Chapter 3

Curve Sketching & Optimization

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 Sections:

3.1   Analyzing Graphs with the First Derivative

 

3.2   Analyzing Graphs with the Second Derivative

 

3.3   Absolute Extrema

 

3.4   Optimization

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Curve Sketching & Optimization

 

In theory and applications, we often want to maximize or minimize some quantity. A manu­ facturer may want to maximize profit or minimize cost. An engineer may want to maximize the speed of a new computer or minimize the amount of heat produced by an appliance. A student may want to maximize their grade in calculus or minimize the number of study hours needed to earn a particular grade.

 

Without calculus, we only know how to find optimal points in a few specific cases. For example, we know how to find the vertex of a parabola, but what if we need to maximize or minimize an unfamiliar function?

 

The best way we have to find an optimal solution without using calculus is to examine the graph of the function, perhaps using technology. However, our view of the graph will depend on the viewing window we choose, and we might miss something important! In addition, we will probably only get an approximation by examining the graph of the function and not an exact answer.

 

Calculus provides ways of drastically narrowing the number of points we need to examine to find the exact locations of maxima and minima, while at the same time ensuring that we

 

1

 

have not missed anything important.

 

In this chapter, we will explore how we can use calculus (specifically, use first and second derivatives) to investigate the behavior of functions, and we will apply this knowledge to locate the extreme values and inflection points of functions. In addition, we will use the information we obtain from analyzing a function’s first and second derivatives to create a rough sketch of the graph of the function (without using technology!). Finally, we will apply our knowledge about first and second derivatives to find optimal solutions to real-world problems using a process called optimization.

 

3.1 Analyzing Graphs with the First Derivative

Recall from Section 2.2 that a function is increasing where its derivative is positive, and it is decreasing where its derivative is negative. The derivative can also provide other useful

information about a function. For instance, looking at the graph of the function f shown in

Figure 3.1.1 below, we see that the largest y-value of f(x) occurs at x = 3. There is also a horizontal tangent line at x = 3, which means f'(3) = 0. This is not a coincidence!

 

Figure 3.1.1: Graph of a function fin    which f'(3) = 0 and the largest y-value of f(x)

occurs at x = 3

 

There is a close relation between the x-values where the derivative equals zero and minimums

and maximums of a function. Be warned, however, that it’s not as simple as you may think. Consider the graph of g(x) = x3 –  9×2 + 27x – 22 shown in Figure 3.1.2:

 

Figure 3.1.2: Graph of a function gin which g'(3) = 0, but there is no largest (or smallest) y-value of g(x) occurring at x = 3

Notice the function g shown in Figure 3.1.2 has a horizontal tangent line at x = 3, but the point (3, 5) is not a local extremum! In other words, the point (3, 5) is not at the bottom of a valley or at the top of a hill. In this section, we will explore the relationship between the derivative and local extrema of a function as well as learn a test for finding local extrema.

Learning Objectives: In this section, you will learn how to apply the Increasing/Decreasing Test to find the intervals where a function is increasing/ decreasing, apply the First Derivative Test to find local extrema, and solve problems involving real-world applications. Upon completion, you will be able to:

•    Define partition number of f’.

•    Determine the intervals where a function is increasing/ decreasing using the Increas- ing/Decreasing Test.

•    Define critical value of f.

•    Find the critical values of a function  f given its rule.

•    Specify the conditions necessary for a function to have a local extremum.

•    Prove that a critical value of a function may not coincide with a local extremum by constructing a graph of a function.

•    Distinguish between critical values of  f and   partition numbers of  f’.

•    Apply the First Derivative Test to find the local extrema of a function f given its rule.

•    Apply the First Derivative Test to find the local extrema of a function f given the rule for its derivative f’ and the domain of f.

•    Analyze the graph of f to determine the partition numbers of f’, the critical values of

f, the  intervals where f is increasing/ decreasing, and the local extrema of  f.

•    Analyze the graph of f’ to determine the partition numbers of f’, the critical values of f, the intervals where f is increasing/decreasing, and the x-values where f has local extrema.

•    Demonstrate knowledge of the First Derivative Test by applying it to real-world ap­ plications including cost, revenue, and profit.

Increasing/Decreasing Intervals

We will begin our journey by recalling that if the graph of a function has a positive slope on an interval, then the function is increasing on the interval. Likewise, if the graph of a function has a negative slope on an interval, then the function is decreasing on the interval:

To find on what intervals a function is increasing and decreasing, we will create a sign chart of f'(x). The sign chart will be partitioned into intervals by “important” x-values. We will then test x-values within each interval to determine whether the derivative of the function is positive or negative on the interval. This, in turn, will tell us if the function is increasing or decreasing on that interval.

 

The big question is, what are the “important” x-values we need to put on our sign chart, and how do we calculate them?

 

The “important” x-values are those where the graph of f might possibly switch from in­ creasing to decreasing or decreasing to increasing, or, restating the problem, the x-values where .f'(x) changes sign. It turns out that if .f'(x) changes sign at a particular x-value, then

either f'(x) = 0 or if f'(x) does not exist at that x-value. We have seen an example where

f’ (x)  = 0 and the function  f switches from increasing to decreasing previously in Figure

3.1.1  (at x = 3). The graph off  shown in Figure 3.1.3 demonstrates the other possibility:

 

 

 

Figure 3.1.3: Graph of a function fin        which f'(3) and f'(5) do not exist and f switches between increasing/ decreasing at both x = 3 and x = 5

 

At x = 3, f has a cusp and switches from increasing to decreasing, and at x = 5, f has a corner with the opposite switch.

 

It is also possible f'(x) does not exist at an x-value because f is undefined at that x-value. However, it is still possible for f to switch from increasing to decreasing, or the other way around. For example, consider the graph of f shown in Figure 3.1.4. The function is increasing on the interval (-oo, 4) and decreasing on the interval (4, oo):

 

 

 

 

Figure 3.1.4: Graph of a function fin        which f'(4) does not exist, but f switches from increasing to decreasing at x = 4

 

 

We claim that the only way a function f can switch between increasing and decreasing at a particular x-value is if f'(x) = 0 or f'(x) does not exist at that x-value. How do we know these are the only two possibilities? Well, the only ways the derivative f’ can switch from negative to positive or positive to negative is to pass through a point where f’ (x) = 0 or change sign by the graph off’ being discontinuous and “jumping across” the x-axis.

 

For a non-mathematical example, consider that the only ways to cross a river are to get wet or teleport. This is a special case of a fact that mathematicians call the Intermediate Value Theorem. If the river is the line y = 0, then the only way for f’ (x) to “cross the river” (i.e., switch from positive to negative or negative to positive) is for it to get wet (in which case f’ (x) = 0 at that point) or for it to teleport (in which case f’ is discontinuous, which also means it does not exist). N For the purposes of this textbook, we assume that the individual

trying to cross the river does not have a horse, a boat, or the ability to build a bridge. The individual also does not have a strong friend to throw them across the river, the ability to grow wings and fly, access to a hot air balloon, or any other number of ways to go over, under, or around the river.

 

Just because we get wet does not mean we cross the river! See Figure 3.1.2 for an example where f'(3) = 0, but the derivative does not change sign at x = 3.

 

Thus, the important x-values where f might switch between increasing and decreasing are the x-values where f'(x) = 0 or f'(x) does not exist. We call these x-values partition num­ bers of f’, and we will use them to partition the sign chart of f’ (x):

Note: For the function f (x) = ln x, the definition above includes all x-values less than or equal to zero as partition numbers of f’ (because the domain of f is x > 0, the derivative f’ certainly does not exist for x ::; 0). However, having all of these x-values as partition numbers will not meaningfully partition a number line into discrete intervals that we can use to create a sign chart of f'(x) (which is the goal). Although, because f'(O) does not exist

and there is a meaningful transition happening at x = 0 (the derivative changes from not existing to existing at x = 0), we consider x = 0 to be a partition number off’. Also, x = 0

would be needed to partition a number line into the proper intervals when creating a sign chart of f'(x). The definition of partition numbers off’ can be made more mathematically precise for cases like this, but it is beyond the scope of this textbook. We will use the defi­ nition above and make judgement calls about what is or is not an “important” x-value when determining the partition numbers of f’.

 

Combining all of this information, we now have a process for finding where a function is increasing/ decreasing:

 

 

We demonstrate using this process in the following example.

 

Example 1 Find the intervals where f(x) = x3 + x2 –  x + 1 is increasing/decreasing.

 

Solution: We will use the three-step process outlined previously to apply the Increasing/Decreas­ ing Test:

1.   Determine the domain of f:

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

2.   Find the partition numbers of f’:

We start by finding f'(x):

f’ (x) = 3×2 + 2x – 1

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:

J'(x) = 0 3×2 + 2x – 1 = 0

(3x – l) ( x + l) = 0

This means we need to solve 3x – l = 0 and x + l = 0. Starting with 3x – 1 = 0, we have

3x – l = 0

3x = l

1

X=-

3

Next,

x+l=0

X = -l

Thus, there are two partition numbers of f’ : x =  – l and x = ½.

3.   Create a sign chart of f'(x): We will place the partition numbers off’ on a number line, with both x = – l and x = ½ having solid dots to indicate that they are in the domain of f.

Now, we need to determine the sign of f’ (x) on the intervals created by the partition numbers off’ : (-oo, -1), (-1, ½), and (½, oo). To do determine the sign of f'(x) on each interval, we will select an x-value in each interval to test. Vve will choose x = -2, 0, and 2, but any x-value in each interval will give the same information:

J’ (x) = 3×2 + 2x – 1

f’ (- 2) = 3(- 2) 2 + 2(- 2) – 1 = 7 > 0

J’ (0) = 3(0)2 + 2(0) – 1 = -1  < 0

.!'(2) = 3(2)2 + 2(2) – 1 = 15 > 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 3.1.5.

 

 

Figure 3.1.5: Sign chart of .f’ (x) with the corresponding information for

f (x) = x3 + x2 –  x + 1

From here, we can determine the answer: The function f (x) = x3 + x2 –  x + 1 is increasing on (-oo, -1) and (½, oo), and it is decreasing on (-1, ½).

We can check our answer by looking at the graph of the function .f(x) = x3 +

x2 –  x + 1 shown in Figure 3.1.6:

Figure 3.1.6: Graph of f(x) = x3 + x2 –  x + 1

Note: When applying the Increasing/Decreasing Test, it is not necessary to include solid dots or open circles on the number line to indicate whether or not the partition numbers of .f’ are in the domain of f when creating the sign chart of .f’ (x) (like we did in step 3 of the previous example). However, it is good practice to do so because we will need to know whether or not the partition numbers of .f’ are in the domain of f when applying future tests.

Example 2 Find the intervals where f(x) = (5 – x)½ is increasing/decreasing.

Solution: We will use the three-step process outlined previously to apply the Increasing/Decreas­ ing Test:

1.   Determine the domain of f:

This function is not a polynomial like the function in the previous example, so we need to be careful. Let’s recall the domain restrictions discussed previously in Section 1.4:

 

 

This function does not contain division, nor does it have a logarithm. There is a root (fractional power), but it is an odd root, not an even root. Thus, the domain is (-00,00).

2.   Find the partition numbers of f’:

We begin by finding the derivative using the Chain Rule:

 

J’ ( X) = 1 (5 – X)-l     ( d  (5 – X))

1               2

= 3(5- x)-3(-1)

= – 1(5 – x)-32

3

 

If we were just asked to find the derivative, we could stop here and leave the function f’ in its current form.

However, because we have to continue using the derivative to find the x-values where f'(x) = 0 or f'(x) does not exist, we must use algebraic manipulation to simplify the derivative. In this particular case, we will use the laws of exponents to ensure f’ does not have any negative exponents:

 

f ‘( X ) = -31 (5 – X )- 3

-1

3(5-x)l

 

To find the partition numbers off’, we find the x-values where f'(x) = 0 or f'(x) does not exist. Because f'(x) is a quotient, or ratio, of two functions, it will equal zero where the function in its numerator equals zero:

 

J'(x) = 0 ==*

-1   = 0

 

Because -1  -/- 0, there are no partition numbers where the derivative equals zero.

 

Similarly, f’ (x) will not exist when the function in its denominator equals zero:

j'(x) DNE =?

2

3(5-x)3 =0

2

(5-x)3 =0

(  (5-x)32)

=023

5-x=0

x=5

Thederivative does not exist at x = 5, and it is the only partition number of f’.

3.   Create a sign chart of f’ (x):

We will place the partition number of f’ on a number line with a solid dot to indicate that x = 5 is in the domain of f.

Now, we need to determine the sign of f'(x) on the intervals (-oo,5) and (5,oo) by selecting an x-value in each interval to test. We will choose x = 0 and x = 13, but any x-value in each interval will give the same information:

f

‘(x) –       -1       =?

–      3(5-x)i

–1

j'(0) = 3(5 _ O)i      -0.114 < 0

–1                  1

!'(13) =                      = —       < 0

3(5 – 13)i          12

Using this information, we can create the sign chart of f'(x). See Figure 3.1.7.

Figure 3.1.7: Sign chart of  f'(x) with the corresponding information for

f(x) = (5 –  x)l/3

In conclusion, the function  f (x) = (5 –  x)½ is decreasing on (-oo, 5) and (5, oo).

Note: The sign chart of f’ (x) will not always have alternating signs! Notice in the previous example that f'(x) did not change signs at x = 5.

 

Local Extrema

We can continue using the ideas developed previously to find where a function has hills or valleys, or more formally, local extrema.

 

Note: The plurals of maximum, minimum, and extremum are maxima, minima, and ex­ trema, respectively. Also, note that some texts may refer to local extrema as relative extrema. Local extrema need to only be the largest (or smallest) function values in some neighborhood of a point. For instance, the David G. Eller Oceanography & Meteorology Building (at 151 feet tall) is the tallest building on Texas A&M University’s College Station campus, so it is locally the tallest building. However, it is not the tallest building in the world, nor is it even the tallest building in the state of Texas. The J.P. Morgan Chase Tower in Houston is the tallest building in Texas at 1002 feet tall!

 

Note: As of the writing of this textbook, the Burj Khalifa in Dubai (at 2717 feet tall) is the tallest building in the world. This serves as an example of an absolute maximum, which is something we will discuss more fully in Section 3.4.

 

Next, we will focus on how to find the local extrema of a function. Let’s look at the graph of f we discussed in the introduction, which is shown again in Figure 3.1.8:

 

Figure 3.1.8: Graph of a function fin    which f'(3) = 0 and the largest 71-value of f(x)

occurs at x = 3

The graph off has a local maximum at x = 3, and x = 3 is a partition number of f'(x) (because f'(3) = 0) that is in the domain off. Notice also that f'(x) switches from positive to negative at x = 3. Based on this information, to find local extrema, we should find the partition numbers of f’ that are in the domain of the function f and investigate whether or not f'(x) changes sign at these x-values.

 

The x-values of this special subset of partition numbers off’ that are in the domain off we will investigate are called critical values of f:

We must be careful, though, because a critical value will only result in a local extremum if f’ (x) changes sign at the critical value. For example, let’s look at the function g we discussed in the introduction, which is shown again in Figure 3.1.9. x = 3 is critical value of g because g'(3) = 0 and x = 3 is in the domain of g, but there is no local extremum at x = 3:

 

Figure 3.1.9: Graph of a function gin which g'(3) = 0, but there is no largest (or smallest) 71-value of g occurring at x = 3

Before we learn how to determine whether or not a critical value results in a local extremum, let’s practice finding critical values.

Example 3 Find the critical values of the following functions, if they exist.

(a)     f (x) = x3 –  6×2 + 9x + 2

 

(b)   J(x) = e-x(x – 8)

(c)    f(x) = (x_:4)2

 

 

Solution:  (a) The critical values are the partition numbers off’ that are in the domain off, so we will start by finding the domain of the function. This function is a polynomial, so its domain is all real numbers. Thus, any partition number of f’ we find will be a critical value of f as well!

Our next step is to find f’ (x):

 

J'(x) = 3×2 –  12x + 9

 

Recall that to find the partition numbers of f’, we need to find the x-values where f'(x) = 0 or .f'(x) does not exist. Because the derivative f’ is a polynomial, it will exist everywhere. So we only need to find the x-values where f'(x) = 0:

 

J'(x) = 0 ==*

3×2 –  12x + 9 = 0

3 (x2 –  4x + 3) = 0

3(x – l)(x – 3) = 0

==* x = l and x = 3

 

Thus, x = l and x = 3 are the partition numbers of .f’. Because the domain of  f

is (-oo, oo), x = l and x = 3 are also critical values of f.

(b)   Again, we will start by finding the domain of f (x) = e-x (x – 8), and then we will find the partition numbers of f’ and compare the two to determine the critical values off.

Considering domain restrictions, we see that the function has no logarithms or even roots. There is division hidden in the function by a negative exponent: e-x = e . However, because ex > 0, there are no x-values that will give division by zero. Therefore, the domain off  is (-oo, oo). Next, we find the derivative of the function in order to find the partition numbers of f’. We start with the Product Rule:

 

J'(x) = (x-   8) (d  (e-x)) + e-x (d (x -8))

 

= (x – 8)e-x ( d  (-x)) + e-x(l)

= (x – 8)e-x(-l) + e-x

 

Again, in the previous chapter, we could stop here and leave the derivative in this form. However, because we have to find the partition numbers off’ by setting the derivative equal to zero and finding where it does not exist, we need to continue

 

and  algebraically manipulate the function by factoring e-x from both terms:

f'(x) = (x – 8)e-x(-1) + e-x

= e-x[(x – 8)(-1) + 1]

= e-x (-x    + 8 + 1)

= e-x(–x + 9)

 

Now, we will find the partition numbers off’ such that f'(x) = 0:

 

J'(x) = 0

e-x(-x + 9) = 0

 

This gives two equations: e-x = 0 and -x  + 9 = 0. Because e-x =  e  > 0, there is no solution to the first equation. The second equation gives x = 9.

Checking the domain restrictions to determine for which x-values f'(x) does not

exist, we again see hidden division with the term e-x = e .     However, because

ex > 0, the denominator will never equal zero so f'(x) exists everywhere.

Thus, x = 9 is the only partition number of f’, and because the domain of f is (-00,00), it is also the only critical value off.

(c)   Again, we start by finding the domain of f (x) = (x_:4)2•  This time the function includes division, so we must ensure the function in the denominator is nonzero:

 

(x – 4)2 =/- 0

J(x    -4)2 =J vo

x-4=/-0

x=/-4

 

The domain of f is every x-value except x = 4, or using interval notation, (-oo, 4) U (4, oo). Next, we find the derivative off                                                                  using the Quotient Rule:

 

,x    _ (x – 4)2 (d (x)) – (x) (d ((x – 4)2))

f (  )-                             ((x–    4)2)2

(x – 4)2(1) – x (2(x – 4) C (x – 4)))

(x – 4)4

(x – 4)2 –  2x(x – 4)(1)

(x – 4)4

(x –  4)2 –  2x(x – 4)

(x – 4)4

 

Because we need to use this function to find the partition numbers of f’, we will factor the x – 4 term from the numerator (remember we factor the lowest power

 

of the common term) and continue simplifying:

f’( )=     (X–        4)2 – 2x( X–       4)

X                              (x – 4)4

(x – 4)[(x – 4) – 2x]

(x – 4)4

(x – 4)[(x – 4) – 2x]

(x – 4)43

(x – 4) – 2x

(x – 4)3

-x-4

(x – 4)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 J'(x) does not exist when the denominator equals zero (remember we are looking at the domain off’ when determining where f'(x) does not exist). First, we will find the x-values where f'(x) = 0:

J'(x) = 0

-x-4 =0

X=       -4

Next, to find the x-values where f'(x) does not exist, we set the denominator of

f'(x) equal to zero:

f'(x) DNE

((x – 4)3)31

 

 

= (0)3

x-4=0

x=4

Thus, the partition numbers of f’ are x = -4 and x = 4. Because the domain off is (-oo, 4) U (4, oo), x = -4 is the only partition number off’ that is in the domain off. Thus, x = -4 is the only critical value of f.

 

 

Example 4 Given a function f that is continuous on its domain of (-oo, oo) and its derivative f’ is defined by J'(x) = -½(x + 4)(x + 2)(x – 5)2, find the critical values off, if they exist.

 

Solution: The domain and derivative of f are given. Furthermore, because the derivative .f’ is a polynomial (of degree 4, which we would see if we multiplied all the terms), it exists everywhere. Thus, any critical values off  will occur at the x-values where .f'(x) = 0:

J'(x) = 0

2

– 1(x + 4)(x + 2)(x – 5)2 = 0

 

To find the x-values where this equation equals zero, we set each factor of the product on the left-hand side of the equation equal to zero. Thus, the partition numbers off’, and by extension the critical values off, arex = -4, x = -2, and x = 5.

As mentioned previously, not all critical values will result in local extrema. In addition to the function g we discussed shown in Figure 3.1.2 and Figure 3.1.2, the function f in the previous example also has a critical value, x = 5, in which there is no local extremum. The graph of f corresponding to the previous example is shown in Figure 3.1.10.

 

Figure 3.1.10: Graph of a function f with critical values x = -4, x = -2, andx = 5

 

It is important to remember that for a function to have a local extremum at a critical value, its derivative must change sign at the critical value.

 

 

First Derivative Test

To determine whether a function has a local extremum at a critical value x = c, we must determine the sign of f’ (x) on the left and right sides of x = c. This result is known as the First Derivative Test:

 

 

To apply the First Derivative Test, as well as determine where a function is increasing/de­ creasing, we use the following four steps:

Example 5 For each of the following functions, find the critical values of f, the  intervals where f

is increasing/decreasing, and the local extrema off  as well as where they occur.

(a)     f(x) = 2×3 –  3×2 –  36x

(b)     f (x) = X:!t

(c)     f(x) = 7xln(x) – x

Solution:      (a) We will use the four steps outlined previously to apply the First Derivative Test:

 

1.   Determine the domain of f:

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

2.   Find the partition numbers off’:

 

J'(x) = 6×2 –  6x – 36

 

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:

J'(x) = 0 6×2 –  6x – 36 = 0

6 (x2 –  X –  6) = 0

6(x + 2)(x – 3) = 0

x = -2  and x = 3

Thus, the partition numbers off’ arex = -2 and x = 3.

3.   Determine which partition numbers off’ are in the domain off           (i.e., find the critical values of f):

Because the domain of f is all real numbers, x = -2 and x = 3 are both critical values of f.

4.   We will place the partition numbers off’ on a number line, with x = -2 and

x = 3 having solid dots to indicate that they are in the domain of f.

Now, we need to determine the sign of f'(x) on the intervals (-oo, -2), (-2, 3), and (3, oo). We will choose the values x = -5, 0, and 5 to test:

f’ (x) = 6×2 –  6x – 36

.f'(-5) = 6(-5)2 –  6(-5) – 36 = 144 > 0

J'(0) = 6(0)2 –  6(0) – 36 = -36  < 0

!'(5) = 6(5)2 –  6(5) – 36 = 84 > 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 3.1.11.

Figure 3.1.11: Sign chart of f’ (x) with the corresponding information for

f(x) = 2×3 –  3×2 –  36x

Using Theorem 3.2, we see that a local maximum occurs at x = -2, and a local minimum occurs at x = 3 (note that we also double check on the number line that  f is defined at both of these x-values).  To find the local

 

maximum and minimum values (i.e., 71-values) associated with these x-values, we substitute the x-values into the original function f:

 

f(x) = 2×3 –  3×2 –  36x ==}

f (-2) = 2(-2)3 –  3(-2)2 –  36(-2) = 44

f (3) = 2(3)3 –  3(3)2 –  36(3) = -81

 

In conclusion, x = -2 and x = 3 are the critical values of f, f is increasing on (-oo, -2) and (3, oo), f is decreasing on (-2, 3), and f has a local maximum of 44 at x = -2 and a local minimum of -81 at x = 3.

We can check our work by looking at the graph of f, which is shown in Figure 3.1.12, but using the First Derivative Test allows us to find exactly where the function is increasing/ decreasing and has local extrema. While using technology to look at a graph of the function would give us roughly the same information, if we want exact answers, calculus is the way to go! Calculus provides more precision than technology allows.

Figure 3.1.12: Graph of f(x) = 2×3 –  3×2 –  36x

(b)

1

Recall  f (x) = :2!° 3.  Again, we follow the same four steps to apply the First

Derivative Test:

 

1.   The function f is a rational function, so we must ensure the denominator does not equal zero:

 

Thus, the domain off    is (-oo, 1) U (1, oo).

2.   Find the partition numbers off’:

 

We start by finding .f’ ( x) using the Quotient Rule:

 

J'(x)=   (x -1) (d! (x2 + 3)) – (x2 + 3) (d!(x – 1))

(x – 1)2

(x – 1)(2x) – (x2 + 3) (1) (x-1)2

2×2 –  2x – x2 –  3

(x – 1)2

x2 – 2x – 3 (x-1)2

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 off’ when determining where f'(x) does not exist). First, we find the x-values where f'(x) = 0:

 

J'(x) = 0 ===}

x2 –  2x – 3 = 0

(x + l)(x – 3) = 0

===} x = -1  and x = 3

 

Next, to find the x-values where f'(x) does not exist, we set the denominator of .f'(x) equal to zero:

J'(x) DNE ===}

(x – 1)2 = 0

1

((x – 1)2)21 = (0)2

x-1=0

x=l

Therefore, the partition numbers off’ arex = -1, x = 1, and x = 3.

3.   Determine which partition numbers off’ are in the domain off (i.e., find the critical values of f):

The only x-value not in the domain off  is x = 1. Thus, the critical values of f are x = -1 and x = 3.

4.   Create a sign chart of f’ (x):

We will place the partition numbers of f’ on a number line, with x = -1 and x = 3 having solid dots and x = 1 having an open circle to indicate which are included in the domain off.

Next, we determine the sign of f’ (x) on each of the four intervals created by the partition numbers off’: (-oo, -1), (-1, 1), (1, 3), and (3, oo). We select x-values to test in each of these intervals. We will choose x = -3, 0, 2, and 5,

 

but remember, any x-values in the intervals will give the same sign of f'(x):

J’( x)= x2          2x-   3

  –

(x – 1)2

==}

I    •           (-3)2 – 2(-3)  – 3      3

f(-3)=         ((-3)-1)2        =4>0

J'(O)=    (0)2-     2(0-)    3 =     -3       0

((0) – 1)2                       <

j'(2)=   (2)2-   2(2-)    3 =    -3 < 0

((2) – 1)2

J’( )   _ (5)2 – 2(5) – 3 _

5 –           ((5) – 1)2  –       4> 0

 

Using this information, we can fill in the sign chart off'(x). Again, we include the corresponding information for f below the number line. See Figure 3.1.13.

Figure 3.1.13: Sign chart of  f'(x) with the corresponding information for

f(x) = :2 13

Using Theorem 3.2, we see that a local maximum occurs at x = -l, and a local minimum occurs at x = 3. To find the local maximum and minimum values (i.e., y-values) associated with these x-values, we substitute the x­ values into the original function f:

 

j ( X)=    X2  +3   ==}

x-l

J(-1)=    (-1)2 +3=    -2

(-1)-1

f (3)=    (3)2 +3 =     6

(3) – 1

In conclusion, x = -l and x = 3 are the critical values of f, f is increasing on (-oo,-1) and (3,oo),f is decreasing on (-1,1) and (1,3), and f has a local maximum of -2 at x = – l and a local minimum of 6 at x = 3.

Note: Notice here that the local minimum is greater than the local maximum! This can happen because a local extremum only needs to be the largest/ smallest y-value on some interval around the corresponding x-value.

(c) Recall f (x) = 7x ln(x) – x. Again, we follow the same four steps to apply the First Derivative Test:

1.      Determine the domain of f:

We need to ensure that the argument of the logarithm is positive, or in other words, x > 0. This means the domain of f is (0, oo).

 

2.       Find the partition numbers of .f’:

We start by finding .f'(x) using the Product Rule:

 

.f'(x) = 7 (x (d (ln(x))) + (ln(x)) (d (x)) )-1

= 7 (x (t) + (ln(x))(1))-1

= 7(1 + ln(x)) – 1

=  7+7ln(x)-1

= 71n(x) + 6

To find the partition numbers of .f’, we 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) = 0:

f'(x) = 0 ===}

7ln(x) + 6 = 0

7ln(x) = -6

7

ln(x) = – 6

eln(x=)

e-¥

 

 

When a function involves logarithms, it is important to check the answer by substituting it back into the function (.f’ in this case):

J’ ( e-¥) = 7 ln ( e-¥) + 6

=7 (- ) +6

= -6 + 6 = 0

Next, let’s consider the x-values where f'(x) does not exist. Because the derivative function .f'(x) = 71n(x) + 6 has a domain of (0, oo), it stands to reason that .f'(x) does not exist on the interval (-oo, 0]. This would mean that every x-value in the interval (-oo, 0] would count as a partition number off’ (according to the definition of a partition number given).

However, having all of these x-values as partition numbers will not meaning­ fully partition a number line into discrete intervals that we can use to create a sign chart of .f'(x) (which is the goal). Although, because .f'(0) does not exist and there is a meaningful transition happening at x = 0 (the derivative changes from not existing to existing at x = 0), we consider x = 0 to be a

partition number of f’.  Also, x = 0 is needed to partition the number line

into the proper intervals when creating the sign chart of f’ (x).

Note: This example shows how sometimes it is necessary to be slightly imprecise with the definition of a partition number. We must use good judgement when determining the partition numbers of f’ in cases like this one.

 

Thus, there are two partition numbers of f’ : x = 0 and x = e–617, which is approximately 0.4244.

3.       Determine which partition numbers off’ are in the domain off (i.e., find the critical values of J):

The only critical value of f is x = e-617.

4.        Create a sign chart of f’ (x):

Because f has a domain of (0, oo), we will indicate this on our number line (remember, we are considering x = 0 to also be a partition number of f’). Also, x = e-617 will have a solid dot because it is in the domain off.

There are only two intervals in the domain of f to check for the sign of the derivative: (0, e-617)  and (e-617, oo). We will select x = 0.3 and x = l to test:

J'(x) = 7ln(x) + 6 ==;,

j'(0.3) = 7 ln(0.3) + 6      -2.43 < 0

J’ (1) = 7 ln(l) + 6 = 6 > 0

 

Using this information, we can fill in the sign chart of f'(x). See Figure 3.1.14.

Figure 3.1.14: Sign chart of  f'(x) with the corresponding information for

f(x) = 7x ln(x) –  x

Using Theorem 3.2, we see that a local minimum occurs at x = e-617. To find the local minimum value (i.e., y-value) associated with this x-value, we substitute x = e-617 into the original function f:

 

In conclusion,  x = e-617  is the only critical value of f, f 1s mcreasmg on

(e-617, oo) , f  is decreasing on (0, e-617),  and f  has a local minimum of

-7e-617  at x = e-617.

 

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’, critical values of f, intervals where f is increas­ ing/ decreasing, and local extrema of f algebraically using the rule of the function. Let’s consider finding this information if we are given the graph of the function, or its derivative, instead.

 

Example 6  Given the graph off    shown in Figure 3.1.15, find each of the following.

 

Figure 3.1.15:  Graph of a function  f

 

(a)   Partition numbers off’

(b)    Critical values off

(c)   Intervals where  f (x) is increasing/ decreasing

(d)  x-values where any local extrema off     occur (specify the type)

 

Solution:  (a) The partition numbers off’ are the x-values where f'(x) = 0 or f'(x) does not exist. Recall from Section 2.2 that the derivative f’ does not exist when the graph off  has a cusp or corner, a discontinuity, or a vertical tangent line. This graph

of f has a discontinuity at x = a and a corner at x = b. That gives us two of the partition numbers off’. We also need to find where f'(x) = 0, or where the graph of f has horizontal tangent lines. This occurs at x = c and x = d. Thus,

the partition numbers of f’ are x = a, x = b, x = c, and x = d.

(b)   The critical values of f will be the subset of partition numbers of f’ that are in the domain off. Looking at the graph off, we see its domain is (-oo, a) U (a, oo). Thus, the only partition number of f’ that is not in the domain of f is x = a. This means x = a is not a critical value. All of the other partition numbers of f’ are in the domain off, so the critical values off  are x = b, x = c, and x = d.

(c)   The intervals we need to check and determine if f is increasing/ decreasing on those intervals are (-oo, a), (a, b), (b, c), (c, d), and (d, oo). Instead of selecting x-values in each of these intervals and algebraically checking the sign of the derivative like we have done previously, we will just look at the graph of f. We see that f is decreasing on (-oo,a), (b,c), and (d,oo), and f is increasing on (a,b) and (c,d).

(d)   Local extrema are the largest (or smallest) y-values in a localized area of a func­ tion. We will first focus on finding the local maxima of the function f. By looking at the graph of J, we see local maxima occur at x = b and x = d. Similarly, f

has a local minimum at x = c. Notice that f does change from decreasing to increasing at x = a, but because x = a is not in the domain of f, it cannot be a local extremum!

 

 

Now, suppose we want to find this same information for a function f, but we are only given the graph of its derivative, f’. Algebraically, we know that partition numbers of f’ are the x-values where the 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 find the critical values of f, we observe which of the partition numbers of f’ are in the domain off.

 

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

 

 

Example 7 Given the graph off’ shown in Figure 3.1.16 and that f is continuous on its domain of (-oo, a) U (a, oo), find each of the following.

 

 

 

Figure 3.1.16: Graph of a derivative function  f’

 

(a)   Partition numbers off’

(b)    Critical values of f

(c)   Intervals where f is increasing/ decreasing

(cl) x-values where any local extrema off  occur (specify the type)

 

Solution: We will use the same ideas from the previous example, 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’ are the x-values where f'(x) = 0 or f'(x) does not exist. Remember that we are given the graph of f’, and the only x-value where f'(x) does not exist is x = a (the function is undefined at x = a). We may be tempted to include x =band x = c because of the vertical tangent line and sharp

turn, respectively, but these x-values would be where the derivative off’ 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 = b and x = d.

Thus, the partition numbers of f’ are x = a, x = b, and x = d.

(b)   The critical values of f are the partition numbers of .f’ that are in the domain of .f. We are given that the domain of .f is (oo, a) U (a, oo). Therefore, the only partition number of f’ outside the domain of .f is x = a, so the critical values of

.f are x = b and  x = d.

 

(c)   To find the intervals where .f is increasing, we need to determine where .f’ (x) > 0, or where the graph of .f’ is above the x-axis. This occurs on the intervals (-oo, a) and (b, d). Similarly, .f decreases when .f'(x) < 0, or where the graph off’ is below the x-axis. This occurs on the intervals (a, b) and (d, oo).

 

 

In summary, f is increasing on (-oo, a) and (b, d), and f is decreasing on (a, b)

and (d, oo).

 

 

 

(d)   We have to be really careful when we are given the graph off’, butwe need to find the local extrema of f. The local extrema of f will not be at the tops of hills or bottoms of valleys on the graph of f’, which is the graph we are given here. To

determine the local extrema of J, we must rely on the relationship between the sign of f’ (x) and the corresponding behavior of the graph of f. To find the local maxima, we look at the critical values off and see if f'(x) changes from positive to negative, or in other words, we see if the graph of f’ switches from being above the x-axis to below it.  The only critical value of f where this occurs is x = d.

For local minima, it is the opposite: we look at the critical values of f and see if f’ (x) changes from negative to positive, or in other words, we see if the graph of f’ switches from being below the x-axis to above it. In this case, that happens at

X = b.

 

 

 

In summary, f has a local maximum at x = d and  a local minimum at x = b.

 

 

 

 

 

 

 

 

 

 

Note: Remember that even if the derivative changes sign at a particular x-value, that x­ value must be in the domain of f for a local extremum to occur. Because x = b and x = d are critical values of f, they are in the domain of f.

 

 

Note: x = c and x =fare thex-values where the local maxima off’ occur, not the function f ! Similarly, x = e does not correspond to a local minimum of f, but it does correspond to a local minimum off’. These x-values do provide important information about f, which we will discuss more in depth in the next section.

 

Applications

We can apply what we have learned about intervals of increase/decrease and local extrema to analyze business functions such as profit, revenue, and cost, as well as their corresponding marginal, average, and marginal average functions.

 

Example 8 An office supply company that specializes in producing high-end office chairs with lumbar support has a cost function given by C(x) = 0.04×2 +15x +144 dollars, where

x is the number of chairs produced in one day. Find the intervals where the average cost function, C, is increasing/ decreasing and the local extrema of the average cost function.

 

Solution:

 

Before we can apply the First Derivative Test to the average cost function, C, we must first find the average cost function:

 

C(x)=  C(x)

X

0.04×2 + 15x + 144

X

0.04×2     15x      144

x

+-+-

X                X

= 0.04x + 15 + 144x-1

 

Now, we use the previous four steps to find where C is increasing/decreasing and apply the First Derivative Test:

1.   Determine the domain of C:

The domain of this function is (-oo, 0) U (0, oo) because we cannot divide by zero. However, because this is a real-world application, we must restrict the domain to (0, oo). The company cannot produce a negative number of chairs!

2.   Find the partition numbers of C’:

First, we take the derivative:

C'(x) = 0.04 + 0 – 144x-2

= 0.04 – 144x-2

-2

144=0.04-

X

 

Now, we must find the x-values where C'(x) = 0 or C'(x) does not exist. Let’s start by finding the x-values where C'(x) = 0:

C'(x) = o

–

1440.04-      =0

x2

0.04 =

144

-2

X

(x2) (0.04) = 1424 (x2)

X

0.04×2 = 144

2       144

X  =–

0.04

x2 = 3600

x = -60 and x = 60

 

Thus, the partition numbers of C’ corresponding to where C'(x) = 0 are x =

-60 and x = 60. However, because we cannot have -60 high-end office chairs (remember we restricted the domain to be (0, oo)), we will only consider the

 

partition number x = 60. Next, let’s consider the x-values where C'(x) does not exist. We can see by looking at the function

–                         144

C'(x) = 0.04- –

X2

 

that C'(x) does not exist at x = 0. However, because we restricted the domain of C to be (0, oo), it stands to reason that C'(x) actually does not exist on the interval (-oo, O]. This would mean that every x-value in the interval (-oo, O] would count as a partition number of C’ (according to the definition of a partition number given).

However, as we discussed previously, having all of these x-values as partition numbers will not meaningfully partition a number line into discrete intervals that we can use to create a sign chart of C'(x) (which is the goal). Although, because C'(O) does not exist and there is a meaningful transition happening at x = 0 (the derivative changes from not existing to existing at x = 0), we consider x = 0 to be a partition number of C’. Also, x = 0 is needed to partition the number line into the proper intervals when creating the sign chart of C’ (x).

In summary, the partition numbers of C’ are x = 0 and x = 60.

3.   Determine which partition numbers of C’ are in the domain of C (i.e., find the critical values of C):

The only partition number of C’ in the restricted domain of (0, oo) is x = 60.

4.   Create a sign chart of C’ (x):

Because C has a restricted domain of (0, oo), we will indicate this on our number line (remember, we are considering x = 0 to also be a partition number of C’). Also, x = 60 will have a solid dot because it is in the domain of the function.

Vve have two intervals on which to determine the sign of C'(x) : (0, 60) and (60, oo). We will select the x-values x = 30 and x = 100 to test in order to determine the sign of C’ (x):

–                         144

C'(x) = 0.04- –

X2

–                           144

==}

C'(30) = 0.04 – (30)2 = -0.12 < O

–                               144

C'(lOO) = 0.04 – (100)2 = 0.0256 > 0

 

We will use this information to fill in the sign chart of C'(x). See Figure 3.1.18.

 

Figure 3.1.18: Sign chart of C'(x), where C(x) = 0.04x + 15 + 144x-1

 

This sign chart tells us that the average cost is decreasing when production is between 0 and 60 high-end office chairs, and the average cost is increasing when production is more than 60 chairs. A local extremum occurs at x = 60; Theorem

3.2 tells us there is a local minimum at x = 60. To find the local minimum (value),

we substitute x = 60 into C(x):

C(x) = 0.04x + 15 + 144x-1 ==;,

0(60) = 0.04(60) + 15 + 144(60)-1

= $19.80 per chair

 

Thus, when 60 high-end chairs are produced, the average cost function has a local minimum of $19.80 per chair.

Using Technology

Most calculators can numerically estimate, or approximate, the instantaneous rate of change (i.e., derivative) of a function at a certain x-value. Not only can we use the calculator to check our answers when we calculate rates of change, we can also use it to quickly find the sign

of f’ (x) when creating sign charts. We can input the original function f into the calculator

and have it find the value off’(x) at the x-value we are testing in an interval. This allows us to quickly see if the derivative is positive or negative on each interval of the sign chart off'(x).

 

The calculator used in this textbook is the TI- 84 Plus CE. Any version of the TI- 83 and TI- 84 has this functionality, but your screen and inputs may look a little different from the ones we show here.

 

Example 9 Using technology, find f'(4) if f(x) = x2 –  3.

 

Solution: Using the TI-84 Plus CE, we press the MATH button and then select command 8: nDeriv( . See Figure 3.1.19.

 

Figure 3.1.19: Location of the nDeriv( command in the TI- 84 Plus CE

 

With newer calculators like this one, we can see the inputs needed: first is the variable, next is the function, and the third is the x-value where we want to find the value of the derivative. Inputting these three items in the correct location gives us the screen shown in Figure 3.1.20.

 

Figure 3.1.20: Inputs of the nDeriv( command: the variable, function, and x-value Pressing ENTER will display the answer, 8. See Figure 3.1.21.

 

 

Figure 3.1.21: Approximation of the derivative after pressing ENTER

 

Example 10 Using technology, find f'(-3) if .f(x) = 5_’.9x.  Round your answer to three decimal places, if necessary.

 

Solution: Pressing the MATH button and then selecting command 8: nDeriv( will lead us to the same screen as before. Then, we input the correct variable, function, and x-value. Pressing ENTER gives the value shown in Figure 3.1.22:

 

5 9

Figure 3.1.22: Using the nDeriv(command to find .f'(-3), where .f(x) =  ! x

 

 

g

This answer, 0.062, is an approximation of the exact answer, which is 1        4. It is much

preferred to have an exact answer (i.e., a decimal that terminates, a fraction, or an expression that can yield the exact number) rather than an approximation. Because we know the exact answer is a fraction, let’s use the calculator to try and convert our approximation to a fraction.

Under the MATH menu, the command 1: Frac converts approximations to fractions (when possible). Selecting this command and pressing ENTER yields the same result. See Figure 3.1.23.

 

Figure 3.1.23: Attempting to use the Frac command to convert an approximate answer to a fraction (an exact answer)

 

It appears as though nothing happened! The calculator was unable to give us the exact answer. However, this answer is good enough for this problem because we can round to three decimal places. But, if we needed an exact answer, we would have to find f'(-3) using the limit definition of the derivative (by hand).

In any case, using this calculator function to check your answers or quickly find the sign of f’ (x) on each of the intervals of your sign chart may prove quite useful!

 

Using a graphing calculator can help you check your answer. However, the calculator can only give an approximation. Even trying to convert the answer in the previous example to a fraction did not help. Using calculus is the only way to guarantee exact answers!

 

Try It Answers

1.    f is increasing on (-oo,4), and f is decreasing on (4,oo).

2.      (a) x = -2; x = 0; x = 6

(b)         X =  – l; X =  l

(c)         x = -ifn; x = 0; x = if27

3.      (a) The critical values off      are x = l and x = 3; f is increasing on (-oo, 1) and (3, oo), and f is decreasing on (1, 3); f has a local maximum of 6 at x =land a local minimum of 2 at x = 3.

(b)        The critical value off     is x = 9; f is increasing on (-oo, 9), and f is decreasing on (9, oo); f has a local maximum of e\ at x = 9 and no local minima.

4.      (a) x = 0; x = 2; x = 7; x = 11; x = 12

(b)        x = 0;x = 2;x = ll;x = 12

(c)           f is increasing on (-oo, 0), (2, 7), and (12, oo), and f is decreasing on (0, 2), (7, 11),

and (11, 12).

(d)           f has a local maximum at x = 0 and local minima at x = 2 and x = 12.

5.   The revenue function, R, is increasing on (0, 140) and decreasing on (140, oo), and it has a local maximum of $3920 when x = 140 web cameras are sold.

Exercises

 

Basic Skills Practice 

For Exercises 1-   3, the graph off is shown. Find (a) the x-values where .f'(x) = 0, (b) the x-values where f'(x) does not exist, (c) the intervals where f is increasing/decreasing, and

(d) the x-values where any local extrema off  occur (specify the type).

 

1.

 

2.

 

 

 

For Exercises 4 – 7, find (a) the partition numbers off’ and(b) the intervals where f is increasing/ decreasing.

4.     f(x) = x3 –  27x + 4

5.     f(x) = 3×4 –  40×3 + 150×2 –  25

6.     f(x) = ¼x4 –  !x3 –  16×2

7.    f(x) = -3×4 –  14×3 –  9×2 + 6

For Exercises 8 – 11, find the critical value(s) of the function.

8.     f(x) = x3 + 15×2 + 27x – 1                          10. h(t) = t4 –  2t2 + 9

9.   g(x) = x5 + 5×4 –  35×3

 

9x

11. f(x) =

4 –  22×3 –  30×2 + 50

For Exercises 12 – 16, f’ and the domain off are given. Assuming f is continuous on its domain, find (a) the partition numbers of f’, (b) the critical values of f, (c) the intervals where f is increasing/decreasing, and (d) the x-values where any local extrema off occur using the First Derivative Test (specify the type).

12.     f'(x) = (x + 5)2(x – 2)(x – 8); domain off  is (-oo, oo)

13.     f'(x) = 3x(x + 4)(x – 7)3; domain off  is (-oo, oo)

 

14• !

‘( )  _     (x-l)(x+2).

6

X –              (x-6)4       ‘

domain off  is (-oo, 6) U (6, oo)

 

15.

!‘( X ) – -2x(x(x++l)8)2., domam•

Off

l•S

(-OO, – 1) U (- 1 , 00)

 

 

For Exercises 17 – 19, find (a) the partition numbers off’, (b) the critical values off, (c) the intervals where f is increasing/ decreasing, and (d) any local extrema of f as well as where they occur using the First Derivative Test (specify the type).

 

3

17.  f(x) = x3-9×2+24x-5       18.  f(x) = x5-5.T4-20×3+       19.  f(x) = 3×4-2 °x3-16×2

100

For Exercises 20 – 22, the graph of f’ is shown. Assuming f is continuous on its domain of (-00,00), find (a) the partition numbers of .f’, (b) the critical values off, (c) the intervals where f is increasing/ decreasing, and (d) the x-values where any local extrema of f occur

(specify the type).

 

20.

 

21.

 

 

 

23.      Party Tune, a company that makes sing-a-long microphones for kids, has a revenue function given by R(x) = l86x – 0.2×2 dollars, where xis the number of microphones sold.

(a)   Determine the intervals where the revenue is increasing and where it is decreasing.

(b)   Find the value of x in which revenue is maximum.

24.      The profit function for a company that sells designer coats is given by P(x) = -0.25×2+

460x – 10, 500 dollars, where x is the number of coats sold.

(a)   Determine the intervals where the profit is increasing and where it is decreasing.

(b)   Find the value of x in which profit is maximum.

25.      A company that sells vacuum cleaners has a weekly cost function given by C(x) 4100 – 26x + 0.2×2 dollars, where x is the number of vacuum cleaners produced each week.

(a)   Determine the intervals where the cost is increasing and where it is decreasing.

(b)   Find the value of x in which cost is minimum.

 

 

 

Intermediate Skills Practice 

For Exercises 26 – 29, find the intervals where f is increasing/decreasing.

 

26.         f(x) = xex

27.         f(x) =xX:-3

 

28.         f(x) = x2 ln(x)

29.         f(x) = (x2 –  25)213

 

For Exercises 30 – 35, find the critical value(s) of the function.

 

30.         f(x) = 2ln(x) –  lOx

31.      g(x) = 3eo.2×2

32.      h(t) = 2 + ½ + b

33.         f(x) = xln(x)

34     ( )      t +1

_   2

. gt    –  t2-4

 

35.      h(x) =

For Exercises 36 – 45, find (a) the critical values off, (b) the intervals where f is increas­ ing/ decreasing, and (c) any local extrema of f as well as where they occur (specify the type).

36.      J(x) = (x – 4)ex

37.         f(x) = x3 ln(x)

38.         f(x) =x :_1

39.         f(x) = i(x2 –  3x)4

40.         f(x) = In;:)

41.         f(x) = ex3-27x

 

42.         f(x) = XJ g

43.         f(x) = ln (x2 + 4)

44.         f(x) =ex+ e-x

 

45.         f(x) = 6 +x

46.      Given the graph off shown below, find (a) the partition numbers off’, (b) the critical values off, (c) the intervals where f is increasing/decreasing, and (d) any local extrema off as well as where they occur (specify the type).

 

For Exercises 47 – 49, the graph off is shown. Find (a) the partition numbers off’, (b) the critical values off, (c) the intervals where f is increasing/decreasing, and (d) the x-values where any local extrema of f occur (specify the type).

 

47.

 

48.

 

49.

 

50.      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 critical values of f, (c) the intervals where f is increasing/decreasing, and (d) the x-values where any local extrema off occur (specify the type).

 

 

 

For Exercises 51 – 53, 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 critical values of f, (c) the intervals where f is increasing/decreasing, and (d) the x-values where any local extrema of f occur (specify the type).

 

51.      domain off  : (-oo, oo)

 

 

 

52.      domain off: (-oo,a) U (a,oo)

 

53.      domain off  : (-oo, d) U (d, f) U (f, oo)

 

54.      The profit function for a lamp manufacturer is given by P(x) = 30x – 0.3×2 – 240 dollars, where x is the number of lamps sold. Find the intervals where P is increas­ ing/ decreasing, and interpret your answer.

55.      A company specializing in custom cell phone cases has a weekly cost function given by C (x) = 5000 – 32x + 0.4×2 dollars, where x is the number of cell phone cases produced each week. Find the intervals where C is increasing/ decreasing, and interpret your answer.

56.      Feeling Fabulous, a local spa, sells an all-inclusive spa package. It has determined its revenue function for the packages to be R(x) = 200x – 0.8×2 dollars, where x is the number of spa packages sold. Find the intervals where R is increasing/decreasing, and interpret your answer.

57.       A company has a revenue function given by R(x) = -25×3 + 225×2 dollars, where xis the number of items, in hundreds, sold.

(a)   Find the intervals where R is increasing/decreasing, and interpret your answer.

(b)   Find the company’s average revenue function, fl.

(c)   How many items must the company sell so their average revenue per item is maximum?

 

 

 

Mastery Practice

 

For Exercises 58 – 60, find the intervals where f is increasing/ decreasing.

 

58.         f(x) = lnt2)

59.         f(x) = xe-0.5×2

60.         f(x) = (x:::)3

For Exercises 61 – 63, find the critical value(s) of the function.

61.         f(x) = (x – 3)3(x – 6)2

62.      g(x) = (x :)2

63.       h(t) = t ln(2t)

For Exercises 64 – 77, find (a) the critical values off, (b) the intervals where f is increas­ ing/decreasing, and (c) the x-values where any local extrema off occur (specify the type).

2

64.         f (x) =e;2

65.         f'(x) = e(x-1l(x-7)(x+2)5(x+6)4;              f is continuous on its domain of (-oo, 7)U(7, oo)

 

66.         f(x) = 2ffe –

67.         f (x) = (x – 3)2e-2x

 

68.         f'(x) =- : ;)-;9\           f is continuous on its domain of (-oo, -4) U (-4, oo)

69.         f(x) = ln; x)

 

 

70.

_ (x+5)4

f( )

X –        (x-l)3

71.         f'(x) = l-  (x); f is continuous on its domain of (0, oo) 72.  f(x) = (x – 1)4(x + 5)6

72.

-f-

j”(x) =

e” +2

73.         f'(x) = 12×5 -26×4 -16×3;         f is continuous on its domain of (-00,00)

74.         f(x) = x ln (x2)

75.         f(x) = (x:_ )2

76.      J(x) = 2×2 (x2 –  4)

77.      Given g'(t) = -t3(t + a)4(t – l)(t – b)15, where 1 <a<  b, and that g is continuous on its domain of all real numbers except t = 1, find (a) the partition numbers of g’, (b) the critical values of g, (c) the intervals where g is increasing/decreasing, and (d) the t-values where any local extrema of g occur (specify the type).

78.       Given the graph of g shown below, find (a) the partition numbers of g’, (b) the critical values of g, (c) the intervals where g is increasing/decreasing, and (d) the x-values where any local extrema of g occur (specify the type).

 

79.      Given the graph of .f’ shown below and that .f is continuous on its domain of (-oo, a) U (a,j) U (j, oo), find (a) the partition numbers of .f’, (b) the critical values off, (c) the intervals where .f is increasing/ decreasing, and (d) the x-values where any local extrema of .f occur (specify the type).

 

80.      The price, in dollars, per flashlight when x flashlights are sold is given by the price­ demand function p(x) = 15 – 0.0Olx.

(a)   Determine the intervals where the revenue is increasing and where it is decreasing.

(b)   What price should be charged for each flashlight so revenue is maximum?

81.      Summer Splashin’ can sell 100 pop-up pools when the price is $141 per pool. If the price per pool decreases by $1, they can sell an additional 50 pop-up pools. The company has fixed costs of $8, 000 and production costs of $75 per pool. Find the intervals where the company’s profit function is increasing/ decreasing (assuming the company’s price-demand function is linear), and interpret your answer.

82.      Big-time Floats makes and sells floats shaped like unicorns. Their daily cost function is given by C(x) = .05×2 + 32x + 180 dollars, where xis  the number of unicorn floats

produced each day. How many floats must be produced each day so the average cost per float is minimum?

83.       A company has a revenue function given by R(x) = -2×3 + 96x dollars, where x is the number of items, in hundreds, sold. Determine the interval(s) where the marginal revenue is decreasing.

84.      The weekly profit function for a bubble machine manufacturer, Pop It Like It’s Hot, is given by P(x) = 50x – 0.3×2 – 270 dollars, where xis the number of bubble machines made and sold each week. Find the manufacturer’s profit when the average profit per bubble machine is maximum.