6 Chapter 6

Chapter 2

The Derivative

Sect-ions:

2.1 Average and Instantaneous Rates of Change

2.2 The Limit Definition of the Derivative

2.3 Introductory Derivative Rules and Marginal Analysis

2.4 The Product and Quotient Rules

2.5 The Chain Rule

• Exercises

Suppose a company has determined that its profit from making and selling 10,000 items is $L 990,000. If the company makes and sells additional items, will its profit increase or decrease?

If we know the rate of change (or slope) of the company’s profit function, we can determine whether or not the company should increase the number of items it makes and sells. For instance, if the rate of change of the company’s profit is positive when it sells 10,000 items, making and selling more items will increase profit.

This is far from the only application of rates of change. Many of them are seen every day: the speedometer in a car, an accelerometer in a video game motion control sensor, the rate of change of a business function such as revenue, cost, or profit, the pressure of fluid in a container, air traffic flight patterns and schedules, and many more.

In this chapter, we will explore average and instantaneous rates of change and their applications. We will also investigate differentiation, the process for finding a function, called the derivative, which represents the rate of change of a given function. We will also learn techniques for finding the derivative including the Product, Quotient, and Chain Rules.

1

2.1 Average and Instantaneous Rates of Change

A common amusement park ride lifts riders to a particular height and then allows them to free-fall a certain distance before safely stopping. Suppose such a ride drops riders from a height of 150 feet. Students of physics may recall that the height, in feet, of the riders t seconds after free-fall (ignoring air resistance, etc.)

can be accurately modeled by J(t) = -16t2 + 150. Using this function, which is called a position function,

we can verify that the riders will hit the ground at t ;::::: 3.06 seconds.

Suppose the designers of the ride decide to begin slowing the riders’ falls after two seconds, which corresponds to a height of 86 feet. How fast will the riders be traveling at that time?

We have the position function, f (t) = -16t2 + 150, but what we wish to compute now is the velocity of the riders at a specific point in time (at exactly two seconds after beginning free-fall). In other words, we want to calculate an instantaneous velocity. We currently do not know how to calculate this precisely, but we can use a somewhat intuitive technique to arrive at a good estimate.

For example, if we traveled 90 miles in 3 hours, we know we had an average velocity of 30 miles per hour. We can take the same approach to calculate the average velocity over any time interval:

average velocity =

change in position change in time

Thus, we can approximate the riders’ instantaneous velocity at t = 2 seconds by considering their average velocity over some time period containing t = 2 seconds. If we make the time interval small, we will get a good approximation of their instantaneous velocity.

Note: This fact is commonly used. For instance, high speed cameras are used to track fast moving objects. Distances are measured over a fixed number of frames to generate an accurate approximation of their in­ stantaneous velocity.

Consider finding the average velocity of the riders on the interval [2,3], which is just before they hit the ground. We can calculate the change in the riders’ height (position) on this interval by finding the difference in their position at t = 2 and t = 3 seconds using the position function f (t) = -16t2 + 150 feet. The change in time would be the difference in time. Thus, on the interval [2,3], the average velocity is

f (3-)  f (2)= 6- 86= -80 feet per second

3-2 1

where the negative sign indicates the riders are moving downward. This gives us an approximation for the instantaneous velocity at t = 2. In order to find the exact value for the instantaneous velocity, or the instantaneous rate of change of position, we need to build upon this idea of average velocity, or average rate of change of position, further.

Learning Objectives: In this section, you will learn how to calculate the average rate of change and the instantaneous rate of change of a function and use these ideas to solve problems involving real-world applications. Upon completion you will be able to:

• Calculate the slope of a secant line given a function, table of function values, or graph of a function.

• Describe the slope of the secant line as the average rate of change, difference quotient, and average velocity.

• Calculate the average rate of change of a function involving a real-world scenario, including cost, revenue, profit, and position.

• Interpret the meaning of the average rate of change of a function involving a real-world scenario, including cost, revenue, and profit.

• Interpret the slope of the tangent line as the limit of slopes of secant lines.

• Describe the slope of a tangent line as the instantaneous rate of change (rate of change), limit of the difference quotient, and instantaneous velocity (velocity).

• Estimate the slope of a tangent line by computing the limit of slopes of secant lines given a function, table of function values, or graph of a function.

• Calculate the slope of a tangent line using the limit definition of the instantaneous rate of change.

• Find the equation of a tangent line using the limit definition of the instantaneous rate of change.

• Calculate the instantaneous rate of change of a function involving a real-world scenario, including cost, revenue, profit, and position, using the limit definition of the instantaneous rate of change.

• Interpret the meaning of the instantaneous rate of change of a function involving a real-world scenario, including cost, revenue, and profit.

Average Rate of Change

In the motivating example, we used the function f (t) = -16t2 + 150 to calculate the average velocity of the riders on the interval from t = 2 to t = 3 seconds. To find the average velocity, we found the slope of the line containing the points (2, 86) and (3, 6). This line is known as the secant line because it passes through

two points on the graph of f. This leads to the formal definition of the slope of the secant line, which we also discovered is the average rate of change of f from t = 2 to t = 3 seconds.

In the motivating example, we were dealing with a position function. When a function represents the position of an object, the average rate of change of that function represents the average velocity.

The graphs of the position function f and the secant line on the interval [2, 3] are shown in Figure 2.1.1.

Figure 2.1.1: Graphs of f(t) = -16t2 + 150 and a secant line

Example 1: Find the slope of the secant line passing through the points on the graph of f (t) corresponding to t = 2 and t = 2.5.

-16t2 + 150

Solution: To find the y-value of the point on the graph off  corresponding tot= 2, we must find f (2):

f(2) = -16(2)2 + 1-50 = 86

Likewise, to find the y-value of the point on the graph of f corresponding to t = 2.5, we must find f(2.5):

f(2.5) = -16(2.5)2 + 150 = 50

Thus, the secant line passes through the points (2, 86) and (2.5, 50) and has a slope of

m=  50-  86

2.5 – 2

-36

0.5

= -72

In the previous example, we were using the function from our motivating example, f(t) = -16t2 +150, which had input units of seconds and output units of feet. Thus, the slope of the secant line, which is equivalent to the average rate of change, has units of feet per second.

This is always true for any rate of change (i.e., slope). For example, if the units of x are years and the units off (:r) are people, then the units of both the average rate of change and instantaneous rate of change are people per year. In general, the units for any rate of change (both average and instantaneous) of f are given by

Um.ts t·or aR ate ot· Change of”•f =  units off(:1:) =  output units

units of:r input units

Example 2: Given the units of :1: and the units of f(x) in Table 2.1, find the units for the average rate of change of

f.

Table 2.1: Units of independent and dependent variables

2.1. AVERAGE AND INSTANTANEOUS RATES OF CHANGE .s

Solution: (a) When calculating the average rate of change, the units for the numerator are the units of f(i:), which in this case is miles. The units for the denominator are the units of i:, which in this case is hours. Thus, our units for the average rate of change are miles per hour.

(b) Again, the units for the average rate of change are the units of f (x) per unit of i:. So here we have automobiles per person.

(c) Again, the units for the average rate of change are the units of J(:r) per unit of ;i:. Thus, the units are pancakes per dollar.

(d) The units for the average rate of change are miles per second per second, or we may write this as miles per second squared ( ).  These are the units for acceleration! The rate of change of position is velocity, and the rate of change of velocity is acceleration.

Note: Acceleration can be thought of as the rate of change of the rate of change of the position. In Chapter 3, we will further discuss the rate of change of a rate of change function.

Example 3: A manufacturer produces pocket-sized monster plushes. The cost of making ;i: plushes is given by the function C(i:) = -x2 + lOOOx + 50,000 dollars. What is the average rate of change of cost when the first 600 plushes are made? Interpret your answer.

Solution: We must calculate the average rate of change of cost on the interval [0, 600]:

C(600) – C(0) dollars 600 – 0 plushes

$290,000 – $50,000

600 plushes

$240,000

600 plushes

= $400 per plush

Thus, the average rate of change of cost is $400 per plush. To interpret our answer, we write a sentence: “When the first 600 plushes are made, cost is increasing at an average rate of $400 per plush.”

These better be very well made plushes!

Example 4: Jenny is a police officer in Viridian City. Her radar gun measures the position of an oncoming car, first when she presses the button, and then every half second after that while she holds down the button. The radar gun measures the position of a car as 1150 feet away, and half a second later it measures the position of the car as 1225 feet away. vVhat is the average velocity, in miles per hour, that Officer Jenny’s radar gun shows for this car? Round your answer to one decimal place.

Solution: We will start our interval at t = 0 seconds indicating the exact moment Officer Jenny pressed the trigger. A half second later (the time of the second reading) corresponds to t = 0.5 second. The output of the radar gun is 1150 feet and 1225 feet, respectively. Using our formula above for average rate of change, or in this case, average velocity, we have

1225 – 1150 feet

0.5 – 0 seconds

75 feet

0..5 second

= 150 feet per second

Be careful! This is not the answer we are looking for! The problem asks for our answer in miles per hour. We need to convert the units:

150 feet 1 mile 60 seconds  60 minutes = . h

1 second 5280 feet

1 minute 1 h our

102.3 m11es per our

Thus, the average velocity of the car is 102.3 miles per hour. While Officer Jenny writes this driver a ticket, we will move on to our next example.

Example 5: Nurse Joy measures a patient’s heart rate, in beats per minute, every thirty seconds for five minutes using a machine. The results are shown in Table 2.3.

Time (in seconds) 0 30 60 90 120 150 180 210 240 270 300

Heart Rate (in BPM) 85 87 90 100 89 10-5 110 90 85 89 87

Table 2.3: A patient’s heart rate, in BPM, over a five minute period Find the average rate of change of the heart rate

(a) over the entire 5 minute period.

(b) over the first 3 minutes.

(c) between 150 seconds and 300 seconds.

Solution: (a) The entire 5 minute period corresponds to the interval [0,:300]. The average rate of change of the heart rate on the interval [0, 300] is given by

87 – 85BPM 2BPM 1

= — = – BPM per second

300 – 0 seconds

(b) The first 3 minutes corresponds to the interval [0, 180]. The average rate of change of the heart rate on [0, 180] is given by

110-85BPM

180 – 0 seconds

25BPM = -5

180 s 36

BPM per second

(c) The average rate of change of the heart rate on the interval [150, 300] is given by

87-105BPM

300 – 150 seconds

-18BPM = –3

150 s 25

BPM per second

Instantaneous Rate of Change

Let’s revisit our motivating example concerning the amusement park ride. Recall that the height, in feet, of the riders t seconds after free-fall was modeled by f(t) = -16t2 + 150. We were interested in obtaining the instantaneous velocity at t = 2 seconds. We approximated the instantaneous velocity at t = 2 by finding the average velocity on the interval [2, 3], which we found was -80 feet per second. In Example 1, we found

the slope of the secant line on the graph off on the interval [2, 2.5] to be -72. As we discussed earlier, this leads us to conclude that the average velocity, or average rate of change of the position function, is -72 feet per second on that interval.

By continuing to narrow the interval we consider, we can improve our approximation for the instantaneous velocity at t = 2 seconds. For instance, on the interval [2,2.1], we see the average velocity is

f(2.l) – f(2)

2.1- 2

79.44 – 86

0.l = -65.6 feet per second

On the interval [2, 2.01], the average velocity is

!(2.01) – f(2)

2.01 – 2

8,5.3584 – 86

0.0l = -64.16 feet per second

On the interval [2, 2.001], the average velocity is

!(2.001) – f(2)

2.001- 2

85.93,5984 – 86 .

0.001 =  -64.016 feet per second

Based on these calculations, it appears that a good estimate for the instantaneous velocity at t = 2 seconds is -64 feet per second. Thus, we know at exactly two seconds after the start of free-fall, the riders are moving in the downward direction at an approximate speed of 64 feet per second.

Note: We could repeat the process above with 2 being the right endpoint of the interval, and we would reach a similar conclusion.

Example 6: The profit for Singing Samuel’s TVs when :r TVs are manufactured and sold is P(:r) dollars. Table 2.4 gives some values of P(x).

X 497 498 499 500 501 502 503

P(;r) $145, 629.91 $14-5, 919.96 $146, 209.99 $146,500 $146, 789.99 $147,079.96 $147,369.91

Table 2.4: Profit values when x TVs are sold

(a) Find the average rate of change of profit on the following intervals.

i. [497, ,500]

ii. [498,500]

iii. [499,500]

iv. [500, 503]

v. [500,502]

vi. [500,501]

(b) Use part a to approximate the instantaneous rate of change of profit when 500 of Singing Samuel’s TV:c; are manufactured and sold, and interpret your answer.

Solution:  (a) In general, the average rate of change of the profit function, P, on the interval [a, b] is given by

P(b) – P(a) dollars

b-a TVs

1. Thus, the average rate of change of profit on the interval [497,500] is

P(500) – P( 497) dollars 500 – 497TVs

$146,-500 – $145, 629.91 3TVs

$870.09

3TVs

= $290.03 per TV

ii. The average rate of change of profit on the interval [498,500] is

P(-500) – P( 498) dollars 500 – 498TVs

$146, 500 – $145, 629.91 2TVs

$580.04

2TVs

= $290.02 per TV

iii. The average rate of change of profit on the interval [499,500] is

P(500) – P( 499) dollars 500 – 499TVs

$146, 500 – $146, 209.99

lTVs

= $290.01 per TV

iv. The average rate of change of profit on the interval [500,503] is

P(503) – P(500) dollars 503 – 500TVs

$147, 369.91 – $146,500 3TVs

$869.91

3TVs

= $289.97 perTV

v. The average rate of change of profit on the interval [500,502] is

P(502) – P(500) dollars 502 – 500TVs

$147,079.96 – $146,500 2TVs

$579.96

2TVs

= $289.98 per TV

vi. The average rate of change of profit on the interval [500,501] is

P(501) – P(500) dollars 501- 500TVs

$146, 789.99 – $146,500

lTVs

= $289.99 per TV

(b) Each average rate of change in part a is an estimate for the instantaneous rate of change when 500TVs are produced and sold, but if we look more closely at the values of these average rates of change, we may be able to obtain a better estimate.

Notice that the first three intervals have 500 as the right endpoint and the last three intervals have 500 as the left endpoint. In each set of three, the intervals are getting smaller and smaller. Also, each set of approximations is approaching $290 per TV.

Thus, it appears that a good approximation for the instantaneous rate of change of profit when 500TVs are manufactured and sold is $290 per TV. To interpret our answer, we write a sentence:

“When 500 TVs are manufactured and sold, profit is increasing at a rate of $290 per TV.”

Example 7: Given f(x) = 10:i:2 –  30:r + 23.5,

(a) find the slope of the secant line passing through the points on the graph off at :r = 1 and :r = 1.5.

(b) find the slope of the secant line passing through the points on the graph off at :r = 1 and x = 1.2.

(c) find the slope of the secant line passing through the points on the graph of f at ;r: = 1 and

:r = 1.01.

(d) estimate the limit of the slopes of these secant lines as the right endpoint of the intervals approaches 1.

Solution:  (a) To find they-value of the point on the graph off  corresponding to x = 1, we must find f(l):

f(l) = 10(1)2 – 30(1) + 23.5 = 3.5

Likewise, to find the y-value of the point on the graph off corresponding to x = 1.5, we must find f(l.5):

f (1.5) = 10(1.5)2 – 30(1.5) + 23.-5 = 1

Thus, the secant line passes through the points (L 3.5) and (1.5, 1) and has slope

1-  3.5

m—-

– 1.5 -1

-2.5

0.5

= -5

(b) Once again, we need to calculate f(l) and f(l.2) to find they-values of the points on the graph off corresponding to ;r: = 1 and :r: = 1.2, respectively. From part a, we know f(l) = 3.5. We now find f(l.2):

f (1.2) = 10(1.2)2 –  30(1.2) + 23.5 = 1.9

Thus, the secant line passes through the points (L 3.5) and (1.2, 1.9) and has slope

1.9 – 3.5

m= 1.2 -1

-1.6

0.2

= -8

(c) Once again, we already have f (1), so we only need to find f (l. 01):

f(l.01) = 10(1.01)2 –  30(1.01) + 23.5 = 3.401

Thus, the secant line passes through the points (L 3.5) and (1.01, 3.401) and has slope

3.401 – 3.5

m= 1.01-1

-0.099

0.01

= -9.9

(d) Recall that we need to estimate the limit of the slopes of these secant lines as the right endpoint of the intervals approaches l. As the intervals get smaller and smaller in parts a through c, the slopes of the secant lines go from -5 to -8 and then from -8 to -9.9. Thus, it appears that these values are approaching -10.

Let’s consider the previous example graphically to investigate what happens when we find slopes of secant lines on smaller and smaller intervals. The function from the previous example, f(x) = 10:r2 – 30x + 23.5, along with three different secant lines whose slopes we found in parts a through c, are shown in Figures

2.1.2, 2.1.3, and 2.1.4, respectively.

Figure 2.1.2: Graphs of f(x) = 10:r2 – 30:r + 23.5, the line tangent to the graph off at x = 1, and the secant line that passes through the points at x = 1 and :r = 1.5

Figure 2.1.3: Graphs of f(:r) = 10×2 – 30:r + 23.5, the line tangent to the graph off at :r = 1, and the secant line that passes through the points at x = 1 and :r = 1.2

Figure 2.1.4: Graphs of f(x) = 10×2 – 30x + 23.,5, the line tangent to the graph off at x = l, and the secant line that passes through the points at :i: = 1 and x = 1.01

Notice that as the intervals get smaller and smaller, the secant lines approach the dotted line. This dotted line appears to touch the graph at only x = l. We refer to this line as the tangent line, and its slope can be approximated by finding slopes of secant lines on smaller and smaller intervals. We define the tangent line below and then discuss how to find the exact value of the slope of the tangent line.

Note that this definition is not mathematically precise. We will now work to formally define the slope of the tangent line at the point (a, f(a)). We start by generalizing the ideas from Figures 2.1.2, 2.1.3, and 2.1.4.

To find the slope of the tangent line at the point (a, f (a)), we will start by finding slopes of secant lines around the point (a, f(a)). The slope of the secant line passing through the points (a, f(a)) and (:r, f(x)) is given by

m.= .f(:r) – f(a)

:r-a

Because we want smaller intervals, we consider secant lines in which the x-values are getting closer to a. See Figure 2.1.5, Figure 2.1.6, and Figure 2.1.7. The dotted line in the figures is the line tangent to the graph of f at :1: = a:

Figure 2.1.5: Graphs off, theline tangent to the graph of f at x = a, and the secant line that passes through the points (a, f (a)) and (x, f(:r))

Figure 2.1.6: Graphs off, theline tangent to the graph of f at x = a, and the secant line that passes through the points (a, f (a)) and (:r, f(:r))

Figure 2.1.7: Graphs of .f, the  line tangent to the graph of f at x = a, and the secant line that passes through the points (a, f (a)) and (x, .f(;]_:))

Mathematically, when we talk about ;J_; “getting close” to a particular value, we are talking about a limit! This allows us to formally define the slope of the tangent line, which is equivalent to the instantaneous rate of change:

Note: Even though the graphs in Figure 2.1.5, Figure 2.1.6, and Figure 2.1.7 showed the secant lines as x approached a from the right, for the above limit to exist, the slopes of the secant lines from both the left and right of ;J_; = a must approach the same number (a good fact to remember from Sections 1.1 and 1.2!).

This definition is very useful conceptually because it shows that the slope of the tangent line is the limit of the slopes of secant lines, but it is difficult to compute. Let’s revisit the amusement park ride example to

help lead us to a definition that is easier to work with.

Recall that we found five estimates for the instantaneous velocity, or instantaneous rate of change, at t = 2 seconds by finding the average velocity on smaller and smaller intervals. These computations can be viewed as finding the average velocity, or average rate of change, on the interval [2, 2 + h] for small values of h. That is, we computed

f(2 + h) – f(2) f (2 + h) – f (2) (2 + h) – 2 h

for small values of h(h-/=- 0). Our previous calculations using this new notation are summarized in Table 2.6.

h Average Velocity on [2, 2 + h] = f(2 + h) – f(2)

h

1 -80 feet per second

0.5 -72 feet per second

0.1 -65.6 feet per second

0.01 -64.16 feet per second

0.001 -64.016 feet per second

Table 2.6: Approximating the instantaneous velocity

Because we want h to become small, we are really wanting to find the limit as h approaches 0. Thus, the instantaneous velocity, or instantaneous rate of change, at t = 2 seconds can be found exactly by finding the following limit:

. f (2 + h) –  f (2)

1m

h-+0 h

This leads us to a definition that is easier to work with than our previous definition:

To visualize this definition, look at Figures 2.1.8, 2.1.9, and 2.1.10, which are identical to Figures 2.1.5, 2.1.6, and 2.1.7 except we replace x with a+ h:

Figure 2.1.8: Graphs off, theline tangent to the graph of f at x = a, and the secant line that passes through the points at x = a and

:r=a+h

Figure 2.1.9: Graphs off, theline tangent to the graph of f at :r = a, and the secant line that passes through the points at :r = a and

:r=a+h

Figure 2.1.10: Graphs of .f, the line tangent to the graph of .f at x = a, and the secant line that passes through the points at :i: = a and

;r=a+h

Note that in this definition of the instantaneous rate of change, the quantity  f(a+hz-f(a), where h -/- 0, is sometimes referred to as the difference quotient. Because this quantity represents the slope of a secant line, it can also be referred to as the average rate of change.

Now, let’s focus on computing this limit algebraically! Recall from Section 1.2 that we should first attempt to find limh-+O  f(a+h2-f(a) using direct substitution. However, because the difference quotient consists of a ratio of two functions, we need to make sure the limits of the numerator and denominator exist and that the limit of the denominator is nonzero. Let’s start by checking the numerator:

lim f(a + h) – f(a) = f(a + 0) –  f(a)

h-t0

= .f(a) –  f(a)

=0

Finding the limit of the denominator we get

lim h = 0

h-+0

These two limits exist, but because the limits of the two functions equal zero, the limit is of the indeterminate form §. This will always be the case when finding the limit of the difference quotient as h —+ 0. Thus, we

must always use algebraic techniques to further investigate this type of limit. The ultimate goal of our algebraic manipulations is to divide the h from the denominator of the difference quotient. Doing so will ensure the limit of the denominator is nonzero. Let’s look at this process for a specific function in the example below.

Example 8: Find the exact value of the instantaneous rate of change of J(x) = ;i:2 –  3 at ;r = 4.

Solution: To find the exact value of the instantaneous rate of change, we must use the limit definition of the instantaneous rate of change. We defined two different limit definitions, but in this textbook, we will always use the second definition, limh-+O  f(a+hz- f(a), in our computations.

As seen previously, direct substitution with the difference quotient leads to the indeterminate form

§. Thus, we must use algebraic techniques to further investigate the limit. Remember, the goal is

to eventually divide the h from the denominator so we can use direct substitution to find the limit.

Because we are trying to find the instantaneous rate of change at :r = 4, we let a = 4 in the limit:

lim  f(4 + hh) – f(4)= lim ((4 + h)2- 3h-)

h O , h O

2

(4- 3)

= lim ((4+h)(4+h)-3)-13

h O h

= lim ((16 + 4h + 4h + h2) – 3) – 13

h O h

=  lim 16 + 8h + h2 – 16

h O h

=  lim 8h + h2

h O h

= lim  ,h(8 + h)

h O h

=  lim 8+h

h O 1

= lim(8 + h)

h O

vVe were able to divide the h from the denominator as we hoped. We now attempt direct substitution again:

lim !(4+ h-) f(4)= lim (8+ h)

h O h h ll

=8+0

=8

Thus, the instantaneous rate of change, or slope of the tangent line, of f(:r) = ;r:2 – 3 at ;r: = 4 is 8.

Some students may be content with finding this type of limit in one step. But for many, the process may feel overwhelming. Also, computing the limit all at once makes it much more likely that algebraic mistakes will be made!

For the remainder of this section, we will use the Suggested Four-Step Process outlined below to evaluate this type of limit using a step-by-step process.

Let’s rework the previous example using the Suggested Four-Step Process.

Example 9: Find the exact value of the instantaneous rate of change of f(x) = x2 – 3 at x = 4.

Solution: We still need to find lim !(4 !(4) as in the previous example, but instead of algebraically

+ h-)

h–+0 h

manipulating the limit in one single (long) step, we will use the Suggested Four-Step Process:

l. Calculate f (4 + h), and algebraically manipulate the resulting expression, if possible:

f(4+h)=(4+h)2-3

= (4+h)(4+h)-3

Using FOIL gives

= 16 + 4h + 4h + h2 –  3

= 13 + 8h + h2

2. Calculate the numerator of the difference quotient, .f(4 + h) – .f(4), and algebraically manipulate the resulting expression, if possible:

We start with .f(4):

f(4) = (4)2 –  3 = 13

So in the numerator of the difference quotient, we have

f(4 + h) – f(4) = 13 + 8h + h2 –  13

= 8h + h2

3. Calculate the difference quotient (i.e., divide the result from the previous step by h, where h -I 0), and algebraically manipulate the resulting expression:

f(4 + h) – f(4)

h

8h + h2

h

Factoring an h from the numerator gives

Dividing the h’s gives

4. Take the limit ash-+ 0:

h(8+h) h

)i(8 + h)

h

=8+h

lim f(4+h)-f(4)=  lim(8+h)

h–+0 h h–+0

=8+0

=8

Thus, the instantaneous rate of change, or slope of the tangent line, of j(J.:) = x2 – 3 at J.: = 4 is 8. If you look at the solution to the previous example, you will see that we reached the same conclusion. Either technique is correct, but we will continue to use the Suggested Four-Step Process for the remainder of this section.

Note: We were able to use direct substitution to find the limit in step 4 when using the Suggested Four-Step Process in the previous example. In other words, we were able to substitute 0 for h to find lim1i–to(8 + h). Recall that we are able to use direct substitution if there are no domain issues.

Example 10: Find the slope of the line tangent to the graph of f(:r)  =6x + 2 at x = l.

Solution: The slope of the tangent line is equivalent to the instantaneous rate of change, which is what we found in the previous two examples. Because we are asked to find the slope of the tangent line at x = l, we let a= l in the limit:

.  f(l + h) – f(l) 1lm h-+0 h

Again, to simplify the process of finding this limit, we will use the Suggested Four-Step Process:

l.  Calculate f (l + h), and algebraically manipulate the resulting expression, if possible:

f (l + h) = j6(l + h) + 2

=6+ 6h+2

 =s+ 6h

2. Calculate the numerator of the difference quotient, f(l + h) – f(l), andalgebraically manipulate the resulting expression, if possible:

We start with f(l):

f(l) = )6(1) + 2 = VS

So in the numerator of the difference quotient, we have:

f(l + h) – f(l)  =s + 6h – VS

Notice there is no algebraic manipulation to perform at this step. When working with functions containing square roots, the algebra will mostly be performed in step 3.

3. Calculate the difference quotient (i.e., divide the result from the previous step by h, where h-=/- 0), and  algebraically manipulate the resulting expression:

f(l + h) – f(l)  8+6h-VS

h h

To algebraically manipulate this expression, we rationalize the numerator by multiplying by the conjugate:

  8+6h-VS= ( 8+6h-VS) ( 8+6h+VS)

h h s+6h+VS

(8+ 6h) – 8

 h(s + 6h + v18)

6h

 h(s + 6h + v18)

Note: Remember that multiplying by the conjugate is extremely useful for calculations where there is a square root in the function because it eliminates the square roots where necessary. It is not without cost, however. The process helps us simplify the numerator, but the denominator becomes more complicated. That’s okay for us, though, because we can calculate the limit with this slightly more complicated denominator.

Dividing the h’s gives

4. Take the limit as h -+ 0:

6h

 h(s + 6h + vs)

6

8 + 6h + v18

lim  f(l + h-)f(l)=  Jim 6

,Ho h ,H os + 6h + v18

We have to be careful before proceeding. Unlike the previous example, we should not immediately substitute O for h to find this limit. Why? You guessed it! The expression we are taking the limit of consists of a ratio of two functions, so we must check that the limits of both the numerator and denominator exist and that the limit of the function in the denominator is nonzero.

Let’s start by checking the limit of the numerator:

lim 6 = 6

h-+0

Next, we check the limit of the denominator (note we can use direct substitution because there will not be any domain issues):

lim( 8 + 6h + v’S) = j8  + 6(0) + v’8

h-+0

=v’S+v’S

= 2v’8

We see both limits exist and the limit of the function in the denominator is nonzero. Thus, we can find the limit using the Properties of Limits:

1. 6

h-+0 8 + 6h + -Js

limh-+0 6

limh-+O(8 + 6h +s)

6

2-Js

3

s

Thus, the slope of the line tangent to the graph of f at x = l is }s.

Note: For the instantaneous rates of change we will find in this textbook using the limit definition, alge­ braically manipulating the difference quotient will ensure that we can substitute O for h to find the limit in step 4 of the Suggested Four-Step Process, even if the resulting expression in step 3 consists of a ratio of two functions (as seen in the previous example). The limits of the resulting numerator and denominator will exist and the limit of the resulting denominator will be nonzero (due to dividing h from the denominator in step 3). Thus, from this point forward, we will use direct substitution and substitute O for h when calculating the limit in step 4 of the Suggested Four-Step Process because this will lead us to the same answer as using the Properties of Limits to divide the limits of the resulting numerator and denominator.

Note: We will discuss the x:-values where the limit definition of the instantaneous rate of change does not exist in Section 2.2.

Example 11: Find the exact value of the instantaneous rate of change of f(x) = 5_!9x at :r = -:3.

Solution: To find the instantaneous rate of change of .f at x = -3, we let a = -3 in the limit definition of the instantaneous rate of change:

. f(-3 + h) – f(-3)

hh-m+0 h

Again, to break down the steps for finding this limit, we will use the Suggested Four-Step Process:

l. Calculate f(-3 + h), and algebraically manipulate the resulting expression, if possible:

f (-3 + h) =

7

5 – 9(-3+ h)

7

5 + 27 – 9h

7

32 – 9h

2. Calculate the numerator of the difference quotient, f(-3+h)-J(-3), andalgebraically manipulate the resulting expression, if possible:

We start with J(-3):

-3) – 7 7

– 5 – 9(-3) 32

So in the numerator of the difference quotient, we have

f(-3+h)-f(-3)= 7 7

32 – 9h 32

To algebraically manipulate this expression further, we will get a common denominator:

32: 9h- ;2= (! ) (32: 9h)- (;2) (=!: )

224 224 – 63h

32(32 – 9h) 32(32 – 9h)

224 – (224 – 63h)

32(32 – 9h)

224 – 224 + 63h

32(32 – 9h)

63h 32(32 – 9h)

3. Calculate the difference quotient (i.e., divide the result from the previous step by h, where h cl 0), and algebraically manipulate the resulting expression:

f(-3 + h) – f(-3)

h

63h  32(32-9h)

h

At first glance, it may not be obvious how to proceed because we have a complex fraction. Anytime we are presented with a complex fraction such as this one, we will rewrite the division by h as

multiplication by ¾.  Recall that we can do this because dividing by a fraction is equivalent

to multiplying by its reciprocal. Thus, we continue algebraically manipulating the difference quotient:

63h 32(32-9h)

h

=

=

63h 32(32-9h)

h

T

32(32 – 9h) h

32(32 – 9h) h

63

32(32 – 9h)

4. Take the limit ash-+ 0 (i.e., substitute O for h):

I. f (-3 + h) –  f (-3) = 1. 63

h-tO h h-tO 32(32 – 9h) 63

32(32 – 9(0))

63

1024

Thus, the instantaneous rate of change of f at x = -3  is 1 g4.

Example 12 Find the exact value of the instantaneous velocity at t = 2 seconds for the amusement park ride example. Recall that the height, in feet, of the riders t seconds after free-fall was given by f(t) = -16t2 + 150.

Solution: Remember, if the function f represents the position of an object, the instantaneous rate of change of

f represents the instantaneous velocity of the object. To find the instantaneous velocity at t = 2, we let a= 2 in the limit:

. f(2 + h) –  f (2)

Im

h-tO h

Again, we will use the Suggested Four-Step Process:

1. Calculate f (2 + h), and algebraically manipulate the resulting expression, if possible:

f(2 + h) = -16(2 + h)2 + 150

Using FOIL and distributing gives

= -16 (4 + 4h + h2) + 150

= -64 – 64h – 16h2 + 150

= 86 – 64h – l 6h2

2. Calculate the numerator of the difference quotient, f(2 + h) – f(2), and algebraically manipulate the resulting expression, if possible:

We start with f(2):

f(2) = -16(2)2 + 150 = 86

So in the numerator of the difference quotient, we have

f (2 + h) –  f (2) = 86 – 64h – 16h2 –  86

= -64h -16h2

3. Calculate the difference quotient (i.e., divide the result from the previous step by h, where h =/= 0), and algebraically manipulate the resulting expression:

f (2 + h) – f (2)

h

Factoring an h from the numerator gives

-64h -16h2

h

h(-64 -16h)

h

Dividing the h’s gives

Ji(-64 – 16h)

Ji

= -64 -16h

4. Take the limit ash-+ 0 (i.e., substitute O for h):

lim f(2+ h-) f(2)= lim (-64 – 16h)

h O h h O

= -64- 16(0)

= -64 feet per second

5. Thus, instantaneous velocity at t = 2 seconds is -64 feet per second. Remember that the negative sign indicates the ride is moving downward!

Recall that we approximated this value earlier in this section and actually guessed that the value was -64 feet per second. However, we can be more confident in our answer when we use the limit definition to find the exact value!

Example 13 Ninvento, a toy company, can sell x Swap gaming consoles at a price of p(a:) = -O.Ola:+400 dollars per Swap. The cost of manufacturing a: of these consoles is given by C(x) = lO0x + 10,000 dollars. Find the instantaneous rate of change of profit when 10,000 Swaps are produced and sold, and interpret your answer.

Solution: First, we need the profit function. Recall profit is equal to revenue minus cost. We have the cost function, so we need to find the revenue function. The revenue function, R, is given by the number of items sold, x, times the price of each item, p:

R(a:) =a:·p(:r)

= a:( -0.0lx + 400)

= -0.0h:2 + 400x

Thus, the profit function is

P(x) = R(a:) – C(x)

= (-0.0la:2 + 400:1:) – (100a: + 10,000)

= -0.0Lr2 + 400x – 100:r – 10,000

= -0.0h:2 + 300a: – 10,000

Now that we have the profit function, P(x) = -0.01:r:2 + 300:r:-10, 000 dollars, where :r: is the number of Swap gaming consoles made and sold, we can find the instantaneous rate of change of the function P at a:= 10,000 by letting a= 10,000 in the limit definition:

]. P(lO, 000 + h) – P(lO, 000)

h O h

Again, we will use the Suggested Four-Step Process:

l. Calculate P(lO, 000 + h), and algebraically manipulate the resulting expression, if possible:

P(lO, 000 + h) = -0.01(10, 000 + h)2 + 300(10, 000 + h) – 10,000

Using FOIL and distributing gives

= -0.01 (100,000,000 + 20, 000h + h?) + 3,000,000 + 300h –  10,000

= -1, 000, 000 – 200h – O.Olh2 + 3,000,000 + 300h – 10,000

= lO0h – O.Olh2 + 1,990,000

2. Calculate the numerator of the difference quotient, P(lO, 000 + h) – P(lO, 000), and algebraically manipulate the resulting expression, if possible:

We start with P(lO, 000):

P(lO, 000) = -0.01(10, 000)2 + 300(10, 000) – 10,000 = 1,990,000.

So in the numerator of the difference quotient, we have

P(lO, 000 + h) – P(lO, 000) = lO0h – 0.01h2 + 1,990,000 – 1,990,000

= lO0h – 0.0lh2

3. Calculate the difference quotient (i.e., divide the result from the previous step by h, where h-=/- 0), and  algebraically manipulate the resulting expression:

P(lO, 000 + h) – P(lO, 000)

h

Factoring an h from the numerator gives

lO0h – 0.0lh2

h

Dividing the h’s gives

h(lO0 – 0.01h)

h

,h(lOO – 0.0lh)

Ji

= 100 – 0.0lh

4. Take the limit ash-+ 0 (i.e., substitute O for h):

11.m P(lO, 000 + h) – P(lO, 000) =11.m(l. OO -O.O1h)

h–+0 h h–+0

= 100 – 0.01(0)

= $100 per Swap

Thus, profit is increasing at a rate of $100 per Ninvento Swap. To interpret our answer, we write a sentence:

“When 10,000 Swaps are produced and sold, profit is increasing at a rate of $100 per Swap.”

Notice we did not include the word “average” in our interpretation because we calculated an instantaneous rate of change at one point (when :r = 10,000), not an average rate of change between two points. When we interpret an average rate of change, we should always include the word “average” to distinguish it from an instantaneous rate of change.

Note: After the previous example and Try It, you should see that the instantaneous rate of change of profit at 10,000 units is the instantaneous rate of change of revenue at 10,000 units minus the instantaneous rate of change of cost at 10,000 units.

Example 14 An office super store sells a laptop stylus pen. The price-demand function for the stylus pen is p(:r) =

-0.2:r + 55, where p(:-c) is the  price, in dollars, per stylus pen when x stylus pens are purchased. Find

the instantaneous rate of change of revenue when the super store sells 225 stylus pens, and interpret your answer.

Solution: First, we need to find the revenue function:

R(:r) = :r • p(x)

= x(-0.2x + 5-5)

= -0.2×2 + 55x

Now that we have the revenue function, R(:r) = -0.2:r2 + 55J: dollars, where :i: is the number of stylus pens sold, we can find the instantaneous rate of change of the function R at :r = 225 by letting a = 225 in the limit definition: r  R(225 + h) – R(225)

h h

Again, we will use the Suggested Four-Step Process:

l. Calculate R(225 + h), and algebraically manipulate the resulting expression, if possible:

R(225 + h) = -0.2(225 + h)2 + 55(225 + h)

Using FOIL and distributing gives

= -0.2 (50,625 + 450h + h2) + 12, 375 + 55h

= -10, 125 – 90h – 0.2h2 + 12,375 + 55h

= 2250 – 35h – 0.2h2

2. Calculate the numerator of the difference quotient, R(225 + h) – R(225), and algebraically ma­ nipulate the resulting expression, if possible:

We start with R(225):

R(225) = -0.2(225)2 + 55(225) = 2250

So in the numerator of the difference quotient, we have

R(225 + h) – R(225) = 2250 – 35h – 0.2h2 – 2250

= -35h – 0.2h2

3. Calculate the difference quotient (i.e., divide the result from the previous step by h, where h =/= 0), and algebraically manipulate the resulting expression:

R(225 + h) – R(225)

h

Factoring an h from the numerator gives

-35h – 0.2h2

h

Dividing the h’s gives

h(-35 – 0.2h)

h

/i(-35 – 0.2h)

Ji

= -35 -0.2h

4. Take the limit ash-+ 0 (i.e., substitute 0 for h):

lim  R(225 +hh-) R(225)= lim (-35- 0.2h)

h O h O

= -35 – 0.2(0)

= -$35 per stylus pen

Thus, revenue is decreasing at a rate of $35 per stylus pen. To interpret our answer, we write a sentence:

“When 225 stylus pens are sold, revenue is decreasing at a rate of $35 per stylus pen.”

Finding the Equation of the Tangent Line

Recall that to find the equation of a line, we need two pieces of information: a point the line passes through and the slope of the line. We saw earlier in this section that we can use the limit definition of the instan­ taneous rate of change to find the slope of the line tangent to the graph of f at :r = a. Thus, in order to find the equation of the line tangent to the graph of f at :r = a, the only question we have left to answer is, “Where do we get the y-value of the point?”

We know that the tangent line passes through the graph off  at :r = a. Thus, they-value of the point can be found by calculating f (a). In other words, the tangent line passes through the point (a, f (a)). Once we have the point and the slope of the tangent line, we can find the equation of the tangent line using either the point-slope form or the slope-intercept form of the equation of a line.

Example 15 For each of the following functions, find the equation of the line tangent to the graph of the function at the indicated value of :r, and graph the function and its tangent line on the same axes.

(a) f(x) = :r2 at :r = 2

(b) f(x) = x -6 + 3 at x = 15

(c) f(x) = 3×2  3 at x = 4

Solution:

(a) We start by calculating f (2) to find the y-value of the point:

f(2) = 22 = 4

Thus, the tangent line passes through the point (2, 4).

Now, we use the limit definition of the instantaneous rate of change at x = 2, limh Of(2+h -f(2), to find the slope of the tangent line at :r = 2. Once again, we use the Suggested Four-Step Process:

1. Calculate f (2 + h), and algebraically manipulate the resulting expression, if possible:

f (2 + h) = (2 + h)2

= 4+4h+ h2

2. Calculate the numerator of the difference quotient, f(2 + h) – f(2), and algebraically manipulate the resulting expression, if possible:

We found above that f(2) = 4, so

f(2 + h) – f(2) = 4 + 4h + h2 –  4

= 4h+ h2

3. Calculate the difference quotient (i.e., divide the result from the previous step by h, where h-=/- 0), and algebraically manipulate the resulting expression:

f (2 + h) – f (2)

h

4h+ h2

h

Factoring an h from the numerator gives

Dividing the h’s gives

h(4+h)

h

h(4+h)

h

=4+h

4. Take the limit as h-+ 0 (i.e., substitute O for h):

lim J(2+ h-) f(2)= lim(4+ h)

h-tO h h-tO

=4+0

=4

Now, we know the tangent line passes through the point (2, 4) and has a slope of 4 (i.e., rn = 4). To find the equation of the tangent line, we can either use the point-slope form of the equation of a line or the slope-intercept form of the equation of a line. In this example, we will demonstrate both.

The point-slope form of the equation of a line that passes through the point (x1, y1) and has a slope of m is given by

Substituting the point (2, 4) and slope of the tangent line m = 4 gives

y – 4 = 4(:r – 2)

This is an equation for the tangent line at :r = 2. Oftentimes, we need the equation of the line written in slope-intercept form: y = m:r + b, where m is the slope and b is the y-intercept. We can achieve this by solving the above equation for y:

y – 4 = 4(:r – 2) y=4x-8+4

=4x-4

Thus, the equation of the line tangent to the graph of f at x = 2 is given by y = 4x – 4.

Instead, we could choose to use the slope-intercept form to find the equation of the tangent line from the beginning. Substituting the point (for J: and y) and slope, we can solve for b:

y = mJ: + b ===;-

4 = 4(2) + b

-4  = b

Thus, because the slope is 4 and the y-intercept is -4, the equation of the tangent line is

y=4x-4

Notice that both methods lead to the same tangent line! For the remainder of this section, we will use the point-slope form to find the equation of a line and then solve for y so it is ultimately in slope-intercept form. You should use whichever method you feel more comfortable with.

The graphs of J(x) = :r2 and the line tangent to the graph off at :r = 2 (i.e., y = 4:r – 4) are shown in Figure 2.1.11.

Figure 2.1.11: Graphs of f(:r) = :r2 and y = 4x – 4

(b) Recall that we must find the equation of the line tangent to the graph of f (:r) = x – 6 + 3 at x = 1-5. We will start by evaluating the function, f(x) =:r – 6 + 3, at :r = 15 to find they-value of the point:

f (15)  =15 – 6 + 3

=3+3

=6

Thus, the tangent line passes through the point (15, 6).

Now, we use the limit definition of the instantaneous rate of change at :i: = 15, limh-+Of(l5+h -f(l5), to find the slope of the tangent line. Once again, we use the Suggested Four-Step Process:

1. Calculate J(15 + h), and algebraically manipulate the resulting expression, if possible:

f(l5 + h) = J(15 + h) – 6 + 3

 =9+ h+3

2. Calculate the numerator of the difference quotient, J(l5+h)- J(l5), and algebraically manipulate the resulting expression, if possible:

Again, we already have f (15) = 6:

f (15 + h) –  f (15)  = 9 +h+3 – 6

 = 9 + h-  3

3. Calculate the difference quotient (i.e., divide the result from the previous step by h, where h-=/- 0), and algebraically manipulate the resulting expression:

f (15 + h) –  f (15)  9 + h-3

h h

Multiplying by the conjugate gives

 9+h-3= ( 9+h-3) ( 9+h+3)

h h   9 +h+3

(9+h)-9

h( 9 + h+  3)

h

h( 9 + h+  3)

Dividing the h’s gives

h

h( 9 + h + 3)

1

 9+ h+3

4. Take the limit as h—+ 0 (i.e., substitute O for h):

lim f (15 + h) –  f (15)= lim 1

IHO h h O9 + h + ;3

1

v’§+o+3

1

3+3

1

6

Now, we know that the tangent line passes through the point (15, 6) and has slope m = fi- To find the equation of the line, we will use point-slope form:

Y – YI = m (:r – X1)  ====}

y-6= 1 (x-15)

6

Solving for y to get the equation in a more useful form (i.e., slope-intercept form), we have

-:r + 6

6 2

1 7

=ir+2

Thus, the equation of the line tangent to the graph off  at x = 15 is given by y = ix+ ; .

The graphs of f(:r) =:r – 6+a3nd the line tangent to the graph off at x = 15 (i.e., y = ½:r+ ) are shown in Figure 2.1.12.

Figure 2.1.12: Graphs of f(:r) = 1 – 6 + 3 and

Y =B1 x+ 27

(c) Recall that we must find the equation of the line tangent to the graph of f (x) = 3x 3 at x = 4. We will start by evaluating the function, f(:r) = 3x 3, at x = 4 to find they-value of the point:

!(4) =

2

3(4) – 3

2

9

Thus, the tangent line passes through the point (4, i) .

Now, we use the limit definition of the instantaneous rate of change at x = 4, limh-+O f(4+h -f(4), to find the slope of the tangent line.

As you can probably guess, we will use the Suggested Four-Step Process:

1. Calculate f (4 + h), and algebraically manipulate the resulting expression, if possible:

f(4

2

+h)= 3(4+h)-3

2

12 + 3h – 3

2

9 +3h

2. Calculate the numerator of the difference quotient, f(4 + h) – f(4), and algebraically manipulate the resulting expression, if possible:

f(4 + h) –  f(4) = — 2 – -2

9 + :3h 9

To algebraically manipulate this expression further, we will get a common denominator:

2 2 ( 9) ( 2 ) ( 2) ( 9 + 3h)

9 + 3h- = 9 + 3h – 9 + 3h

18 2(9 + 3h)

9(9 + 3h) 9(9 + 3h)

18 – 2(9 + 3h)

9(9 + 3h)

18 -18  – 6h

9(9 + 3h)

-6h

9(9 + 3h)

3. Calculate the difference quotient (i.e., divide the result from the previous step by h, where h-=/- 0), and  algebraically manipulate the resulting expression:

f(4 + h) –  f(4)

h

-6h

h

-6h

h

T

= (9(9-: h)) (¼)

= (9(9-: h)) (¼)

-6

9(9 + 3h)

Notice when we divided by h in this step, we created a complex fraction because the numerator

contained a fraction. We saw this in a previous example and stated that anytime we have this situation, we will rewrite the division by has multiplication by ¾.

4. Take the limit as h -+ 0 (i.e., substitute 0 for h):

1. f(4+h)-f(4) -6

h-tO h

h-tO 9(9 + 3h)

-6

9(9 + 3(0))

-6

81

2

27

Now, we know that the tangent line passes through the point (4, i) and has slope m = -:}7.

Using the point-slope form to find the equation of the line gives

2 2

y–=–(:r-4)

9 27

Or, in slope-intercept form, we have

y= –:r- -+-

27 27 9

2 14

— –2:7r•+-27

Thus, the equation of the line tangent to the graph off  at :r = 4 is given by y =  – 227:r +  i.

In Figure 2.1.13, the function and the tangent line are graphed on the same set of axes.

Figure 2.1.13: Graphs of f(:r) = 3x 3 and

2 14

y -27X 27

We learned (and used) a lot of new terms in this section! To help summarize the important terminology (thus far), we have included the following for your reference:

Try It Answers

1. (a) trout per day

(b) gallons per second

(C) seconds per gallons

(d) test points per study hour

2. -$800 per plush; When the number of plushes made is between 600 and 1200 plushes, cost is decreasing at an average rate of $800 per plush.

3. (a) -23/120 BPM per second

(b) -1/30 BPMpersecond

4. (a) i. -$1.57 per coffee cup

ii. -$1.58 per coffee cup

iii. -$1.59 per coffee cup

IV. -$1.63 per coffee cup

V. -$1.62 per coffee cup

vi. -$1.61 per coffee cup

(b) -$1.60 per coffee cup; When 530 coffee cups are sold, revenue is decreasing at a rate of $1.60 per coffee cup.

5. (a) -67.75

(b) -73.51

(c) -74.8-501

(d) -75

6. 4

7. 1/6

8. -1/8

9. -32 feet per second

10. Instantaneous Rate of Change of Cost: $100 per Swap; When 10,000 Swaps are produced, cost is increasing at a rate of $100 per Swap.

Instantaneous Rate of Change of Revenue: $200 per Swap; When 10,000 Swaps are sold, revenue is increasing at a rate of $200 per Swap.

11. -$50 per stylus pen; When 225 stylus pens are produced and sold, profit is decreasing at a rate of $50 per stylus pen.

12. y = 8:r – 26

Exercises

Basic Skills Practice

1. Use the table of function values below to answer each of the following.

:J: -8 -6 -4 -2 0 2 4 6 8 10 12

j(J:) -6 -17 -18 16 -17 12 22 3 0 3 26

(a) Find the average rate of change off  on the interval [-8, 12].

(b) Find the average rate of change off  on the interval [-6, 0].

(c) Find the slope of the secant line passing through the points at x = 4 and J: = 10.

2. Use the graph off  below to answer each of the following. Round your answers to two decimal places, if necessary.

(a) Find the slope of the secant line passing through the points at :r = 0 and :r = 2.

(b) Find the average rate of change of g on the interval [-0.5, 0.5].

(c) Find the slope of the secant line passing through the points at x = 1.5 and x = 2.

For Exercises 3 – 5, find the average rate of change of the function on the given interval. Round your answer to three decimal places, if necessary.

4. y = 4x + 16 on the interval [-4, 1]

5. g(x) = -x2 + 3J: – 1 on the interval [5, 7]

6. J(t) = on the interval [-2, 2]

For Exercises 6 – 8, find the instantaneous rate of change of the function at the given J:-value.

6. y = -8x  – 12 at x = 2

7. y = x2 –  4 at x = 4

8. f(x) = -J:2 + J: + 5 at J: = 0

For Exercises 9 and 10, find the slope of the line tangent to the graph of .f at the given :r-value.

9. J(x) = 7J: at x = 6

10. J(x)=x2-2J:+4atx=0

For Exercises 11 – 13, the instantaneous rate of change and the value of f (i:) at a particular x-value are given. Use this information to find the equation of the line tangent to the graph off at the given i:-value.

11. The instantaneous rate of change at :r = 2 is 3, and f(2) = 7.

12. The instantaneous rate of change at x = 14 is -7, and f(14) = 0.

13. The instantaneous rate of change at x = -5 is 0, and f (-5) = 9.

14. Given the units for x and the units for f (:r) in the table below, find the units for the average rate of change off.

15. A small fast-food restaurant specializing in selling chicken sandwiches has a weekly price-demand function given by p(:1:) = 12.5 – 0.0lx, where :1: is the number of chicken sandwiches sold at a price of p( x) dollars each.

(a) Find the average rate of change of price when the number of chicken sandwiches sold each week is between 800 and 900.

(b) Find the instantaneous rate of change of price when 700 chicken sandwiches are sold each week.

16. The restaurant referred to in the previous exercise has a weekly revenue function given by R(x)

12.5.r – 0.0l:r2, where R(.r) is the revenue, in dollars, from selling :r chicken sandwiches.

(a) Find the average rate of change ofrevenue when the number of chicken sandwiches sold each week is between 900 and 1000.

(b) Find the instantaneous rate of change of revenue when 800 chicken sandwiches are sold each week.

17. Brown & Co. sells flux capacitors. They have determined their weekly cost function to be C(;r) 75:l: + -50, 000 dollars, where x is the number of flux capacitors produced each week.

(a) Find the average rate of change of cost when the number of flux capacitors produced each week is between 125 and 175.

(b) Find the instantaneous rate of change of cost when 150 flux capacitors are produced each week.

18. The position of an object moving along a horizontal line after t seconds is given by s(t) = 4t + 2 feet.

(a) Find the average velocity of the object between 2 and 6 seconds.

(b) Find the instantaneous velocity of the object after 8 seconds.

Intermediate Skills Practice

19. When x of Magnus’s hand carved wooden ducks are sold, it costs him C(x) = 0.008×2 + 0.261x + 159.2 dollars to make them.

(a) What is the average rate of change of cost when the first 350 hand carved ducks are made? Interpret your answer.

(b) What is the average rate of change of cost when the number of hand carved ducks made is between 500 and 700? Interpret your answer.

20. The position of an experimental supersonic car driving down a straight track is shown in the table below, where t is measured in seconds since starting the engine and f (t) is measured in feet from the starting position.

ItI o I 10  I2030405060

I J(t) I o I 1130 I 18,113 21,213 39,642 50,935 62,818

(a) Find the average velocity of the car over the minute measured. Round your answer to two decimal places, if necessary.

(b) Find the average velocity of the car from time t = 20 to t = 40 seconds. Round your answer to two decimal places, if necessary.

21. The table below gives function values of g(x) for certain values of x.

I J’. I -4 I -3 I -2 I -1 I o I 1 I 2

I g(J:) I 9 I 3 I o I -1 I o I 1 I 26

(a) Find the slope of the secant line on the following intervals.

I.  [-4,-1] IV. [-1,2]

ii. [-3,-1] V. [-1, 1]

lll.  [-2,-1] VI. [-1,0]

(b) Use part (a) to approximate the slope of the line tangent to the graph of g at J: = -1.

For exercises 22 – 24, (a) find the average rate of change of the function on the given intervals, and (b) use that information to estimate the instantaneous rate of change of the function at the given x-value, if it exists. Round your answers to three decimal places, if necessary.

22. f (:r) = :r3 + 5:r:

(a) i. [1, 2]

ii. [1.5, 2]

iii. [1.9, 2]

(b) :r = 2

23. f(:r) = <,IJ: –  2 + 7

(a) i. [7,8]

ii. [7.5, 8]

iii. [7.9, 8]

(b) :r = 8

24. f(x) = x¥ 4

(a) i. [-0.15, 0]

ii. [-0.1,0]

iii. [-0.05,0]

(b) :r = 0

iv. [2, 3]

v. [2,2.5]

vi. [2, 2.1]

iv. [8, 9]

V.  [8,8.5]

vi. [8,8.1]

iv. [O, 0.15]

V.  [0,0.1]

vi. [0, 0.05]

2-5. The profit for Ugly Mug brand coffee cups when ;r coffee cups are manufactured and sold is P(J:)

dollars. The table below gives some values of P(:r).

X 720 721 722 723 724 725 726

P(x) $815.89 $812.34 $810.60 $810.18 $809.57 $807.94 $806.29

(a) Find the average rate of change of profit on the following intervals.

i. [720, 723]

ii. [721, 723]

iii. [722, 723]

iv. [723, 726]

V. [723, 725]

vi. [723, 724]

(b) Use part (a) to approximate the instantaneous rate of change of profit when 723 coffee cups are sold.

26. The daily cost function for Stinking in the Rain wet active deodorant for rainy days is given by C(:r) = 10ft + 20, where C(x) is the cost, in dollars, of making x deodorant sticks.

(a) Find the average rate of change of cost on the following intervals. Round your answers to three decimal places, if necessary.

1.  [97,100]

ii. [98, 100]

iii. [99, 100]

IV. [100, 103]

v. [100, 102]

vi. [100,101]

(b) Use part (a) to approximate the instantaneous rate of change of cost when 100 sticks of deodorant are made. Round your answer to two decimal places, if necessary.

27. Barnabus Noble’s Bookstore has a weekly revenue function given by R(:r) = 50:r • (0.958?  dollars when

x books are sold.

(a) Find the average rate of change of the weekly revenue on the following intervals.

i. [27, 30]

ii. [28, 30]

iii. [29, 30]

iv. [30, 33]

v. [30, 32]

vi. [30, 31]

(b) Use part (a) to approximate the instantaneous rate of change of the weekly revenue when 30 books are sold.

28. A laboratory experiment measures a runner’s velocity for 15 seconds. The scientists have determined their velocity to be J(t) = 4t – it; meters per second, where t is time in seconds.

(a) Find the average rate of change of the runner’s velocity on the following intervals. Round your answers to three decimal places, if necessary.

i. [5, 6]

ii. [-5.5, 6]

iii. [5.9, 6]

iv. [6, 7]

v. [6, 6.5]

VI.  [6, 6.1]

(b) Use part (a) to approximate the runner’s instantaneous acceleration (i.e., instantaneous rate of change of velocity) after six seconds have passed. Round your answer to one decimal place, if necessary.

For Exercises 29 – 31, compute the difference quotient for the function at the given :r-value using the indicated value of h. Round your answer to four decimal places, if necessary.

29. f(x) = 2x+4 at x = 3 with h = 0.2

30. f(x) = 5 – x2 at x = -2 with h = O.l

31. f(x) = 3 +x at x = l with h = O.l

For Exercises 32 – 34, find the instantaneous rate of change of the function at the given x-value.

32. f(x) = 3;, at x = 2

33. g(;i:) = 5;i5 at X = 1

34. h(x) = at x = 4

For Exercises 35 – 37, find the slope of the line tangent to the graph of the function at the given x-value.

35. g(:1:) = t:}2 at :1: = 10

36. f(x) = 3ft  + 5 at x = 9

37. h(x)  = x +1a3;ti: = 3

For Exercises 38 – 40, find the equation of the line tangent to the graph of the function at the given x-value.

38. g(x) = x2 –  4 at x = 6

39. h(:r) = -:r2 + 3:r – 1 at x = -3

40. J(:1:) = x:3 at:1: = 2

41. The amount of sales for Walter Green’s pharmacy is given by S(t) = 22t + 0.3t2 thousand dollars after

t months. What is the instantaneous rate of change of sales after one year? Interpret your answer.

42. A car’s value depreciates according to the function V(t) = 20,000 – 400t2, where t represents the number of years since the car was purchased and V(t) is its value in dollars. Find the instantaneous rate of change of the car’s value seven years after it is purchased, and interpret your answer.

43. The profit function for a ghost busting business is given byP(:1:) = -:1:2 + 100:1: – 2000 thousand dollars when :1: ghosts are busted. vVhat is the instantaneous rate of change of profit when 40 ghosts are busted? Interpret your answer.

44. The position of a car relative to an observer t seconds after they notice the car is given by f (t)

t2 – 16t + 85 feet. Find the instantaneous velocity of the car 12 seconds after the observer notices the car.

Mastery Practice

4-5. Given g(x) = 3×2-1, find the slope of the secant line passing through the points at x = 4 and ;i: = 4+h.

46. Given the units for x and the units for f(:1:) in the table below, find the units for the instantaneous rate of change of f.

47. The position, in feet, of an object traveling along a horizontal line after t seconds is shown in the table below. Find the average velocity of the object between 3 and 9 seconds. Round your answer to two decimal places, if necessary.

I t I  o 1 3 4 6 9 12

I J(t) I  o 2.5 4.4 5.8 7.4 8.5 11.3

48. The table below gives function values of g(x) for certain values of :r. Approximate the slope of the line tangent to the graph of g at :r = -0.l. Round your answer to one decimal place, if necessary.

:r -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4

g(:r) 0.09 0.04 0.01 0 0.01 0.04 0.09 0.16

49. A company has a revenue function given by R(x) = 20x · (0.997)x dollars when x items are sold. Approximate the instantaneous rate of change of revenue when 100 items are sold, and interpret your answer.

50. The position, in meters, of an object after t seconds is given by s(t) = 2y’t+l – 2. Approximate the object’s velocity after 15 seconds.

For Exercises 51 – 53, compute the difference quotient for the function at the given ;1:-value using the indicated value of h. Round your answer to four decimal places, if necessary.

51. f(:i:) = -J:3 + ;1: at ;1: = 1 with h = 0.1

52. f(x) =x2′:+-1 at x = 3 with h = 0.02

53. f(x) = x +8 -x at x = -l  with h = 0.01

For Exercises 54 – 58, find the instantaneous rate of change of the function at the given value.

54. f(;1:) = -7;1: + 8 at ;1: = 14

55. f(x) = -2×2 + 3x + 12 at x = -l

56. g(;1:) = :r:3 – 2;1:2 + 1 at :r: = -2

57. g(t) = 2vr;=t: at t = 4

58. h(x) =  + at ;1: = -9

For Exercises 59 and 60, find the equation of the line tangent to the graph of the function at the given value.

59. t (t) = A’=-t\ at t = 3

60. g(:r)  =8:r + 9 – 2;1: at :r = 2

61. A company has a revenue function given by R(:1:) = 50:i:e-·002x dollars, where ;1: is the number of items sold. Find and interpret the slope of the secant line of the graph of this function passing through the points at x = 600 and ;1: = 675.

62. Suppose that the profit, in dollars, obtained from the sale of x fish-fry dinners each week is given by P(:r) = -0.03:r2 + 6:r- 50. Find and interpret the slope of the line tangent to the graph of this function at

(a) X = 75.

(b) X = 100.

(C) X = 135.

63. The height of a ball thrown upward is given by s(t) = -18t2 + 150t feet, where t is time in seconds. Find the slope of the line tangent to the graph of this function at t = 15 seconds.

64. The cost of producing :r cameras each week is given by C(;r:) = 600 + 100ft: dollars.

(a) Find and interpret the average rate of change of cost when the number of cameras produced each week is between 20 and 25.

(b) Find and interpret the average rate of change of cost when the first 1-5 cameras are produced each week.

(c) Find and interpret the instantaneous rate of change of cost when 17 cameras are produced each week.

65. The position of a car relative to an observer t seconds after they notice the car is given by f (t) = t2 – 16t + 80 feet. Find the average velocity of the car between 2 and .5 seconds after the observer notices the car.

66. A company’s profit when x gift baskets are made and sold is P(x) dollars. If P(35+ -P(35l = 10 – h, find the instantaneous rate of change of profit when 35 gift baskets are made and sold, and interpret your answer.

67. The equation of the  line tangent to the graph of  f at  the  point (2, 7) is y

. f(2+h)-7

linh-+0 h .

-3:r + 13.  Find

68. (a) Find the average rate of change of f(:r) = 14:r + 81 on each of the following intervals. Round your answers to three decimal places, if necessary.

i. [-2.1, -2] iv. [-2, -1.9]

ii. [-2.01, -2] V.  [-2, -1.99]

lll. [-2.001, -2] Vl.  [-2, -1.999]

(b) Use part (a) to approximate the instantaneous rate of change off, if it exists, at :r = -2.

69. Given f(x) = x2, find (a) the slope of the line tangent to the graph off at :r = a, where a is any real number, and (b) the equation of the line tangent to the graph of f at the point (a, f (a)).

Communication Practice

70. Briefly explain the difference between slopes of secant and tangent lines.

71. Are the terms “average rate of change” and “slope of the tangent line” equivalent? Why or why not?

72. If R(J.:) is the revenue of a company, in dollars, when it sells ;1; items per week, interpret

 R(500); R( 450)= 20, 000

73. If C(:r) is the cost of a company, in dollars, when it makes J.; items per week, interpret

. 0(800 + h) – 0(800)I 650

h-+0 h

74. If the average rate of change of revenue when a company sells between 100 and 125 items each week is

$85/item, does that mean when the company sells 150 items the revenue will increase on average by

$85 if it sells one more item? Explain.