12 Chapter 12

3.2 Analyzing Graphs with the Second Derivative

Thus far, we have learned that the derivative f’ provides a great deal of information about the graph of f. In this section, we will investigate the information we can obtain about the graph of f from taking the derivative of the derivative.  In other words, we will find

d  (f’ (x)), which is called the second derivative of f. It turns out that not only does the second derivative provide useful information about the graph of f, it also provides relevant information about the graph of the first derivative, f’. Why? Because the second derivative

is the derivative of the first derivative!

 

Learning Objectives: In this section, you will learn how to apply the Concavity Test to find where a function is concave up, concave down, and has inflection points as well as solve problems involving a real-world application: the point of diminishing returns. Also, you will learn how to apply the Second Derivative Test to find local extrema. Upon completion, you will be able to:

• Find the second derivative of a function.
• Examine the graph of a function to determine where its slopes are increasing and where they are decreasing.
• Define partition number of f”.
• Find the partition numbers of f” for a function f given its rule.
• Specify the conditions necessary for a function to have an inflection point.
• Apply the Concavity Test to determine where a function f is concave up/down and has inflection points given its rule.
• Apply the Concavity Test to determine where a function f is concave up/down and has inflection points given the rule for its derivative f’ and the domain of f.
• Apply the Concavity Test to determine where a function f is concave up/down and has inflection points given the rule for its second derivative f” and the domain of f.
• Analyze the graph of f to determine the partition numbers of f”, the intervals where f  is concave up/ down, and the inflection points of f.
• Analyze the graph of f’ to determine the partition numbers of f”, the intervals where f is concave up/down, and the x-values where f has inflection points.
• Analyze the graph of f” to determine the partition numbers of f”, the intervals where f is concave up/ down, and the x-values where f has inflection points.
• Investigate the relationships between f, f’, and f”, and construct a table indicating these relationships.
• Analyze the graph of f, f’, or f” to determine pertinent information about the other two functions.
• Calculate the point of diminishing returns and explain its significance.
• Compare and contrast the Concavity Test and the Second Derivative Test.
• Compare and contrast the First Derivative Test and Second Derivative Test.
• Specify the type(s) of critical values we may attempt to use with the Second Derivative Test.
• Describe the conditions in which the Second Derivative Test fails.
• Perform the Second Derivative Test, if possible, to find and classify the local extrema of a function.

Finding the Second Derivative

We begin by formally defining the second derivative of a function f:

 

 

 

 

Example 1 Find f”(x), where f(x) = 7×4 –  3×3 + 44×2 + 22x – 83.

Solution: Before we can find f”(x), we must find f'(x):

J'(x) = 7 • 4×3 –  3 • 3×2 + 44 • 2×1 + 22 – 0

= 28×3 –  9×2 + 88x + 22

 

Taking the derivative of this function to obtain f” (x) gives

J” (x) = 28 • 3×2 –  9 • 2×1 + 88

= 84×2 –  l8x + 88

 

There are several ways to write the second derivative:

Second Derivative Notation   Language notation: f”(x) and g” Leibniz notation: d^2 f / dx^2 and d^2y / dx^2

 

Example 2 Find each of the following.

 

(a)     f”(x) if .f(x) = 4s-2×3

(

b) d2g if . (x)=    15 ln(x)

dx2        Y           –2x7 +14

(c) h”(x) if h'(x) = (-9×11 –  8×31) Jx2 –  5×17

Solution:      (a) The first step when finding the second derivative is to find the first derivative of the function. Starting with the Chain Rule to obtain .f’ ( x) gives

.f'(x) = 48-2×3(ln(4)) d  (8 – 2×3)

3

= 4s-2x (ln(4)) (0 – 6×2)

= 48-2×3 (ln(4)) (-6×2)

Remember our rule on whether or not to algebraically manipulate our result: If we need to use the function to find another quantity or solve an equation, then we should algebraically manipulate the function. In this case, we are going to take the derivative of this function, so we should algebraically manipulate it, or in this

8 2                       2

case, rewrite it:                                          3

.f'(x) = 4 – x (ln(4)) (-6x )

= -6(ln(4))x24s-2×3

Now, we take the derivative of this function using the Product Rule:

.f”(x) = d  (-6(ln(4))x24s-2×3)

 

= -6(ln(4)) (d  (x24s-2×3))

 

=                                                  3                    3

-6(ln(4)) ((x2) (;    (48-2x )) + (48-2x ) (d  (x2)))

)))                              )

= -6(ln(4)) ( (x2) ( 4s-2×3 (ln(4)) ( d  (8 – 2×3         + (4s-2×3  (2x))

2          2 3                    2                2 3

= -6(ln(4)) ((x ) (4s- x (ln(4)) (-6x )) + (4s- x ) (2x))

Note: \i\Then finding f”(x), notice we had to find d ( 4s-2×3). We already found this

derivative previously; it is .f’ (x) ! Because we found it already, we could have substituted .f'(x) = -6(ln(4))x248-2×3 at that step instead of finding the derivative again.

(b)    Recall g(x) =     ; Jt. To find the first derivative, we start with the Quotient

Rule:

dg      (-2×7 + 14) C (15ln(x))) – (15ln(x)) (d (-2×7 + 14))

dx                                           (-2×7 + 14)2

(-2×7 + 14) (1;) – (15ln(x)) (-14×6) (-2×7 + 14)2

-30×6 + 210x-1 + 210×6(ln(x))

(-2×7 + 14)2

taking the derivative of this function will be slightly easier. We take the derivative again starting with the Quotient Rule. Because this function is more complicated, we will start by finding the derivative of the numerator only:

 

d

dx (-30×6

 

+ 210x-1

+ 210×6(ln(x))) = -180×5

 

– 210x-2

d

+ dx (210×6

 

ln(x))

 

–

= -180×5      210x-2 + (210×6) (d (ln(x)))

+(ln(x)) (d (210×6))

–

= -180×5       210x-2 + ( (210×6) 1+ (ln(x)) (1260×5))

= -180×5 –  210x-2 + 210×5 + 1260×5(ln(x))

=30×5 –  210x-2 + 1260×5(ln(x))

 

 

Next, we find the derivative of the denominator:

 

 

) )

d  ( (- 2×7 + 14) 2           = 2 (- 2×7 + 14) ( d  (- 2×7 + 14))

= 2 (-2×7 + 14) (-14×6)

= -28×6 (-2×7 + 14)

= 56×13 –  392×6

 

 

Using the Quotient Rule and placing the two derivatives we found in the proper places gives

FORMATTING ISSUE

 

d2g =!!:_ (-30×6 + 210x-1 + 210×6(ln(x)))

dx2     dx                     (-2×7 + 14)2

(-2×7 + 14)2 (30x5 – 210x-2 + 1260×5(ln(x))) ((-2×7 + 14)2/

(-30x6 + 210x-1 + 210×6(ln(x))) (56×13 –  392×6)

((-2×7 + 14)2) 2

 

(c)   Recall that h'(x) = (-9×11 – 8×31) Jx2 – 5×17.  We must be careful!  h”(x) is the derivative of h'(x). We are already given h'(x), so we just need to take the derivative of this function. We start with the Product Rule and then incorporate

 

the Chain Rule:

 

–

h”(x) = ( – 9×11

 

8×31(d (Jx2

 

 

–

5×17)) + ( Jx2

 

5×17) (d (-9×11       8×31))

–

–

= ( – 9×11 – 8×31 (}  (x2 – 5×17)-½ ( d  (x2 – 5xl7)))

+ ( Jx2 –  5×17) (-99×10-  248×30)

= ( – 9×11 –  8×31 (}  (x2 – 5×17)-½(2x – 85×16))

+ ( Jx2 – 5×17) (-99×10- 248×30)

 

Note: \i\Then finding the second derivative, all of our previous derivative rules apply. We will still use the Introductory Derivative Rules, the Product Rule, the Quotient Rule, and the Chain Rule (in all its forms).

 

There is nothing stopping us from taking higher-order derivatives (third, fourth, and so on), except that the information we glean from these is beyond the scope of this textbook. The notations for higher-order derivatives are given below:

 

 

 

Concavity

Similar to the way the first derivative f’ provides information about where the graph of a function f is increasing and where it is decreasing, as well as where it has local extrema, the second derivative also provides important information about the graph of f. The second derivative .f” informs us of the concavity off.

 

Before we investigate this idea from a calculus standpoint, let’s define concavity pictorially:

 

This is not a very precise definition, but we can use calculus to find a more precise definition. Let’s look at the graph of a function that is concave up. In other words, the graph of the function “bends upward” like a bowl or part of a bowl. If we also look at lines tangent to the graph of the function, we see the slopes of the tangent lines are increasing. This will always be true for graphs that are concave up! See Figures 3.2.1, 3.2.2, and 3.2.3.

 

Figure 3.2.1: Graphs of a function that is concave up and a line tangent to the function at x = -2.5

Figure 3.2.2: Graphs of a function that is concave up and a line tangent to

the function at x = 1.5

 

 

Figure 3.2.3: Graphs of a function that is concave up and a line tangent to the function at x = 4

Let’s take a more in-depth look at why the slopes of the graph of the function shown in Figures 3.2.1, 3.2.2, and 3.2.3 are always increasing. The slopes on the interval (-oo, 0) are negative because the function is decreasing. Because the function is getting less steep, its slopes are getting less negative. Thus, its slopes are increasing (numbers that are getting less negative are increasing numbers). This will always be true for graphs that are concave up!

 

Likewise, the slopes on the interval (0, oo) are positive because the function is increasing. Because the function is getting steeper (i.e., “more” steep), its slopes are getting more posi­ tive. Thus, its slopes are increasing (numbers that are getting more positive are increasing numbers). This will always be true for graphs that are concave up!

 

The same analysis holds for graphs of functions that are concave down (i.e., “bend down­ ward” like an upside-down bowl or part of a bowl). The slopes of the graphs of such functions will be decreasing regardless of whether the function itself is increasing or decreasing. If the function is increasing and concave down, the slopes are getting less positive, and if the func­ tion is decreasing and concave down, the slopes are getting more negative. In either case, the slopes are decreasing. See Figures 3.2.4, 3.2.5, and 3.2.6.

 

 

Figure 3.2.4: Graphs of a function that is concave down and a line tangent to the function at x = -2.5

Figure 3.2.5: Graphs of a function that is concave down and a line tangent to the function at x = 1.5

 

 

 

 

 

Figure 3.2.6: Graphs of a function that is concave down and a line tangent to the function at x = 4

 

 

 

Recall from Section 3.1 that the slopes of the graph of a function fare given by its derivative function, f’. Thus, we can now state a formal definition of concavity:

 

 

 

Notice that for concavity, we are looking at the rate of change of the rate of change (slope). In other words, we are looking at the second derivative! This allows us to interpret concavity from a calculus perspective:

 

 

 

In other words, (c, f(c)) is an inflection point off  if f”(x) changes sign.

 

If we have the graph of a function, we can just look at the graph to find the intervals of concavity and identify inflection points. But, how can we determine the concavity and find inflection points if we are only give the rule of the function and not its graph? We will answer

this question by asking another question: If f has an inflection point where it switches from concave up to concave down (or vice versa), what does that imply about f” at the inflection point?

 

Well, if the concavity off changes, then J”(x) must change sign. There are only two possi­ bilities for f” (x) to change sign: f” (x) = 0 or if f” (x) does not exist (for further explanation, see the analysis in Section 3.1 about f'(x) changing sign and there being only two ways to cross a river). Each function whose graph is shown in Figures 3.2.8 and 3.2.9 has an inflection point at x = 4 because the concavity of f changes:

 

 

Figure 3.2.8: Graph of a function f that has an inflection point at x = 4

Figure 3.2.9: Graph of a function f that has an inflection point at x = 4

In Figure 3.2.8, we see the function f is concave down on the interval (-oo, 4), so f”(x) < 0 on this interval. On the interval (4, oo), the function is concave up, so f”(x) > 0 on this interval. And, in this case, f” (4) = 0.

 

In Figure 3.2.9, we see the function f is concave up on (-oo, 4), so f”(x) > 0 on this in­ terval. On the interval (4, oo), f is concave down, so f” (x) < 0 on this interval. Looking at the behavior at x = 4, we see the function has a vertical tangent line. Thus, f’ (4) does not exist, and so f”(4) does not exist either.

 

This may seem somewhat familiar. In Section 3.1, we learned that if a function has a local extremum, it must occur at an x-value in which f'(x) = 0 or f'(x) does not exist (among other conditions). The analysis here is similar for inflection points, only we are dealing with the second derivative instead of the first. Thus, if a function has an inflection point at x = c, it must occur where f”(c) = 0 or f”(c) does not exist. However, just like with local extrema, if there is an x-value such that f”(x) = 0 or f”(x) does not exist, that does not necessarily mean there is an inflection point at that x-value.

 

For f to have an inflection point at x = c, the second derivative must also change sign (switching between positive/negative) as shown in Figures 3.2.8 and 3.2.9. Also, x = c must be in the domain of f. There cannot be an inflection point if there is no point on the graph!

 

Thus, to find the intervals of concavity and any inflection points of a function f, we will create a sign chart of f”(x). We will partition the sign chart again with “important” x-values. As discussed previously, these “important” x-values are where f”(x) = 0 and f”(x) does not exist. We call these x-values partition numbers off”. They will be used to partition the sign chart of f” (x):

 

Note: Similar to partition numbers off’, we will need to make judgement calls on which x-values that meet the conditions of the definition are “important enough” to be considered

 

partition numbers of .f” based on how meaningfully the x-values partition the number line.

 

Combining all of this information, we now have a process for finding where a function is concave up/down and has inflection points.

 

 

Concavity Test

To determine the intervals where a function is concave up and where it is concave down, which will determine if there are any inflection points, we will use the following three steps to apply the Concavity Test:

 

 

We will demonstrate using this process in the following example.

Example 3 For each of the following, find the intervals of concavity and any inflection points of f.

(a)     f(x) = ½x4 + ½x3 –  14×2

(b)   f (x) =x2x  9

(c)    f(x) = xe-x2

Solution:      (a) We will use the three steps outlined previously to apply the Concavity Test:

1    Determine the domain of f:

f is a polynomial, so its domain is (-oo, oo).

2   Find the partition numbers of .f”:

 

First, we must find f'(x):

J'(x) = 2x3 + x2 –  28x

Now, we can find f”(x):

J”(x) = 6×2 + 2x – 28

 

To find the partition numbers off”, we find the x-values where f”(x) = 0 or f” (x) does not exist. Because f” is a polynomial and has a domain of all real numbers, it will exist everywhere. Thus, we only need to find the x-values where f”(x) = 0:

f”(x) = 0 6×2 + 2x – 28 = 0

2 (3×2 + X –  14) = 0

2(3x + 7)(x –  2) = 0

7

x=–andx=2.

3

Thus, the partition numbers off” are x = -i     and x = 2.

3. Create a sign chart of f” (x):

We will place the partition numbers of f” on a number line, with x = -i and x = 2 both having solid dots to indicate they are in the domain of f.

Next, we need to determine the sign of f”(x) on the intervals (-oo, -D,(-i, 2),

and (2, oo). We will choose the x-values x = -4, 0, and 3 to test:

J”(x) = 6×2 + 2x – 28

J”(-4) = 6(-4)2 + 2(-4) – 28 = 60 > 0

J” (0) = 6(0)2 + 2(0) – 28 = -28  < 0

J” (3) = 6(3)2 + 2(3) – 28 = 32 > 0

Using this information, we can fill in the sign chart of f” (x). Because we are also interested in the information this yields for f, we include that information below the number line. See Figure 3.2.10.

Figure 3.2.10: Sign chart of  f”(x) with the corresponding information for

J(x) = ½x4 + ½x3 –  14×2

 

Looking at the sign chart, we see that f”(x) changes sign at both x = -i  and x = 2. Now, we must check to see if the function is defined at these x-values, which it is. Thus, f has inflection points at both of the partition numbers of

 

.f”. To find the 71-values associated with the inflection points, we substitute the x-values into the original function f:

 

–

f(x) = }x4 + }x3      14×2 ==}

3         2       3          3        3                     3                162

f (- ) =! (- )4 + ! (- )3-14 (- )2   10,633      -65.6J58

1            1

f(2) =      4                    3

2              136

2(2)  +3(2) –       14(2)

= -3        -45.3333

 

3′                                                                                ‘          3’   162                           ‘   3      •

In conclusion, f(x) is concave up on (-oo, -i) and (2, oo). It is concave down on (-1 2) and has inflection points at (-1 -10,633) and (2 -136)

We can check our work by looking at the graph off shown in Figure 3.2.11, but using the Concavity Test allows us to determine the concavity and find the inflection points completely algebraically! Again, this is why we study calculus: to calculate with more precision than technology allows, which is to say we can get exact answers! With this function, technology would give us roughly the same information we calculated, but if we want exact answers, then calculus is the way to go!

Figure 3.2.11: Graph of the function f(x) = ½x4 + ½x3 –  14×2

(b)   Recall f (x) =xL9.  Again, we will use the three steps outlined previously to apply the Concavity Test:

1.   Determine the domain of f:

Considering the three domain restrictions listed previously, the function does not have any logarithms or even roots. However, there is division, so we need to make sure the denominator does not equal zero:

x2 – 9 =I= 0

x2 =I= 9

==}  x =/= 3 and    x =/= -3

 

Thus, the domain of .f is (-oo, -3) U (-3, 3) U (3, oo).

 

2.   Find the partition numbers off”:

 

First, we must find f'(x). Using the Quotient Rule gives

 

 

 

 

 

 

 

X                   2                                                             2

1           _ ( X –        9) c  (X)) –        X c  (X –        9))

.f ( )-                          ( x-2     9)2

(x2 – 9) (1) – x(2x)

(x2 – 9)2 x2 – 9 – 2×2

 

(x2 –  9)2

-x2 –  9

(x2 – 9)2

 

 

 

 

Now, we can find f”(x). We start with the Quotient Rule again and incor­ porate the Chain Rule:

 

 

 

 

(x2 – 9)2 (-4_ (-x2 – 9)) – (-x2 – 9) (-4_ ((x2 – 9/))

f”(x)=                               dx                                                .                   dx                                   

((x2 – 9)2)2

(x2 – 9)2 (-2x) – (-x2 – 9) (2 (x2 – 9) c  (x2 – 9)))

(x2 – 9)4

(x2- 9)2(-2x) – (-x2 – 9) (2 (x2 – 9) (2x))

(x2 – 9)4

 

 

 

 

Because we need to use this function to find the partition numbers of f”, we need to algebraically manipulate it. We will factor the term x2 – 9 from the numerator (remember we factor the lowest power of the term) and continue

 

simplifying:

 

,,           (x2 –  9)2 (-2x) – (-x2 –  9) (2 (x2 –  9) (2x))

f   ( X=)

( x2-    9)4

(x2 –  9) [(x2 –  9) (-2x) – (-x2 –  9) (2)(2x)]

(x2 – 9)4

(x2 –  9) [(x2 –  9) (-2x) – (-x2 –  9) (2)(2x)]

(x2 – 9)43

(x2 –  9) (-2x) – (-x2 –  9) (2)(2x)

(x2 – 9)3

(-2×3 + l8x) – (-x2 –  9) (4x)

 

 

(x2 – 9)3

(-2×3 + l8x) – (-4×3 –  36x)

 

 

(x2 – 9)3

-2×3 + l8x + 4x3 + 36x

 

 

(x2 – 9)3

2×3 + 54x

(x2 – 9)3

To find the partition numbers off”, we find the x-values where .f”(x) = 0 or f”(x) does not exist. f”(x) = 0 when the numerator equals 0, and f”(x) does not exist when the denominator equals zero (remember we are looking at the domain of f” when determining where it does not exist). First, we will find the x-values where f”(x) = 0:

f”(x) = 0 ==;:-

2×3 + 54x = 0

2x (x2 + 27) = 0

This gives us two equations to solve: 2x = 0 and x2 + 27 = 0.2x = 0 gives x = 0, but x2 + 27 = 0 has no solution because x2 -/:- -27  due to the fact that x2 will be always be nonnegative.

Now, to find the x-values where f”(x) does not exist, we set the denominator equal to zero:

f”(x) DNE  ==;:-

(x2 –  9)3 = 0

((x2 –  9)3)½ = (o)½

x2 –  9 = 0

x2 = 9

==;:- x = 3 and x = -3

Thus, the partition numbers of f” are x = -3, x = 0, and x = 3.

 

3.   Create a sign chart of f” (x):

We will place the partition numbers off” on a number line, with x = 0 having a solid dot and x = -3 and x = 3 having open circles to indicate which are included and which are not included in the domain off, respectively.

Next, we need to determine the sign of f”(x) on the intervals (-oo, -3), (-3, 0), (0, 3),

and (3, oo). We will choose the x-values x = -5, -1, 1, and 5 to test:

 

J”(x)=	3 2x + 54x ==} (x2 – 9)3
J”(-5)=	2(-5)3 + 54(-5) = ((-5)2 – 9)3	–        < 0 512
J”(-1)=	2(-1)3+54(-1)= ((-1 )2 – 9)3	!._> 0 64

3

J”(l)=  2(1)  + 54(1)=     _!._< 0

((1)2 – 9)3            64

3

!”(5)=  2(5)  + 54(5)=         > 0

((5)2 – 9)3         512

 

Using this information, we fill in the sign chart of f” (x). Again, we include the corresponding information this yields for f below the number line. See Figure 3.2.12.

Figure 3.2.12: Sign chart of  f”(x) with the corresponding information for

f(x) = x2 9

Looking at the sign chart, we see that f” (x) changes sign at all three partition numbers. We now check to see if the function is defined at these x-values. Because f is only defined at x = 0, there is only one inflection point.  Re­

member, even if f” (x) changes sign, that does not necessarily mean there is an inflection point. We must check to see if the x-value is in the domain off. To find the y-value associated with the inflection point, we substitute x = 0

into the original function f:

 

X

j ( X) = x2-  9   ==}

 

f(0)=      (0)20_ 9 =     9 =    0

 

In conclusion, f is concave down on (-oo, -3) and (0, 3), and it is concave up on (-3,0) and (3,oo). Also, f has an inflection point at (0,0).

Again, we can check our work by looking at the graph off. See Figure 3.2.13.

 

 

Figure 3.2.13: Graph of the function J(x) =x2 9

(c)

•

Recall f (x) = xe-x2        Again, we will use the three steps outlined previously to apply the Concavity Test:

1.   Determine the domain of f:

Considering the three domain restrictions listed previously, the function does not have any logarithms or even roots. There is division hidden in the function

due to a negative exponent: e-x2 = e 2. However, because ex2 > 0, this

denominator will never equal zero. Therefore, the domain off    is (-oo, oo).

2.   Find the partition numbers of f”:

First, we must find f'(x). Using the Product Rule and then incorporating the Chain Rule gives

j’ ( x) = x (

e-x

) )      + e-x

(  d ( x))

s( 2                          2

= xe-x2 ( d  (-x2)) + e-x2 (1)

= xe-x2 (-2x) + e-x 2

Because we are going to use this function to find f” (x), we need to alge­ braically manipulate it by factoring the term e-x2

:

xe-x

(-2x)

e-x

J’ (x) =         2           +     2

2

= e-x2[x(-2x) + 1]

= e-x

(- 2×2

+ 1)

Now, we can find f”(x). We start with the Product Rule again and incorpo-

 

rate the Chain Rule:

 

))

J”(x) = e-x2 (d  (-2×2 + 1)) + (-2×2 + 1) (d   (e-x2

 

=       2                                    2                 2                         2

e-x (-4x) + (-2x  + 1) (e-x  (d  (-x )))

= -4xe–x2 + (-2×2 + 1) (e-x2(-2x))

 

Because we need to use this function to find the partition numbers of f”, we need to algebraically manipulate it. We will factor the term e-x2 again:

 

J”(x) = -4xe-x2 + (-2×2 + 1) (e-x2(-2x))

[

= e-x2  -4x + (- 2×2 + 1) (- 2x)]

= e-x2 (-4x + 4×3 – 2x)

(

= e-x2  4×3 –  6x)

2

To find the partition numbers off”, we find the x-values where f”(x) = 0 or f”(x) does not exist.  f”(x) does not have an even root or a logarithm,

eX

but it does have implied division as a result of the term e-x  = -4. But,

recall from our previous discussion that because ex2 > 0, the denominator will never equal zero. Hence, there are no domain issues and f” (x) will exist for all values of x. Thus, we only need to find the x-values where f”(x) = 0:

f”(x) = 0 ==?

e-x2 ( 4×3 –  6x) = 0

 

This gives us two equations to solve: e-x2 = 0 and 4×3 – 6x = 0. Because e-x2 > 0, there are no solutions to the first equation. Solving the second equation gives

4×3 –  6x = 0

2x (2×2 –  3) = 0

Setting 2x = 0 gives the partition number x = 0, and setting 2×2 –  3 = 0 gives

2×2 –  3 = 0

3

2×2 = 3

x2 = –

2

(3                (3==? x = y 2 and x = -y 2

Thus, the partition numbers of f” are x = -ji,x = 0, and x = ji.

3.   Create a sign chart of f” (x):

 

We will place the partition numbers of f” on a number line. All of the partition numbers will have solid dots on the number line because they are all in the domain of f.

Now, we need to determine the sign of f”(x) on the intervals (-oo, -j’i), (-j’i,0), (0, y

and  ( j’i, oo).

We will choose the x-values x = -2, -1, 1, and 2 to test:

e-x (4x

J” (x) =        2       3 –  6x) ==}

J”(-2)  = e-(-2)2  (4(-2)3-6(-2))       -0.3663 < 0

J”(-1) = e-(-1)2  (4(-1)3 –  6(-1))       0.7358 > 0

 

J”(l) = e-(1l2 (4(1)3 –	6(1))	-0.7358 < 0
J”(2) = e-(2l2  4(2)3 – (	6(2))	0.3663 > 0

Using this information, we can fill in the sign chart of f” (x). Because we are also interested in the corresponding information this yields for f, we will include that below the number line. See Figure 3.2.14.

Figure 3.2.14: Sign chart of  f”(x) with the corresponding information for

f(x) = xe-x2

Looking at the sign chart, we see that f” (x) changes sign at all three partition numbers. Vve now check to see if the function is defined at these x-values, which it is. Thus, f has inflection points at all three partition numbers. To find the y-values associated with the inflection points, we substitute the x-values into the original function f:

 

In conclusion, f is concave down on ( -oo, -j’i) and ( 0, j’i), and it is concave up on ( -j’i,0)and ( j’i, oo). The function has inflection points

at (-j’i,-j’ie- ),(o,o),and (j’i,j’ie- )-

Again, we can check our work by looking at the graph off shown in Figure 3.2.15:

 

 

Figure 3.2.15: Graph of the function f(x) = xe-x2

Note: Do not assume that the sign chart off”(x) will always have intervals with alternating signs. This just happened to be the case for the functions in the previous example. We will see that it is not always the case in the next example!

 

 

Example 4 Given f is continuous on its domain of (-oo, 2) U (2, oo) and  f”(x) = -4x(x- 2)3(x +

5)2(x – 7), find the intervals of concavity and the x-values of any inflection points of

f.Solution: Even though we were not given the original function, f, we still proceed with the steps to apply the Concavity Test because we need to find the intervals of concavity and location of any inflection points:

1.   Determine the domain of f:

Vve are given the domain off  is (-oo, 2) U (2, oo).

2.   Find the partition numbers of f”:

Because we are given f”(x) = -4x(x – 2)3(x + 5)2(x – 7), we can go straight to finding the x-values where f”(x) = 0 or f”(x) does not exist.

Let’s first consider the x-values where f”(x) does not exist.  f” is a polynomial (it is just in factored form), so it will exist everywhere that f is defined. Thus,

 

f”(x) does not exist at x = 2, so there is a partition number off” at x = 2. Now, we need to find the x-values where J”(x) = 0:

 

J”(x) = 0

–4x(x – 2)3(x + 5)2(x – 7) = 0

x = 0, 2, -5, and 7

 

Thus, the partition numbers of f” are x = -5, x = 0, x = 2, and x = 7.

3.    Create a sign chart of f” (x):

We will place the partition numbers of f” on a number line, with x = -5, x = 0, and x = 7 having solid dots and x = 2 having an open circle to indicate which are included and which are not included in the domain off, respectively.

Next, we need to determine the sign of f”(x) on the intervals (-oo, -5), (-5, 0), (0, 2), (2, 7),

and (7, oo). We will choose the x-values x = -6, -1, 1, 3, and 8 to test:

 

J”(x) = –4x(x – 2)3(x + 5)2(x – 7)

J”(-6) = -4(-6)((-6) – 2)3((-6) + 5)2((-6) – 7) = 159, 744 > 0

J”(-1) = -4(-1)((-1) – 2)3((-1) + 5)2((-1) – 7) = 13,824 > 0

J”(l) = -4(1)((1) – 2)3((1) + 5)2((1) – 7) = -864 < 0

!”(3) = -4(3)((3) – 2)3((3) + 5)2((3) – 7) = 3072 > 0

!”(8) = -4(8)((8) – 2)3((8) + 5)2((8) – 7) = -1, 168,128 < 0

 

Using this information, we can fill in the sign chart of f” (x). Because we are also interested in the information this yields for f, we will include that information below the number line. See Figure 3.2.16.

 

 

Figure 3.2.16: Sign chart of J”(x) with the corresponding information for f

 

Looking at the sign chart, we see that f”(x) changes sign at x = 0, x = 2, and x = 7. We are given the domain of f is (-oo, 2) U (2, oo), so f has inflection points at x = 0 and x = 7. Note that we cannot determine the 71-values of the inflection points because we were not given the original function f.

Furthermore, f is concave up on (-oo, -5), (-5, 0), and (2, 7). It is concave down on (0, 2) and (7, oo).

Graphical Interpretation

Remember, there is more than one way a function may be given to us. Thus far, we have focused on finding partition numbers of f”, intervals where f is concave up/down, and inflection points off algebraically using the rule of the function. Let’s consider finding this information if we are given the graph of the function, the graph of its first derivative, or the graph of its second derivative instead.

Example 5 Given the graph off    shown in Figure 3.2.17, find the intervals of concavity and the x-values of any inflection points off.

 

Figure 3.2.17: Graph of a function  f

 

Solution: Looking at the graph off, we see that it is concave up on (-oo, a), (a, b), and (b, d),

and it is concave down on (d, oo). f has an inflection point at x = d only.

Now, suppose we want to find this information, as well as the partition numbers of

f”, for a function f, but we are only given the graph of its second derivative, f”.

Algebraically, we know that the partition numbers of f” are the x-values where the second derivative equals zero or does not exist. Correspondingly, we need to find the x-values where the graph off” touches the x-axis or does not exist (i.e., is undefined).

To determine where f is concave up when looking at the graph off”, we find where the graph of f” is above the x-axis because f is concave up where f” (x) > 0. Similarly, to determine where f is concave down, we find where the graph of f” is below the x-axis. To find the x-values of any inflection points off, we need to know where f”(x) changes sign, or where the graph of f” switches from above the x-axis (positive) to below the x-axis (negative), or vice versa. In addition, we need to check that such x-values are actually in the domain off, which should be stated.

Example 6 Given the graph off” shown in Figure 3.2.18 and that f is continuous on its domain of (a, c) U (c, oo), find each of the following.

 

 

 

Figure 3.2.18: Graph of a second derivative function  f”

 

(a)   Partition numbers of  f”

(b)   Intervals of concavity of f

(c)   x-values of any inflection points off

 

Solution: We will use the same ideas discussed previously, but we have to be more careful now because we are given the graph of f” instead of the graph of f.

(a)   The partition numbers off” arethex-values where f”(x) = 0 or f”(x) does not exist. Let’s start by finding the x-values where f”(x) does not exist. Remember that we are given the graph off”, and the only x-values where f”(x) does not exist are x = a and x = c (the function is undefined at these x-values). Technically, f” (x) does not exist for any x :S a, in addition to x = c, but as discussed in Section 3.1, we would not include all of these x-values as partition numbers because they would not meaningfully partition a number line if we needed to create a sign chart of f” (x). We do include x = a as a partition number of f” because there is a meaningful transition happening at x = a (the derivative goes from not existing to existing). Also, x = a would be needed to partition a number line into the

appropriate, discrete intervals if we made a sign chart of f” (x).

We may also be tempted to say f”(x) does not exist at x = b because there is a vertical tangent line, but x = b would be where the derivative of f” does not exist, not where f” itself does not exist!

Next, we need to find the x-values where f”(x) = 0. Graphically, this means we need to find the x-values where the graph of f” touches (but not necessarily crosses) the x-axis. This occurs at x = d, x = e, and x = h.

Thus, the partition numbers of f” are x = a, x = c, x = d, x = e, and x = h.

 

(b)   To find the intervals where .f is concave up, we need to determine where .f”(x) > 0, or where the graph of .f” is above the x-axis. This occurs on the intervals (a, c), (d, e), and (h, oo). Similarly, to find the intervals where f is concave down, we need to determine where f”(x) < 0, or where the graph off” is below the x-axis. This occurs on the intervals (c, d) and (e, h).

In summary, f is concave up on (a, c), (d, e), and (h, oo), and f is concave down on (c, d) and (e, h).

(c) Be careful! The inflection points of f are not where the concavity changes on the graph off’. To find the inflection points off, we must find the x-values where f” (x) changes from positive to negative, or vice versa. Recall this amounts to

finding where the graph of f” switches from being above to below the x-axis, or vice versa. The partition numbers of f” where this occurs are x = c, x = d, x = e, and x = h. Checking to see if these x-values are in the domain of f, which is (a, c) U (c, oo), we see that x = c is not in the domain. Thus, the inflection points

of f occur at x = d, x = e, and x = h.

 

Note: The inflection points that occur on the graph off” at x =band x = e are inflection points of J”!

Now, let’s consider how we might find this information (partition numbers of f”, intervals of concavity off, andthex-values of any inflection points off) when we are given the graph of f’ and the domain of f.

The partition numbers off” arethex-values where f”(x) = 0 or f”(x) does not exist. To find the x-values where f”(x) does not exist, we need to look at the graph off’ and find where its derivative does not exist (because its derivative is f”). Recall from Section 2.2 that the derivative of a function does not exist when the graph of the function has a cusp or corner, a discontinuity, or a vertical tangent line. To find where f” (x) = 0, we need to determine the x-values where the graph off’ has horizontal tangent lines because these are the x-values where its derivative equals zero (i.e., the x-values where f”(x) = 0).

To find where f is concave up given the graph of f’, remember from our previous discussion that a function is concave up when the slopes of its graph are increasing (i.e., when .f’ is increasing). This is equivalent to finding where f” (x) > 0, or where the graph of f” is above the x-axis, but because we are given the graph off’ and not the graph off”, we must look for where the function f’ is increasing. Likewise, we look for where f’ is decreasing to find where f is concave down. To find where f has inflection points given the graph off’, we must think

about what the concavity of f changing means in terms of f’.  If a function changes from

concave up to concave down, then that means the slopes of its graph switch from increasing to decreasing. In other words, f’ switches from increasing to decreasing. One way this can happen is if f’ has a local maximum! Similarly, if a function changes from concave down to concave up, then that means the slopes of its graph switch from decreasing to increasing. In other words, f’ switches from decreasing to increasing. Again, one way this could happen is

if f’ has a local minimum! So to find where f has inflection points when we are given the graph of f’, we need to find the x-values where f’ switches between increasing/ decreasing and then check that these x-values are in the domain off, which should be stated.

 

Example 7 Given the graph of f’ shown in Figure 3.2.20 and that f is continuous on its domain of (-oo, c) U (c, oo), find each of the following.

 

Figure 3.2.20: Graph of a first derivative function  f’

 

(a)   Partition numbers off”

(b)   Intervals of concavity of  f

(c)   x-values of any inflection points off

Solution:  (a) To find the partition numbers off”, we must find the x-values where f”(x) = 0 or f”(x) does not exist. Let’s start by finding the x-values where f”(x) does not exist. Because f” is the derivative off’, thex-values where f”(x) does not exist correspond to the x-values where the derivative off’ does not exist. So we need to look at the graph off’ andsee where its derivative does not exist. The derivative of .f’ does not exist at x = a and x = c because there are discontinuities. The

derivative does not exist at x = b because there is a vertical tangent line. Thus,

f”(x) does not exist at x = a, x = b, and x = c.

Now, we need to find the x-values where f”(x) = 0, which means the derivative of f’ equals zero. The derivative equals zero where the function, f’, has a horizontal tangent line. Looking at the graph of f’, we see this occurs when x < a and at x = e and x = g. As discussed previously, we would not consider all of the x-values less than a to be partition numbers because they would not partition a

number line in a meaningful, necessary way if we created a sign chart of f” (x). Thus, we consider the “important” x-values where f”(x) = 0 to be x = e and x = g, and so these are also partition numbers of f”.

In summary, the partition numbers of  f” are x = a, x = b, x = c, x = e, and

X =g.

(b)   To find where f is concave up, we need to find where f”(x) > 0, or in this case, where f’ is increasing.                                           Looking at the graph of f’, we see it is increasing on

 

(a, b), (b, c), and (e, g). Thus, .f is concave up on (a, b), (b, c), and (e, g). To find where f is concave down, we need to find where f”(x) < 0, or in this case, where f’ is decreasing. Looking at the graph of f’, we see it is decreasing on (c, e) and (g, oo). Thus, .f is concave down on (c, e) and (g, oo).

In summary, f is concave up on (a, b), (b, c), and (e, g), and .f is concave down on

(c, e) and (g, oo).

(c)   To find the inflection points of f, we need to find the x-values where the graph of f’ switches between increasing/ decreasing, as discussed previously, and then check to see if these x-values are in the domain of .f. Looking at the graph of f’, we see it switches between increasing/decreasing at x = c, x = e, and x = g. Because the domain off is (-oo, c) U (c, oo), which we were given, f has inflection points at x = e and x = g only.

Note: The inflection points that occur on the graph of f’ at x = b and x = f are inflection points of f’ and not necessarily inflection points of f !

 

Figure 3.2.22 summarizes the relationships between .f, f’, and .f” we have discussed through­ out this section and the last in a handy dandy chart. It may be helpful to use this chart

 

when working problems similar to the last two examples where we are given the graph of the first or second derivative, as well as when we are given the graph of any one of the three functions and asked to find pertinent information about the other two.

Figure 3.2.22: Chart showing relationships between f, f’, and f”

Example 8 Use Figure 3.2.22 and the graph shown in Figure 3.2.23 to answer each of the following.

 

Figure 3.2.23: Graph of a function

(a)   If the graph shown is of f, find where f” (x) < 0.

(b)   If the graph shown is of f, find where f’ is increasing.

(c)   If the graph shown is off, find where f'(x) > 0.

(d)   If the graph shown is off’ and f is continuous on its domain of (-oo, 2) U (2, 7) U

(7, 11) U (11, oo), find where f”(x) > 0.

(e) If the graph shown is off’ and f is continuous on its domain of (-oo, 2) U (2, 7) U

(7, 11) U (11, oo), find where f is decreasing.

(f)   If the graph shown is off’ and f is continuous on its domain of (-oo, 2) U (2, 7) U

(7, 11) U (11, oo), find where the slopes of the graph off     are decreasing.

(g)   If the graph shown is off”, f is continuous on its domain of (-oo, 2) U (2, 7) U (7, 11) U (11, oo), and f’ is continuous on its domain of (-oo, 2) U (2, 7) U (7, 11) U (11, oo), find where f’ is decreasing.

Solution:  (a) To use the chart shown in Figure 3.2.22, we find the entry that states the question (i.e., we locate f”(x) < 0 in the chart). Then, we move across the row to the column corresponding to the function we are given, which is f in this case. Moving across the row to the column corresponding to f, we see that we need to find where the graph is concave down. Looking at the graph, we see that f is concave down on (7, 11) and (11, oo). Thus, f”(x) < 0 on (7, 11) and (11, oo).

 

(b)   We first locate f’ increasing in the chart. Then, because we have the graph off, we move across the row to the column corresponding to f and see that we need to look for where the graph is concave up. f is concave up on (2, 7), so f’ is increasing on (2, 7).

 

 

 

(c)   We proceed as before and locate the question in the chart: f’ (x) > 0. We move across the row to the column corresponding to f and see that we need to look for where the graph is increasing. f is increasing on (4, 7) and (11, oo). Therefore, f'(x) > 0 on (4, 7) and (11, oo).

 

 

 

(d)   Again, we locate the question in the chart: f”(x) > 0. Now, because we are assuming the graph is of f’, we move across the row to the column corresponding to f’. We see f” (x) > 0 corresponds to where f’ is increasing. Because we are assuming the graph is off’, we look for where the graph is increasing. This occurs on the intervals (4, 7) and (11, oo). Hence, f”(x) > 0 on (4, 7) and (11, oo).

 

 

 

(e)   We locate f decreasing in the chart and then move across the row to the col­ umn corresponding to f’ because we are assuming the graph is of f’.  We see that we need to look for where f'(x) < 0. The graph is negative on the intervals

(0, 2), (2.6, 5.4), (7, 11), and (11, 11.8). Thus, f is decreasing on (0, 2), (2.6, 5.4), (7, 11),

and (11, 11.8).

 

 

(f)   We want to know where the slopes of the graph of f are decreasing, and this translates to where the graph of f’ is decreasing because f’ (x) gives the slope of the graph of f at x. While locating f’ decreasing in the chart, we should notice that we do not need to move across the row. We are assuming the graph we have

is of f’, so all we need to do is look at the graph and see where it is decreasing. The graph is decreasing on (-oo, 2), (2, 4), and (7, 11), so the slopes of the graph off are decreasing on (-oo, 2), (2, 4), and (7, 11).

 

 

 

(g)   We locate f’ decreasing in the chart. Now, because we are assuming the graph is of f”, we move across the row to the column corresponding to f”. We see that we need to look for where f”(x) < 0. Because we are assuming the graph

is of f”, we look for where the graph is below the x-axis. This occurs on the intervals (0, 2), (2.6, 5.4), (7, 11), and (11, 11.8).  Therefore, f’ is decreasing on

(0, 2), (2.6, 5.4), (7, 11), and (11, 11.8).

 

Point of Diminishing Returns

One business application involving concavity and inflection points is the point of diminishing returns.

 

Let’s say a company that sells a particular item is trying to increase their sales. The com­ pany’s marketing team decides to increase the amount they spend on advertising. They pur­ chase television and radio commercials, billboard advertisements, podcast advertisements, airplane fly-by banners and more. We would expect sales to increase when the company increases its advertising. At first, the company’s sales will increase at an increasing rate.

 

Eventually though, if the company continues its advertising campaign, sales will continue to increase, but they will increase at a decreasing rate.

 

In other words, the point where the first derivative of a sales function has a local maximum is the point of diminishing returns. As seen previously, this local maximum on the graph of the first derivative corresponds to an inflection point on the original sales function.

Example 9 .Joseph .Joeson owns a local coffee shop, A Cup of .Joe, that specializes in making one­ size flavored coffee drinks. Mr. Joeson expects that if he spends h hundred dollars per week in advertising, his shop will sell J(h) = -0.2h4 + 2h3 + 250 cups of Joe (coffee drinks), where O :::; h::; 7.5.

(a)   On what interval is the rate of change of sales increasing? On what interval is the rate of change of sales decreasing?

(b)   Find the point of diminishing returns.

(c)   Verify your answers by graphing the sales function, J, and the rate of change of sales function, J’, on the same axes.

Solution:  (a) To find where the rate of change of sales, J’, is increasing and where it is decreas­ ing, we must find its derivative, J”. Then, we can use the techniques discussed in Section 3.1 to create a sign chart of J”(h) to determine where J”(h) is posi­

tive (i.e., where J’ is increasing) and where J”(h) is negative (i.e., where J'(h) is decreasing). For J(h) – 0.2h4 + 2h3 + 250, we have

J ‘( h ) = -0.8h3 + 6h2  ==;,

J”(h) = -2.4h2 + 12h

 

To create a sign chart of J”(h), we need to partition a number line. Recall that the partition numbers of J” are the h-values where J” (h) = 0 or J” (h) does not exist. Because J” is a polynomial, it will always exist. Thus, we only need to determine the h-values where J” (h) = 0:

J”(h) = 0 ==;,

-2.4h2 + 12h = 0 h(-2.4h + 12) = 0

12

==;, h =  0 and h =  –      = 5

2.4

Next, we need to determine the sign of J”(h) on the intervals (0, 5) and (5, 7.5); note that the function is defined for O :::; h ::; 7.5. We must select h-values to test

 

in each interval. vVe will test h = l and h = 6:

J”(h) = -2.4h2 + 12h ==}

J”(l) = -2.4(1)2 + 12(1) = 9.6 > 0

J”(6) = -2.4(6)2 + 12(6) = -14.4 < 0

 

Now, we can create a sign chart with this information for J”, as well as the corresponding information for J’ and J. Also, we will remember to mark that the original function, J, has a domain of [0, 7.5] on the chart. See Figure 3.2.25.

 

Figure 3.2.25: Sign chart of J”(h) with the corresponding information for both

J’ and J

 

Thus, the rate of change of sales, J’, is increasing on (0, 5) and decreasing on (5, 7.5). In other words, when the company spends up to $500 on advertising, its coffee sales will increase at an increasing rate. If it spends more than $500 (and less than $750), sales will still increase, but at a decreasing rate.

(b)   Looking at the sign chart of J” (h) in Figure 3.2.25, we see that the point of diminishing returns, which corresponds to where the rate of change of sales, J’, switches from increasing to decreasing, occurs at h = 5. In other words, the point of diminishing returns occurs when the company spends $500 per week on advertising.

Note: As seen in the sign chart of J” (h), the point of diminishing returns of a func­ tion corresponds to an inflection point on the graph of the function and a local maximum on the graph of its first derivative.

To find the y-value (number of cups of Joe sold) associated with the point of diminishing returns at h = 5, we calculate J(5):

 

J(h) = -0.2h4 + 2h3 + 250 ==}

J(5) = -0.2(5)4 + 2(5)3 + 250

= 375

 

Therefore, the point of diminishing returns is (5,375), meaning $500 spent on advertising and 375 cups of Joe sold.

(c)   Figure 3.2.26 shows the graphs of both J and J’ on the same axes:

 

 

 

 

Figure 3.2.26: Graphs of J and J’ showing the point of diminishing returns at

h=5

 

 

We can use these graphs to verify the information in the sign chart of J” (h). We see that J’ has a local maximum at h = 5, and J has an inflection point at h = 5.

 

 

 

 

 

 

 

The Second Derivative Test

 

Up until now, we have used the First Derivative Test (Theorem 3.2) to find local extrema. However, there is an alternative test we can attempt to use to find local extrema called the Second Derivative Test. The reason we say attempt is because there are circumstances in which this test cannot be applied (we will get to those shortly!). However, it may be helpful to try to use the Second Derivative Test (as opposed to the First Derivative Test) because us­ ing the Second Derivative Test is usually less computationally intensive, which saves us time.

 

We have already learned that local extrema occur at critical values. Recall that critical values off  are the x-values where f'(x) = 0 or f'(x) does not exist, and they are in the domain of the function f. Now, consider the critical value x = c of a function f such that f’ (c) = 0, and the critical value is contained on an interval in which f” (x) > 0 (and therefore, f” (c) > 0). This means that f is concave up on the interval and has a horizontal tangent line at x = c. Thus, the graph of f would have roughly the same shape as the graph shown in Figure 3.2.27. Notice that f has a local minimum at x = c!

 

 

 

 

Figure 3.2.27: Graph of a function f that is concave up and has a horizontal tangent line at X = C

 

 

Likewise, if x = c is a critical value of f such that f’ (c) = 0 and f” (x) < 0 on an interval containing x = c (and therefore, f” (c) < 0), then f is concave down on the interval and has a local maximum at x = c. See Figure 3.2.28.

 

 

 

Figure 3.2.28: Graph of a function f that is concave down and has a horizontal tangent line at x = c

 

 

We now formally state the Second Derivative Test:

 

One of the limitations regarding the Second Derivative Test is that we can only attempt to use it for critical values off  such that f'(x) = 0. If there are critical values off  such that f'(x) does not exist, we cannot use the Second Derivative Test, and we must go back and

use the First Derivative Test to determine if there are any local extrema at those critical values. Why? If f’ does not exist at a particular x-value, then the derivative of f’, which is f”, cannot possibly exist at that x-value either. If the second derivative does not exist, we cannot use the Second Derivative Test. The other issue that can arise is if after taking the second derivative and evaluating it at a critical value x = c, we get that J”(c) = 0. In this case, we say the Second Derivative Test fails, meaning it is inconclusive. We would then have to go back and use the First Derivative Test to determine if a local extremum occurs at the critical value x = c.

 

Another similar possibility is that we apply the Second Derivative Test to a critical value x = c in which f'(c) = 0 and the result is f”(c) does not exist. Because the Second Derivative Test assumes f is twice-differentiable at x = c, we would not be able to conclude whether or not f (c) is a local maximum, local minimum, or neither. We would, again, have to go back and use the First Derivative Test to determine if there is a local extremum at x = c.

 

If the Second Derivative Test fails at a critical value, that does not mean the function does not have a local extremum at that critical value. It just means we cannot determine if there is a local extremum using this test. Likewise, if there are critical values such that f’ (x) or J”(x) does not exist, it does not mean there are no local extrema. It just means we cannot use the Second Derivative Test to determine the behavior. In any of these cases, we have to go back and use the First Derivative Test to determine if there are local extrema.

 

Because the First Derivative Test (Theorem 3.2) and the Second Derivative Test (Theorem 3.4) can both be used to find local extrema, we will compare and contrast them here to help us learn which one to apply in which situation:

 

Many students like the Second Derivative Test because it is often easier to use than the First Derivative Test. If the function is a polynomial, its second derivative will probably be a sim­ pler function that its first derivative. However, if you need to use the Product, Quotient, or Chain Rules to find the first derivative, finding the second derivative could be a lot of work. In this case, you might choose to apply the First Derivative Test to find local extrema instead.

 

Remember that even if the second derivative is easy to find, the Second Derivative Test does not always give an answer. Also, it can only be used for critical values of f such that f'(x) = 0. These are reasons why some students prefer the First Derivative Test.

 

 

Example 10 The domain of f is (-oo, oo), and f is twice-differentiable on its domain. Use the Second Derivative Test, if possible, to classify whether f has a local minimum, local maximum, or neither at each of the following x-values. If it is not possible, explain why.

 

(a)    j(6) = -22, f'(6) = 0, and f”(6) = -12

(b)     f(O) = 15, f'(0) = 20, and f”(0) = 10

(c)    J(-2) = 179, f'(-2) = 0, and J”(-2) = 0.07

(d)    j(13) = 13, f'(13) = 0, and !”(13) = 0

 

Solution:  (a) First, note that because f is defined at x = 6 and f'(6) = 0, x = 6 is a critical value of f. Furthermore, we can attempt to use the Second Derivative Test because x = 6 is a critical value off  in which f'(x) = 0. We are given J”(6) = -12, which means f is concave down at x = 6 because its second derivative is negative. Using Theorem 3.4, we see that f has a local maximum at the point (6, -22).

(b)   Even though f is defined at x = 0, f'(0) = 20, which means that x = 0 is not a critical value of f! Recall that the critical values of f are the x-values where f'(x) = 0 or f'(x) does not exist (although we cannot use the Second Derivative Test with critical values such that f'(x) does not exist). Therefore, because x = 0 is not a critical value off, there cannot possibly be a local extremum at x = 0.

(c)   Unlike in part b, x = -2 is a critical value off. Also, because f'(-2) = 0, we can attempt to use the Second Derivative Test. Because f”(-2) = 0.07, which is positive, f is concave up at x = -2. Using Theorem 3.4, we see that f has a local minimum at the point (-2, 179).

(d)   Again, x = 13 is a critical value off. However, because f”(13) = 0, the Second Derivative Test fails. This means we cannot determine whether or not there is a local extremum at x = 13. Without more information to conduct the First

Derivative Test, we cannot conclude anything more than x = 13 is a critical value

off.

 

 

To apply the Second Derivative Test when we are given a function f, we use the following three steps:

 

Example 11 Use the Second Derivative Test, if possible, to find any local extrema of the following functions. If it is not possible, explain why.

(a)     f(x) = 2×3 –  15×2 + 24x – 7

(b)    f ( x) = 3×5 –  8×4

Solution:      (a) We will use the three steps outlined previously to attempt to use the Second Derivative Test:

1.   Determine the domain of f:

f is a polynomial, so its domain is (-oo, oo).

2.   Find the critical values off:

J'(x) = 6×2 –  30x + 24

f’ is a polynomial, so it exists everywhere. Thus, there are no critical values such that f'(x) does not exist. Now, we find the x-values where f'(x) = 0:

J'(x) = 0 6×2 –  30x + 24 = 0

6 (x2 –  5x + 4) = 0

6(x – l)(x – 4) = 0

x =land x = 4

 

Because both of these x-values are in the domain of .f, they are both critical values of .f. Furthermore, because they are both critical values such that f'(x) = 0, we can continue and attempt to use the Second Derivative Test with each critical value.

3.   Find .f”(x) and evaluate it at the critical values:

.f”(x) = 12x – 30

J”(l) = 12(1) – 30 = -18 < 0

.!”(4) = 12(4) – 30 = 18 > 0

Using Theorem 3.4, we see that f is concave down at x = l, so there is a local maximum at x = l. At x = 4, .f is concave up, so there is a local minimum at X = 4.

To find the 71-values associated with these x-values, we substitute the x = l

and x = 4 into the original function f:

f (x) = 2×3 –  15×2 + 24x – 7

j(l) = 2(1)3 –  15(1)2 + 24(1) – 7 = 4

f(4) = 2(4)3 –  15(4)2 + 24(4) – 7 = -23

Thus, f has a local maximum of 4 at x = l and a local minimum of -23 at

X = 4.

We can check our work by looking at the graph of f shown in Figure 3.2.29, but applying the Second Derivative Test allows us to find the local extrema completely algebraically and get exact answers!

Figure 3.2.29: Graph of the function f(x) = 2×3 –  15×2 + 24x – 7

(b)   Recall f (x) = 3x5 – 8×4.            We will use the three steps outlined previously to attempt to use the Second Derivative Test:

1.   Determine the domain of f:

 

.f is a polynomial, so its domain is (-oo, oo).

2.   Find the critical values of .f:

 

.f’(x) = 15×4 –  32×3

 

f’ is a polynomial, so it exists everywhere. Thus, there are no critical values such that .f'(x) does not exist. Now, we find the x-values where .f'(x) = 0:

 

J'(x) = 0 ===} 15×4 – 32×3 = 0 x3(15x – 32) = 0

32

===} x = 0 and x = –

15

 

Because both of these x-values are in the domain off, they are both critical values of f.                        Furthermore, because they are both critical values such that

.f'(x) = 0, we can continue and attempt to use the Second Derivative Test

with each critical value.

3.   Find f”(x) and evaluate it at the critical values:

 

J” (x) = 60×3 –  96×2 ===}

j”(O) = 60(0)3 –  96(0)2 = 0

3                              2

225

J” (3125) = 60 (3152) – 96 (3152)  = 32768 ,-:::;145.6356 > O

 

Using Theorem 3.4, we see that the Second Derivative Test fails at x = 0 because  f”(O) = 0.  Thus, we cannot draw any conclusions about a local

extremum at x = 0 (there may or may not be one). At x = f;,f is concave up, so there is a local minimum at x = f;.

To find the y-value associated with x = f;, we substitute this x-value into the original function f:

 

f(x) = 3×5 –  8×4 ===}

5                           4

f (3-2) = 3 (3- 2) – 8 (3- 2) ,-:::; -33.1402

15              15                15

 

Thus, f has a local minimum of -33.1402 at x = f;.

We can check our work by looking at the graph of f shown in Figure 3.2.30:

 

 

Figure 3.2.30: Graph of the function f(x) = 3×5 –  8×4

Note: Even though the Second Derivative Test was inconclusive regarding a local ex­ tremum at x = 0, we can see from the graph off  in Figure 3.2.30 that it appears as though the function does have a local maximum at x = 0. To be sure, however, we would need to use the First Derivative Test.

 

Try It Answers

(

a)    11 X =     (10×3-27×2+15)2(-160×3+180x-42)-(  -40×4+90×2-42x) (2(10×3-27×2+15) (30×2-54x))

1.              f    (. )                                                       (10×3-27×2+15)4

(b)

 

y” =     (-9×11 –  8×31)-1/3 (-990×9 –  7440×29) +

+ (–99x10 – 248×30) ( –         ( -9x11 –  8x31r4/3

 

 

 

 

( –99x

 

 

 

10 –  248×30))

 

2.     (a) f is concave up on (0,oo), and f is concave down on (-oo,0);f has an inflection point at (0, 0).

(b)    g is concave up on (e-5/5, oo), and g is concave down on  has an inflection point at (e-5/5, – }8e-5/2).

3.     (a) X = -7;    X = -6; X = -2; X = 0; X = 2; X = 3; X = 4; X = 7

(b)   fisconcaveupon(-oo,-7), (-6,-2), (0,2), (3,4),and(7,oo),andfisconcave

down on (-7, -6), (-2, 0), (2, 3), and (4, 7).

(c)   X = -7; X = -6; X = -2; X = 0; X = 2; X = 4; X = 7

4.     (a) X = 2; X = 5; X = 7; X = ll

(b)   f is concave up on (2,5) and (11,oo), and f is concave down on (-oo,2), (5, 7), and (7, 11).

(c)   x = 2; x = 5; x = ll

5.     (a) (-12,-3)and(3,oo)

(b)   (-3,3)

(c)    (-12, -7), (0, 3), and (3, 7)

(cl) (-7,0) and (7,oo)

(e) (-12,-10),  (-4,2), (3,4), and (10,oo)

6.     (a)  f has a local minimum at x = 8.

(b)   The Second Derivative Test fails at x = -15 because f” (-15) = 0, so we cannot determine whether or not the function has a local extremum at x = -15.

7.     (a) f has a local minimum of -e-1 at x = e-1.

(b)     f has a local maximum of -20 at x = -10 and a local minimum of 20 at x = 10.

Exercises 

Basic Skills Practice

 

For Exercises 1 – 6, find .f” (x).

1.   J(x) = 7×3 + 8ln(x) – x – x-5

2.   J(x) = (2x + 13)9

3.

2

f ( X) = ex +x4 –       21

 

4.     f'(x) = e5x+9

5.     f'(x) = x44 – 0.3×12 +VX – x-4/5

6.    J'(x) = ln (8×3 –  22)

For Exercises 7 – 9, the graph of .f is shown. Find (a) the intervals of concavity and (b) the x-values of any inflection points off.

7.

 

8.

 

 

 

For Exercises 10 – 12, f” and the domain off are given. Assuming f is continuous on its domain, find (a) the partition numbers of f”, (b) the intervals of concavity of f, and (c) the x-values of any inflection points off  using the Concavity Test.

10.      f” (x) = (x + 6)4(x – 3)(x – 7); domain off  is (-oo, oo)

11.      J”(x) = 2x(x + 4)(x – 8)5; domain off  is (-oo, oo)

12.

f“(x ) —

3(x(-x1+0)()x+2)., domam•

of j

I•S ( -oo, – 5) U (-

5, oo)

 

5 7

For Exercises 13 – 15, find (a) the partition numbers off”, (b) the intervals of concavity of

f, and (c) any inflection points off    using the Concavity Test.

13.      f(x) = x2 –  1

14.      f(x) = 4×3 + 24×2 –  384x + 3

15.      J(x) = x4 –  54×2

 

For Exercises 16 – 18, the graph off” is shown, and .f is continuous on its domain of (-oo, oo). Find (a) the partition numbers off”, (b) the intervals of concavity off, and(c) the x-values of any inflection points of f.

 

16.

 

17.

 

 

 

 

19.   Dark Owl Records sells vinyl records. Michelle Nguyen, the owner, has determined if she spends x hundred dollars per quarter in advertising, she can sell V(x) = -0.002×5+ 0.4x3 + 0.75x + 120 hundred records in one quarter, where O :S x :S 15.5. Find the amount of money spent by Michelle at the point of diminishing returns. Round your

answer to the nearest cent.

 

20.   Innocent Until Proven Quilt-y makes quilts. Rebekah Shapiro, the owner, has found that she can sell Q(x) = -0.009x3 + 0.25×2 + 0.34x + 22 quilts when she spends x

dollars in advertising per week. Find the amount of money spent by Rebekah at the point of diminishing returns.

21.   Angels and Angles is a company that sells protractors blessed by holy people of various religions. The marketing team has found that if they spend x dollars on advertising each day, they sell

A(x) = -0.0003x3 + 0.04×2 + 2.36x + 35.7

protractors. Find the amount of money spent by the company at the point of dimin­ ishing returns.

 

For Exercises 22 – 26, the domain off is (-oo, oo), and f is twice-differentiable on its domain. Use the Second Derivative Test, if possible, to classify whether f has a local minimum, local maximum, or neither at the indicated x-value. If it is not possible, explain why.

22.   x = l, f'(l) = 0, and f”(l) = 9                    25. x = 12.5, f'(12.5) = 0, and !”(12.5) =

827

23.   x = -8, f'(-8) = 0, and f”(-8) = -13

 

26. X                    -127, f'(-127)

 

0,    and

24.   x = 14,!'(14) = 0, and !”(14) = -7.93               f”(-127) = -e

Intermediate Skills Practice

 

.                             d2y

For Exercises 27 – 31, find dx2.

27.

+

!J = x7 – xl/9           13

X

30. !J = log7(8x + 5) – 10×3

28.       y = xex

29.      Y = x :4

For Exercises 32 – 37, find (a) the intervals of concavity and (b) any inflection points of f.

 

32

_ j”(x) = 2x-18

3x+l8

33.   Y = x: / 34. J(x) = 2: /

34.     f(x) = (8x – 16)7

 

35.     f(x) = ex (x2 – 4x – 94)

 

 

36.     f(x) = e64-2×2

 

 

37.   Given the graph off      shown below, find (a) the intervals of concavity and (b) any inflection points of f.

 

 

 

For Exercises 39 and 40, the graph off     is shown. Find (a) the intervals of concavity and

(b)   the x-values of any inflection points off.

 

39.

 

 

40.

 

 

 

 

 

 

 

 

 

X

 

41.   Given the graph off” shown below and that f is continuous on its domain of (-oo, oo), find (a) the partition numbers of f”, (b) the intervals of concavity of f, and (c) the x-values of any inflection points off.

 

 

 

For Exercises 42 and 43, the graph of f” and the domain of f are given. Assuming f is continuous on its domain, find (a) the partition numbers of f”, (b) the intervals of concavity of f, and (c) the x-values of any inflection points of f.

 

 

42.   domain of f : (-oo, oo)

 

 

 

43.   domain off: (-oo, e) U (e, oo)

 

 

 

44.   Given the graph off’ shown below and that f is continuous on its domain of (-oo, oo), find (a) the partition numbers of f”, (b) the intervals of concavity of f, and (c) the x-values of any inflection points off.

 

For Exercises 45 and 46, the graph of f’ and the domain of f are given. Assuming f is continuous on its domain, find (a) the partition numbers of f”, (b) the intervals of concavity of f, and (c) the x-values of any inflection points of f.

45.   domain off: (-oo,a) U (a,oo)

 

46.   domain of f : (-oo, d) U (d, f) U (f, oo)

 

47.   Tom Nook has an island getaway package. He has determined that if he spends x thousand dollars on advertising each month, he sells A(x) = -¼x4 + 8×3 – 25×2 +75x + 300 getaway packages. Find the x coordinate of the inflection point of this function to five decimal places, and interpret the meaning of the value.

48.   Greyt Toys, Inc. sells toys tailored specifically to greyhound dogs. The marketing team found that they sell T(x) = -0.lx3 + 9.8×2 + 2.5x + 1755 toys when they spend x thousand dollars on advertisements each quarter. Find the x-coordinate of the in­

flection point of this function to five decimal places, and interpret the meaning of the value.

 

49.   Sister Elga’s House of Curses offers haunted house tours in the month of October. The owner, Elga, spends x dollars in advertising in September and sells H(x) = -0.0002×3+ 0.055×2 + 0.025x + 300 tours in October. Find the x-coordinate of the inflection point of this function to two decimal places, and interpret the meaning of the value.

For Exercises 50 – 52, the domain off is (-oo, oo), and f is twice-differentiable on its domain. Use the Second Derivative Test, if possible, to classify whether f has a local minimum, local maximum, or neither at the indicated x-value. If it is not possible, explain why.

 

50.    x = 7, f'(7) = 0, and f”(7) = 9

51.    x = 100, f'(lO0) = 0, and !”(100) = 0

52.    x = -37, f'(-37) = 0, and f”(-37) =

-7

For Exercises 53 – 55, use the Second Derivative Test, if possible, to find any local extrema of the function. If it is not possible, explain why.

53.     f(x) = 2×3 + 3×2 –  36x + 16

54.     f(x) = ex (x3 + 5×2 –  3x + 3)

 

 

 

Mastery Practice

 

56.    Given the graph of g shown below, find (a) the intervals of concavity and (b) the x-values of any inflection points of g.

3.2.  ANALYZING GRAPHS WITH THE SECOND DERIVATIVE                                       99

 

For Exercises 57 – 62, find J”(x).

57.     f(x) = x12 –  3×10 + 4×6 + 19

 

58.     f(x) = log (7 + x2)

 

59.

(x) =

f            8x+14 (x-3)4

 

 

 

 

60.     j(x) = 97xs-33x

61.     f(x) = J44x3 –  8×2 + 1

63.    Given the graph off” shown below and that f is continuous on its domain of (-oo, a) U (a,j) U (j, oo), find (a) the partition numbers off”, (b) the intervals of concavity off, and (c) the x-values of any inflection points off.

 

64.    Big Box Store sells cardboard boxes in bulk to moving companies. The moving com­ panies buy B(x) = -0.05×3 + 2.2×2 + 3.8x + 828 thousand boxes from Big Box Store

when it spends x thousand dollars advertising during a two month period. Find the amount of money spent by Big Box Store at the point of diminishing returns. Round your answer to the nearest cent.

 

For Exercises 65 – 75, find (a) the intervals of concavity and (b) the x-values of any inflection points off.

 

65.     f(x) = 5×4 –  9.r3

66.   J(x) = (o.5y

67.   J(x) = (x – l6)e-2x

68.     j(x) = ln x)

69.     f(x) = (/: i)2

71.     f”(x) = (x – 3)(x + 2)3(x + 7)4;        f is continuous on its domain of (-oo, 3) U (3, oo)

72.     f'(x) = x3 –  3×2 + 1;     f is continuous on its domain of (-oo, oo)

73.     f”(x) = (x(x1l t6l;    f is continuous on its domain of (-oo,-2) U (-2,oo)

74.

3

f’ (x) = (x2! )2;       .f  is continuous on its domain of (-oo, oo)

 

75.   .f” (x) = 3×4 –  2×3 –  5×2;       .f is continuous on its domain of (-oo, oo)

76.    Given the graph of .f’ shown below and that .f is continuous on its domain of (-oo, d) U

(d, g) U (g, oo), find (a) the partition numbers of .f”, (b) the intervals of concavity of

.f, and (c) the x-values of any inflection points of .f.

 

77.   Given g”(t) = (t – a)(t + b)4(t – c)3, where O <a<   b < c, and that g is continuous on its domain of all real numbers except t = c, find (a) the partition numbers of g”, (b) the intervals of concavity of g, and (c) the t-values of any inflection points of g.

 

For Exercises 78 – 80, use the Second Derivative Test, if possible, to find any local extrema of the function. If it is not possible, explain why.

 

78.     f(x) = 1 + 9x + 3×2 –  x3

79.     f(x) = x3e-x/2

80.     f(x) = 4::s

 

 

80.   The graph off  is shown below. Use the graph to find (a) the x-values where f'(x) = 0,

(b) the x-values where f'(x) does not exist, (c) the intervals where f'(x) is positive and the intervals where f’ (x) is negative, (d) the intervals where f” (x) is positive and the intervals where f”(x) is negative, and (e) the x-values where f’ has any local extrema (specify the type).

 

81.   The graph off’ is shown below, and f is continuous on its domain of (-oo, oo). Use the graph to find (a) the intervals where f is increasing/decreasing, (b) the x-values where f has any local extrema (specify the type), (c) the intervals where f” (x) is positive and the intervals where f”(x) is negative, (d) the x-values where f”(x) = 0, (e) the x-values where f”(x) does not exist, and (f) the x-values of any inflection points off.

 

82.   The graph of f” is shown below, .f is continuous on its domain of (-oo, oo), and .f’ is continuous on its domain of (-oo, oo). Use the graph to find (a) the intervals where .f’ is increasing/ decreasing, (b) the x-values where .f’ has any local extrema (specify the

type), (c) the intervals of concavity of f, (d) the x-values where f’ has any inflection points, and (e) the x-values where f”(x) = 0.

 

84.   Staples Depot sells office furniture. The marketing team determines that they sell C(x) = -0.033×3 + 19×2 + 0.12x + 4096 chairs when they spend x thousand dollars marketing them each year. Find the x-coordinate of the inflection point of this function to five decimal places, and interpret the meaning of the value.

85.    Civet Coffee, Inc sells high end coffee beans. Their marketing team has determined that when they spend x thousand dollars in advertising, C(x) = -0.03×3+5×2+x+120 pounds of coffee beans are sold in a year. Find the x-coordinate of the inflection point of this function to five decimal places, and interpret the meaning of the value.

 

For Exercises 86 – 89, the domain off is (-oo, oo), and f is twice-differentiable on its domain. Use the Second Derivative Test, if possible, to classify whether f has a local minimum, local maximum, or neither at the indicated x-value. If it is not possible, explain why.

 

86.    x = -24, .f'(-24) = 0, and f”(-24) =

-2.887

87.    x = 0, f'(0) = 5, and f”(0) = 19

88.    x = 3, f'(3) = 0, and f”(3) = 0

 

89.    x = 11, f'(ll) = 0, and J”(ll) = -8

 

 

Communication Practice

 

90.    Describe how to find the partition numbers off”.

91.   What does the Concavity Test enable us to find?