript Section 3.4 * Concavity and the Second Derivative Test

Section 3.4 – Concavity and
the Second Derivative Test
The Second Derivative and the
Function
The first derivative tells us where a function is increasing or
decreasing. But how can we tell the manner in which a
function is increasing or decreasing?
For example, if f '(x) = 3x2 +3 then f(x) is always increasing
because f '(x) is always positive. But which graph below
represents f(x)?
Concavity
If the graph of a function f lies above all of its tangents on
an interval I, then it is said to be concave up (cupped
upward) on I.
If the graph of a function f lies below all of its tangents on
an interval I, then it is said to be concave down (cupped
downward) on I.
CONCAVE DOWN
CONCAVE UP
Test for Concavity
a) If f ''(x) > 0 for all x in I, then the graph of f
is concave upward on I.
b) If f ''(x) < 0 for all x in I, then the graph of f
is concave downward on I.
Slopes are
CONCAVE UP
decreasing.
Slopes are
CONCAVE
DOWN
increasing.
Procedure for Finding Intervals on which a
Function is Concave Up or Concave Down
If f is a continuous function on an open interval
(a,b). To find the open intervals on which f is
concave up or concave down:
1. Find the critical numbers of f ' and values of x
that make f '' undefined in (a,b).
2. These numbers divide the x-axis into
intervals. Test the sign (+ or –) of the second
derivative inside each of these intervals.
3. If f '' (x) > 0 in an interval, then f is concave up
in that same interval. If f ''(x) < 0 in an interval,
then f is concave down in that same interval.
Find where the first derivative is increasing and decreasing.
Example 1
Use the graph of f '(x) below to determine when f is
concave up and concave down.
f is concave
down when
the
derivative is
decreasing.
A critical point
for the first
derivative
(f''=0)
Concave Up: (1, ∞)
f ' (x)
x
f is concave
up when the
derivative is
increasing.
Concave Down:(-∞,1)
Example 2
Find where the graph f  x   x3  3x  1 is concave up and
Domain of f:
where it is concave down.
All Reals
Find the critical numbers of f '
Find the first derivative.


nd derivative
Find
where
the
2
x  dxd x 3  3x  1
is 0 or undefined
Find the second derivative.
' x  3x2  3
0  6x
f ' 
f  
f ''  x   dxd  3 x 2  3
x0
f ''  x   6 x
Find the sign of the second derivative on
Answer the question
each interval.


f ''  x 
The function is concave
0
x  1
x 1
down on (-∞,0) because
f '' 1  6
f ''<0 and is concave up on
f ''  1  6
(0,∞) because f ''>0
The Change in Concavity
If a graph changes from concave upward to concave
downward (or vice versa), then there must be a point
where the change of concavity occurs. This point is
referred to as an inflection point.
CONCAVE DOWN
CONCAVE UP
Inflection Points
A point P on a curve is called an inflection point if the graph
is concave up on one side of P and concave down on the
other side.
In calculus terms, (if f is continuous on an interval that
contains c) c is an inflection point if f '' changes from
positive to negative or vice versa at c. Thus, f '' (c) must
equal 0 or be undefined.
Example
Determine the points of inflection of f  t   t  6  t  .
Domain of f:
Find the critical numbers of f '
All Reals
3

Find the derivative.
f ' t   t 
f ' t  
6  t 
2
3
185t
13
3 6 t 
NOTE: 6 is not an
inflection point since
the 2nd derivative does
not change sign.
1 3
  6  t 
Find where the 2nd derivative
is 0 or undefined
23
Find the second derivative.
f ''  t  
3 6 t 
13
2
 5 185t 
9 6  t 
23
3 31
 6 t 
2 3
0  10t  72
 10t  72 The 2nd
t  7.2 derivative is
f ''  t   9106tt 724 3
undefined at
 
t=6.
Find the sign of the second derivative on
each interval.
Find the value of the function:



f ''  x 
f  7.2   8.131
6
7.2
x 5
x7
x 8
f ''  5  2.44
f '' 8  0.35
Answer the question
f ''  7   0.22
Example: Answer
The point of Inflection is
(7.2,8.131) because the
second derivative changes
from negative to positive
values at this point.
White Board Challenge
Find where the inflection point(s) are for f
2
if f ''(x) = 4cos x – 2 on [0,π]. Justify.
x  34
or

4
because the second derivative changes
from negative to positive at these points.
How the concavity is connected to
Relative Minimum and Maximum
When a critical point is a relative maximum, what are the
characteristics of the function?
The function is concave downward at B.
When a critical point is a relative minimum, what are the
characteristics of the function?
The function is concave upward at C.
D
f(x)
B
A
C
x
The Second Derivative Test
Let f be a function such that f '(c) = 0 (a critical number of a
continuous function f(x) ) and the second derivative
exists on an interval containing c.
(a) If f ''(c) < 0, there is a relative maximum at x = c.
f(x)
Relative Maximum
f '(c) = 0
f ''(c) < 0
c
x
The Second Derivative Test
Let f be a function such that f '(c) = 0 (a critical number of a
continuous function f(x) ) and the second derivative
exists on an interval containing c.
(b) If f ''(c) > 0, there is a relative minimum at x = c.
f(x)
f '(c) = 0
f ''(c) > 0
Relative Minimum
c
x
The Second Derivative Test
Let f be a function such that f '(c) = 0 (a critical number of a
continuous function f(x) ) and the second derivative
exists on an interval containing c.
(c) If f ''(c) = 0, then the second-derivative test fails (either a
maximum, or a minimum, or neither may occur).
Here is a
common example
of neither a
minimum or
maximum
f(x)
f '(c) = 0
f ''(c) = 0
c
x
When the SecondDerivative Test fails, the
critical point can often be
classified using the FirstDerivative Test.
Example 1
Domain of f:
Find the relative extrema of f  x   3x5  5x3  2 .
All Reals
Find the critical numbers
Apply the 2nd
Find the first derivative.
Find where the derivative
is 0 or undefined
Derivative Test
f '  x   dxd 3x5  5 x3  2
4
2
3
0

15
x

15
x
f
''
x

60
x
 30 x


4
2
2
2
f '  x   15x  15x
0  15 x x  1




0  15x  x 1 x  1
x  0, 1,  1
2
Apply the 1st Derivative Test where the 2nd
Derivative Test Fails.

-1
x  0.5
f '  0.5  2.8125

0
1
f ' x
x  0.5
f '  0.5  2.8125
No sign change means x=0 is not a relative extrema.
f ''  0  0 FAIL
f '' 1  30 MIN
f ''  1  30 MAX
Find the value of the
function:
f 1  0
f  1  4
Answer the question
Example 1: Answer
The function has a relative
maximum of 4 at x = -1 because
the second derivative is negative
at this point.
The function has a relative
minimum 0 at x = 1 because the
second derivative is positive at
this point.
Example 2
The Second Derivative Test is less popular than the First
Derivative Test for two reasons: the Second Derivative Test does
not always work and a question often requires a student to find
where a function is increasing/decreasing before asking for the
relative min/max (Thus, the sign chart for the First Derivative
Test is done already).
Yet, the AP Test will have questions, like the one below,
specifically about the Second Derivative Test:
Example: If the function f has a horizontal tangent line
at x = 4 and f ''(4) = 3, what is true about f(4)?
f  4  is a relative minimum