Download Lesson 15 Concavity and the Second Derivative Test

LESSON 15 CONCAVITY AND THE SECOND DERIVATIVE TEST
Definition Let f be a function that is differentiable at c. Then
1.
the graph of f is concave upward at the point P ( c , f ( c ) ) if there exists an
open interval ( a , b ) containing c such that on the interval the graph of f is
above the tangent line through the point P,
2.
the graph of f is concave downward at the point P ( c , f ( c ) ) if there exists
an open interval ( a , b ) containing c such that on the interval the graph of f
is below the tangent line through the point P.
Theorem (Test for Concavity) Suppose that a function f is differentiable on an
open interval containing c and that f ( c ) exists. Then
1.
if f ( c )  0 , then the graph of f is concave upward at the point
P(c, f (c)) ,
2.
if f ( c )  0 , then the graph of f is concave downward at the point
P(c, f (c)) .
Definition A point on the graph of a function at which the concavity changes is
called a point of inflection or an inflection point.
Example Find the interval(s) on which the following functions are concave
upward and concave downward. Also, find the point(s) of inflection ( or inflection
point(s).)
1.
f ( x )  x 3  10 x 2  25x  50
f ( x )  3x 2  20 x  25
f ( x )  6 x  20  2 ( 3x  10 )
Copyrighted by James D. Anderson, The University of Toledo
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Sign of f ( x ) :

+

10

3
 10

,   , then the
NOTE: Since f ( x )  0 when x is in the interval  
 3

graph of the function f is concave upward on this interval. Since
10 

 , then the graph of the
f ( x )  0 when x is in the interval    , 
3 

function f is concave downward on this interval.
10
is in the domain of the function f, then an inflection
3
10
point occurs when x  
. Now, we will calculate the y-coordinate of
3
this point:
NOTE: Since 
1000
1000
250
 10 
f 


 50 =

27
9
3
 3 
250  4
4
250  4
12
9

 1   50 =
   50 =
 
 
3  9
3
3  9
9
9

250  1 
250
250 1350
1600
 50 = 

    50 = 
= 
3  9
27
27
27
27
Answer:


Concave Upward:  
10

, 
3



Concave Downward:    , 


Inflection Point(s):  
10 

3 
10
1600 
, 

3
27 
Copyrighted by James D. Anderson, The University of Toledo
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2.
y  3x 5  5 x 3
dy
 15 x 4  15 x 2
dx
d2y
 60 x 3  30 x = 30 x ( 2 x 2  1)
2
dx
2
NOTE: 2 x  1  0  2 x  1  x 
2
d2y
Sign of
:
dx 2
2


+


2
1
1
 
 x
2
2
2

2
0
2
+

2
2


 2

2
d2y



,

,
0

,


0
NOTE: Since
when x is in the set 
2



2
dx


 2

then the graph of the function y is concave upward on this set. Since


2 
d2y


 0,


,



0
when x is in the set 
2


2 
dx


of the function y is concave downward on this set.
2
2 
 , then the graph
2 
2
, 0, and
are in the domain of the
2
2
2
x


function y, then inflection points occurs when
, x  0 , and
2
2
x 
. Now, we will calculate the y-coordinate of these points:
2
NOTE: Since the numbers 
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5
3



2 
2 
2 
2

  3 
  5 

y
x


For
: 




 =
2
2
2
2






 4 2 
 2 2


3 
 5 



32 
8



3 2
8

10 2


 2 2
2 



3 
 5 
 = 


8 
8



7 2
=
8
5
NOTE: ( 2 ) 
8
2
2
2
2  22
2
5
For x 
8

2  4 2
3
 2 
 2 
 2 
2

  3
  5

y
:  2 



 =
2


 2 
 2 
4 2 
2 2
  5
3
 32 
 8



3 2




10 2
8
= 

 2 
2 2


  5
3
 =  8 
 8








7 2
8
For x  0 : y ( 0 )  0  0  0
Answer:

Concave Upward:  


 2

2
, 0  
, 

 2

2





2 


 0,


,


Concave Downward: 


2 



2 7 2
,
Inflection Point(s):  
2
8

2 

2 
 2

7 2
 ; (0, 0); 
, 

 2
8


Copyrighted by James D. Anderson, The University of Toledo
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



3.
s( t )  3 t 5  20 t 4  24 t  12
s( t )  15 t 4  80 t 3  24
s( t )  60 t 3  240 t 2 = 60 t 2 ( t  4 )
Sign of s ( t ) :

+

4
+

0
NOTE: Since s ( t )  0 when t is in the set (  4 , 0 )  ( 0 ,  ) and the
number 0 is in the domain of the function s, then the graph of the function s
is concave upward on the interval (  4 ,  ) . Since s ( t )  0 when t is in
the interval (   ,  4 ) , then the graph of the function s is concave
downward on this interval.
NOTE: Since  4 is in the domain of the function f, then an inflection point
occurs when x   4 . Now, we will calculate the y-coordinate of this point:
Using a calculator, we have that s(  4 )  3 (  1024 )  20 ( 256 )  96  12
=  3072  5120  96  12 = 2132
Not using a calculator, we have that
s(  4 )  3 (  4 ) 5  20 (  4 ) 4  24 (  4 )  12 =
 3  45  20  4 4  24  4  12 = 45 (  3  5  1)  4 ( 24  3 ) =
45 ( 2 )  4 ( 21) = 4[ 4 4 ( 2 )  21] = 4 [ 256 ( 2 )  21] =
4 ( 512  21) = 4 ( 533 ) = 2132
Answer:
Concave Upward: (  4 ,  )
Copyrighted by James D. Anderson, The University of Toledo
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Concave Downward: (   ,  4 )
Inflection Point(s): (  4 , 2132 )
4.
y  ( 64  u 3 ) 3
y   3 ( 64  u 3 ) 2 (  3u 2 ) =  9u 2 ( 64  u 3 ) 2
NOTE: In order to obtain the second derivative of the function y, you could
use the Product Rule. However, using properties of exponents, we have that
u 2 ( 64  u 3 ) 2 = [ u ( 64  u 3 ) ]2 = ( 64u  u 4 ) 2
2
3 2
4 2
Thus, y    9u ( 64u  u ) =  9 ( 64u  u ) . Using the Chain Rule
again, we have that
y    9 [ 2 ( 64u  u 4 ) ( 64  4u 3 ) ] =  18[ u ( 64  u 3 ) ][ 4 (16  u 3 ) ] =
 72 u ( 64  u 3 ) (16  u 3 )
3
3
NOTE: 64  u  0  64  u  u  4 and
16  u 3  0  16  u 3  u  3 16 = 2 3 2
Sign of y  :

+

0

+

2
3
2

4
NOTE: Since y   0 when u is in the set (   , 0 )  ( 2 3 2 , 4 ) , then
the graph of the function y is concave upward on this set. Since y   0
when u is in the set ( 0 , 2 3 2 )  ( 4 ,  ) , then the graph of the function y
is concave downward on this set.
Copyrighted by James D. Anderson, The University of Toledo
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NOTE: Since the numbers 0, 2
3
2 , and 4 are in the domain of the function
y, then inflection points occurs when x  0 , x  2 3 2 , and x  4 . Now,
we will calculate the y-coordinate of these points:
3
For x  0 : y ( 0 )  64
For x  2 3 2 
3
3
3
16 : y ( 2 3 2 )  ( 64  16)  48 =
3
For x  4 : y( 4 )  ( 64  64 )  0
Concave Upward: (   , 0 )  ( 2 3 2 , 4 )
Answer:
Concave Downward: ( 0 , 2 3 2 )  ( 4 ,  )
3
3
Inflection Point(s): ( 0 , 64 ) ; ( 2 3 2 , 48 ) ; ( 4 , 0 )
5.
f ( x )  4 x (15  x)
NOTE: The domain of the function f is the interval [ 0 ,  ) .
f ( x )  x1 / 4 (15  x) = 15 x1 / 4  x 5 / 4
f ( x ) 
15  3 / 4 5 1 / 4
x
 x
4
4
f ( x )  
5( x  9)
45  7 / 4
5
5  7/4
x
 x  3/ 4 = 
x
(9  x) = 
16
16
16
16 x 7 / 4
NOTE: f ( x )  0  x   9 . However,  9 is not in the domain of the
function f .
Copyrighted by James D. Anderson, The University of Toledo
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
Sign of f ( x ) :

0
NOTE: Since f ( x )  0 when x is in the interval ( 0 ,  ) , then the graph
of the function f is concave downward on this interval. Since f ( x ) is not
positive anywhere, then the graph of the function f is not concave upward
anywhere.
Answer:
Concave Upward: Nowhere
Concave Downward: ( 0 ,  )
Theorem (The Second Derivative Test for Local Maximums and Minimums) Let
c is a critical number of a function f. If f is differentiable on an open interval
containing c, then
1.
if f ( c )  0 , then f has a local maximum occurs at x  c and the local
maximum is f ( c ) ,
2.
if f ( c )  0 , then f has a local minimum occurs at x  c and the local
minimum is f ( c ) ,
3.
if f ( c )  0 , then the test fails and you will need to use the First Derivative
Test to classify whether a local maximum, local minimum, or neither occurs
at x  c .
NOTE: You would use the Second Derivative Test if it was the second derivative
of the function can be calculate quickly and you didn’t want the information about
when the function is increasing and decreasing. Remember, if you have the sign of
the first derivative, then the First Derivative Test will give you the information
about local maximums and local minimums.
Examples Find the critical numbers for the following functions. If the second
derivative of the following functions can be calculated quickly, then find the local
Copyrighted by James D. Anderson, The University of Toledo
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maximum(s) and local minimum(s) using the Second Derivative Test. If the
second derivative can not be calculated quickly, then state this and do not find the
local maximum(s) and local minimum(s).
1.
f ( x )   2 x 2  12 x  19
NOTE: The domain of the function f is the set of all real numbers.
f ( x )   4 x  12  4 ( 3  x )
Critical Number: 3
f ( x )   4 
f ( 3 )   4  0  loc max occurs at x  3
f ( 3 )   18  36  19   1
Answer:
Local Maximum(s):  1
Local Minimum(s): None
2.
y
6
4t 2  3t  22
NOTE: The domain of the function y is the set of all real numbers such that
11
t
and t  2 .
4
We differentiated this function in Lesson 9 obtaining that
 6 ( 8t  3 )
dy

dt ( 4t 2  3t  22 ) 2
Critical Number: 
3
8
Copyrighted by James D. Anderson, The University of Toledo
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The second derivative can not be calculated quickly. It would be faster to use
the First Derivative Test to classify the local maximum(s) and local
minimum(s). See this work in Lesson 14.
3.
g( w) 
3w  22
4w  3w  22
2
NOTE: The domain of the function g is the set of all real numbers such that
11
w
and w  2 .
4
We differentiated this function in Lesson 9 obtaining that
g ( w ) 
4w ( 44  3w )
( 4w 2  3w  22 ) 2
Critical Numbers: 0 and
44
3
The second derivative can not be calculated quickly. It would be faster to use
the First Derivative Test to classify the local maximum(s) and local
minimum(s). See this work in Lesson 14.
4.
y
5 x  17
9  x2
NOTE: The domain of the function y is the set of all real numbers such that
x   3 and x  3 .
We
y 
differentiated
( x  5 ) ( 5x  9 )
( 9  x2 )2
Critical Numbers:
this
function
in
Lesson
9
,5
5
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9
obtaining
that
The second derivative can not be calculated quickly. It would be faster to use
the First Derivative Test to classify the local maximum(s) and local
minimum(s). See this work in Lesson 14.
5.
h( z )  10 3 ( z 2  3z  18 ) 2
NOTE: The domain of the function h is the set of all real numbers.
We differentiated this function in Lesson 10 obtaining that
h( z ) 
20 ( 2 z  3 )
20 ( 2 z  3 )
=
3 ( z 2  3z  18 )1 / 3
3[ ( z  6 ) ( z  3 ) ]1 / 3
Critical Number(s):  6 , 
3
,3
2
The second derivative can not be calculated quickly. It would be faster to use
the First Derivative Test to classify the local maximum(s) and local
minimum(s). See this work in Lesson 14.
6.
s( t )  2 t 3  13t 2  20 t  18
NOTE: The domain of the function s is the set of all real numbers.
s( t )  6 t 2  26 t  20 = 2 ( 3 t 2  13 t  10 ) = 2 ( t  5 ) ( 3 t  2 )
Critical Number(s): 
2
,5
3
s ( t )  12 t  26 = 2 ( 6 t  13 )
2
 2
s      (  ) (  )  (  )  loc max occurs at t  
3
 3
s ( 5 )  (  ) (  )  (  )  loc min occurs at t  5
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16
52
40
16
156
360
486
674
 2
s    


 18 = 



=
27
9
3
27
27
27
27
27
 3
s( 5 )  250  325  100  18   157
Answer:
Local Maximum(s):
674
27
Local Minimum(s):  157
7.
y  x 4  8x 3  12
NOTE: The domain of the function y is the set of all real numbers.
dy
 4 x 3  24 x 2 = 4 x 2 ( x  6 )
dx
Critical Numbers:  6 , 0
d2y
 12 x 2  48 x = 12 x ( x  4 )
2
dx
x   6:
x  0:
d2y
dx 2
d2y
dx 2
 (  ) (  ) (  )  (  )  loc min occurs at x   6
x6
 0  the test fails
x0
Using the First Derivative Test, we have that
Sign of
dy
:
dx
X

6
Inc
Inc
+
+

0
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Thus, neither a local maximum nor a local minimum occurs at x  0 since
the function y is increasing on the left-hand side of x  0 and it is still
increasing on the right-hand side of x  0 .
dy
NOTE: We do not need the sign of
on the interval (   ,  6 ) in order
dx
to classify what is happening at the critical number of x  0 .
Using a calculator: y (  6 )  1296  1728  12   444 or
y(  6 )  6 4  8  63  12 = 6 ( 6 3  8  6 2  2 ) = 6[ 6 2 ( 6  8 )  2 ] =
6 [ 36 (  2 )  2 ] = 6 (  72  2 ) = 6 (  74 ) =  444
Answer:
Local Maximum(s): None
Local Minimum(s):  444
8.
g ( x )  5x 
4
x2
NOTE: The domain of the function g is the set of all real numbers such that
x  0.
5x 3  8
3
3
3
g ( x )  5  8 x
= x (5 x  8 ) =
x3
Critical Numbers: 
2
3
5
g ( x )   8 x  4 
g ( x )  5  8 x  3 

2 
8
g   

 3 5 



 2
 3 5





4

8
x4
()
 (  )  loc max occurs at x  
()
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2
3
5
 8
5
   4


2
5x 3  4
4
   5
g 
Since g ( x )  5 x  2 
,
then
=
 3 5 
4
x
x2


3 25
8  4
= 
4
3
25
Answer:
12
4
3
= 
12
3
25
4
=  3 3 25
25
Local Maximum(s):  3 3 25
Local Minimum(s): None
9.
f ( x )  4 x (15  x)
NOTE: The domain of the function f is the interval [ 0 ,  ) .
f ( x )  x1 / 4 (15  x) = 15 x1 / 4  x 5 / 4
f ( x ) 
5  3/ 4
5( 3  x )
15  3 / 4 5 1 / 4
x
 x
x
(3  x) =
=
4
4
4
4 x 3/ 4
Critical Number(s): 3
f ( x ) 

15  3 / 4 5 1 / 4
x
 x

4
4
f ( x )  
45  7 / 4
5
x
 x  3/ 4 =
16
16
5( x  9)
5  7/4
x
(9  x) = 
16
16 x 7 / 4
f ( 3 )  
5 ( 12 )
()


 (  )  loc min occurs at x  3
7/4
16 ( 3 )
()
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Since f ( x )  4 x (15  x) , then f ( 3 )  12 4 3
Answer:
Local Maximum(s): None
Local Minimum(s): 12 4 3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850