#### Transcript Applications of the Second Derivative

```Applications of the Second
Derivative
BY
DR. JULIA ARNOLD
USING TAN’S 5TH EDITION APPLIED
CALCULUS FOR THE MANAGERIAL ,
LIFE, AND SOCIAL SCIENCES TEXT
I recommend that you view the power point as a slide
presentation first. After viewing then print the slides as
notes according to the instructions below. Under view, click
on slide show.
To print the slides as notes go to the file command and
select print. On the window that comes up toward the
bottom it may say print what and next to that is a drop
down menu. Select handouts and the number you want per
page.
Determining the Intervals of Concavity
A function is concave up on an interval (a,b) if f’ is
increasing on (a,b)
Slope is -.5
Slope is .5
Since the slopes are
Slope is 0
-.5,0,.5
and since these numbers are increasing
the function is concave up.
Associate concave up with a bowl shape and it is right side up
so everything is stays in the bowl. Hence “positive”
+
A function is concave down on an interval (a,b) if f’(x) is decreasing
on (a,b).
Slope is 0
Slope is 2
Since the slopes are
2,0,-2
and since these numbers are decreasing
the function is concave down.
Slope is -2
Associate concave down with a bowl shape and it is
upside down so everything is falling out of the bowl.
Hence “negative”
-
Since the first derivative can tell us where a function is
increasing and where a function is decreasing, then if that
function is f’, the second derivative will tell us where f’(x) is
increasing or decreasing.
Concave Down
First Derivative is
decreasing.
Second Derivative is negative.
Since the first derivative can tell us where a function is increasing
or decreasing, then if that function is f’, the second derivative will
tell us where f’(x) is increasing or decreasing.
The function is concave up.
The first derivative is increasing.
The second derivative is positive.
The Rule:
1. If f”(x) > 0 for each value of x in (a,b) then f is concave
upward on (a,b).
2. If f”(x) < 0 for each value of x in (a,b) then f is concave
downward on (a,b).
Example 1: f(x)= x2 is a function which is only concave up on
the entire x axis and f(x)= -x2 is concave down on the entire
x axis.
Concave up
Concave down
f’(x)= 2x
f”(x)= 2 which is always a
positive number
f’(x)= -2x
f”(x)= -2 which is always a
negative number
Example 1: Determine the intervals where the function is
concave up and concave down
f(x)  x3  3x2  24x  32
Step 1: Find the second derivative of f
f(x)  x3  3x2  24x  32
f(x)  3x2  6x  24
f(x)  6x  6
Step 2: Set the second derivative equal to 0 and solve for x.
f(x)  6x  6
0  6x  6
6  6x
1x
1
Draw a number line and insert 1 on it.
Two intervals are
formed.
 ,1
1,  
Test Points
 ,1 0
2
1,  
1
Substitute in f”(x)=6x-6
Resulting Sign
-6
6
+
Interpretation of resulting signs is that the function is
concave down on  ,1 and concave up on 1,   .
The red function is concave
down, then when it reaches
x = 1 it switches concavity
to upward.
1
Example 2: Find the intervals where the function f(x)  x 
x
is concave up and concave down.
Remember that the domain does not contain 0.
Thus the intervals over which the function is defined are
 ,00, 
Next, find f’(x).
1
x
1
f(x )  1  2
x
f( x )  x 
Now find f”(x)
f(x)  1 
1
2

1

x
x2
f(x)  2x
3
f(x) 
2
x3
Since f”(x)  0,
we look at where it
is undefined or
discontinuous, which
would be at x = 0.
 ,0
Thus the only intervals are:
0, 
Intervals Test Point Substitution into f(x) 
 ,0
0, 
-1
1
2/-1 = -2
2/1 =2
2
x3
Result
+
Interpretation of results:
On 0,  f”(x) > 0
f(x) is concave up.
On  ,0 , since
f”(x) < 0,
f(x) is concave down.
Example 3: Find the intervals where the function f(x)  x
is concave up and concave down.
Find f’(x).
Now find f”(x)
1
f(x)  x
2
2
f(x)  x 3  3
3
3 x
2
3
2
f(x)  x
3
1
3
2
 3
3 x
4
2 3
2


f (x ) 
x 
4
9
9x 3
Since f” can not be 0, we choose x = 0 where f”(x) is
undefined or discontinuous.
Thus the intervals under consideration are:  ,0
0, 
2
3
Choose a test point:
Test Point
 ,0
0, 
Substitution into f’(x)
-1
1
f(1) 
2
4
3

9(1)
2
2
f(1) 

4
9
913
2
9
Result
-
Interpretation of results:
 ,0
0, 
On this interval, f”(x) < 0 so f(x) is concave down.
On this interval, f”(x) < 0 so f(x) is concave down.
f(x) is concave down
f(x) is concave down
The Rule:
1.Find all values of x for which f”(x)=0 or f” is discontinuous
and identify the open intervals determined by these points.
2.Select a test point c in each interval found in step 1 and
determine the sign of f”(c) in that interval.
A. If f”(c) >0 in that interval, f is concave upward on that
interval.
B. If f”(c) <0 in that interval, f is concave downward on that
interval.
Practice problem: Find the interval where the function is concave
up or concave down.
g(x)  x  2
The Rule:
1.Find all values of x for which f”(x)=0 or f” is discontinuous and
identify the open intervals determined by these points.
2.Select a test point c in each interval found in step 1 and determine
the sign of f”(c) in that interval.
A. If f”(c) >0 in that interval, f is concave upward on that interval.
B. If f”(c) <0 in that interval, f is concave downward on that interval.
Which is the correct first step?
g( x )  x  2
g(x)  x  2
1
g( x )   x  2 2
1
1
2
1
2
1
1
 x  2 1   x  2
2
2
3
1
1
g ( x ) 
 x  2 2 
3
4
4  x  2 2
g ( x ) 
0
1
4  x  2
3
2
_ has _ no _ solutions
x  2 _ is _ a _ discontinuity
g(x)  x  22
1
1
1
x  2 2  1  1 x  2 2
2
2
1
0
_ has _ no _ solutions
1
2
x  2
g(x) 
x  2 _ is _ a _ discontinuity
Practice problem: Find the interval where the function is concave
up or concave down.
g(x)  x  2
The Rule:
1.Find all values of x for which f”(x)=0 or f” is discontinuous and
identify the open intervals determined by these points.
2.Select a test point c in each interval found in step 1 and determine
the sign of f”(c) in that interval.
A. If f”(c) >0 in that interval, f is concave upward on that interval.
B. If f”(c) <0 in that interval, f is concave downward on that interval.
This is the correct first step.
g( x )  x  2
g( x )   x  2
1
2
1
1
1
1
2
x

2

1

x

2



2
2
2
3
1
1
g ( x ) 
 x  2 2 
3
4
4  x  2 2
g ( x ) 
0
1
4  x  2
3
2
_ has _ no _ solutions
x  2 _ is _ a _ discontinuity
1.Find all values of x for which f”(x)=0 or f” is
discontinuous and identify the open intervals
determined by these points.
 ,2 is not in the domain of the function
(2,  )
is the only interval to consider
Test point x = 3, g”(3)=-1
The function is concave down on (2, )
Rule 2B
y
8
7
6
5
4
3
2
1
9 8 7 6 5 4 3 2 1
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
x
10
This is a typical
square root
function. It is
concave down from
(2, )
Inflection Points
A point on the graph of a differentiable function f at which
the concavity changes is called an inflection point.
Finding Inflection Points
1. Compute f”(x)
2. Determine the points in the domain of f for which f”(x) = 0
or f”(x) does not exist.
3. Determine the sign of f”(x) to the left and right of each
point x = c found in step 2. If there is a change in the sign of
f”(x) as we move across the point x = c, then (c,f(c)) is an
inflection point of f.
Determine where the function is concave up and concave down
and any inflection points for
1
f(x)  2
x 1
Step 1: Find f’ and then f” of x.
1
1
2

x

1
x2  1
f(x)  1(x2  1) 2 2x

f(x) 

simpli fy
f(x)  2x(x2  1) 2
f(x)  2x[2(x2  1) 3 2x]  (x2  1) 2  2
simpli fy
f(x)  2(x2  1) 3 [4x2  x2  1]

 2  3x2  1
f(x)  2(x  1) [3x  1] 
(x2  1)3
2
3
2


 2  3x2  1
f(x) 
(x2  1)3

Step 2: Set f”(x) = 0 and also look for values where x is undefined.
 2 3x2  1
0
(x2  1)3
0  2 3x2  1
0  6x2  2
2  6x2
2
 x2
6
1
 x2
3
1
x
3
The numerator provides us with zeroes. x   1
3

 2  3x2  1
f(x) 
(x2  1)3

The denominator provides us with undefined places.

 2  3x2  1
f(x) 
(x2  1)3

(x2  1)3  0
3
(x2  1)3  3 0
x2  1  0
x2  1
no _ real _ solutions
So our only candidates for inflection points are: x   1
3
Step 3: Draw a number line and place the x values on it, then
choose test points to the right and left of the values.
Result
+
1
3
1
3
 1 is approximately -.57
3
3 Intervals
1 

  ,

3

 1 1 
,


3
3


 1

,

 3 
+
Test Points
-1
0
1

 2  3x2  1
f(x) 
Substituting into
(x2  1)3
 2(3(1)2  1)  2(3  1)  2(2) 4 1


 
((1)2  1)3
23
8 2
1  13
 2(3(0)2  1)  2(0  1)  2


 2
((0)2  1)3
0  13 13
 2(3(1)2  1)  2(3  1)  2(2) 4 1


 
((1)2  1)3
23
8 2
1  13

Interpreting the results
On
On
On
1 

  ,

3


 1 1 
,


 3 3
 1

,

 3 
f is concave up.
f is concave down.
f is concave up.
Let’s look at the graph of f.
The blue area is where f is concave up, and
the red area is where we are concave down.
Since there is a sign change over each x value below, the two
points,
are inflection points.
1
x
Result
3
+
1
3
+
1
3
The Second Derivative Test
Because the second derivative tests concavity of a function, it
seems like a natural extension to let the second derivative tell
us whether we have a relative maximum or relative minimum for
a function.
Here is how we will do that:
1. Compute f’(x) and f”(x).
2. Find the critical points of f at which f’(x) = 0.
3. Compute f”(c) for each critical point c found in step 2.
A. If f”( c) < 0 then f has a relative maximum at c.
B. If f”( c) > 0 then f has a relative minimum at c.
C. If f”( c) = 0 then f this test fails and you need to
go back and use the first derivative test.
Why is this true?
A. If f”( c) < 0 then f has a relative maximum at c.
If f”( c) < 0 then f is concave down which creates a maximum
B. If f”( c) > 0 then f has a relative minimum at c.
If f”( c) > 0 then f is concave up which creates a minimum
Determine the relative extrema for the function:
f(x)  x4  2x2  4
1. Compute f’(x) and f”(x).
f(x)  x 4  2x2  4
f(x)  4x3  4x
f(x)  12x2  4
2. Find the critical points of f at which f’(x) = 0.
f(x)  4x3  4x
0  4x 3  4x
0  4x(x2  1)
0  4x(x  1)(x  1)
x  0, x  1, x  1
3. Compute f”(c) for each critical point c found in step 2.
c1 = 0, c2 = -1, c3 = 1
Three Critical points.
f(x)  12x2  4
2
f(0)  120   4  4 ""
2
f(1)  12 1  4  8 ""
2
f(1)  121  4  8 ""
Interpretation of results:
A. If f”( c) < 0 then f has a relative maximum at c.
B. If f”( c) > 0 then f has a relative minimum at c.
C. If f”( c) = 0 then f this test fails and you need to go back
and use the first derivative test.
f(x)  12x2  4
2
f(0)  120   4  4 ""
2
f(1)  12 1  4  8 ""
2
f(1)  121  4  8 ""
Since f”(0) < 0, f has a relative maximum at x = 0. The point is
(0, 4)
Since f”(-1) > 0, f has a relative minimum at x = -1. The point is
(-1, 3)
Since f”(1) > 0, f has a relative minimum at x = 1. The point is
(1, 3)
The graph of
f(x)  x4  2x2  4
(0,4)
rel max
(-1,3)
rel min
(1,3)
rel min
Go to the HW
```