12.5: Derivatives

Learning Objectives

In this section, you will:

• Find the derivative of a function.
• Find instantaneous rates of change.
• Find an equation of the tangent line to the graph of a function at a point.
• Find the instantaneous velocity of a particle.

The average teen in the United States opens a refrigerator door an estimated 25 times per day. Supposedly, this average is up from 10 years ago when the average teenager opened a refrigerator door 20 times per day ².

It is estimated that a television is on in a home 6.75 hours per day, whereas parents spend an estimated 5.5 minutes per day having a meaningful conversation with their children. These averages, too, are not the same as they were 10 years ago, when the television was on an estimated 6 hours per day in the typical household, and parents spent 12 minutes per day in meaningful conversation with their kids.

What do these scenarios have in common? The functions representing them have changed over time. In this section, we will consider methods of computing such changes over time.

Finding the Average Rate of Change of a Function

The functions describing the examples above involve a change over time. Change divided by time is one example of a rate. The rates of change in the previous examples are each different. In other words, some changed faster than others. If we were to graph the functions, we could compare the rates by determining the slopes of the graphs.

A tangent line to a curve is a line that intersects the curve at only a single point but does not cross it there. (The tangent line may intersect the curve at another point away from the point of interest.) If we zoom in on a curve at that point, the curve appears linear, and the slope of the curve at that point is close to the slope of the tangent line at that point.

Figure 1 represents the function f(x)=x3−4x.f(x)=x3−4x. We can see the slope at various points along the curve.

• slope at x=−2x=−2 is 8
• slope at x=−1x=−1 is –1
• slope at x=2x=2 is 8

Figure 1 Graph showing tangents to curve at –2, –1, and 2.

Let’s imagine a point on the curve of function ff at x=ax=a as shown in Figure 2. The coordinates of the point are (a,f(a)).(a,f(a)). Connect this point with a second point on the curve a little to the right of x=a,x=a, with an x-value increased by some small real number h.h. The coordinates of this second point are (a+h,f(a+h))(a+h,f(a+h)) for some positive-value h.h.

Figure 2 Connecting point aa with a point just beyond allows us to measure a slope close to that of a tangent line at x=a.x=a.

We can calculate the slope of the line connecting the two points (a,f(a))(a,f(a)) and (a+h,f(a+h)),(a+h,f(a+h)), called a secant line, by applying the slope formula,

slope = change in ychange in xslope = change in ychange in x

We use the notation msecmsec to represent the slope of the secant line connecting two points.

msec=f(a+h)−f(a)(a+h)−(a) =f(a+h)−f(a)a+h−amsec=f(a+h)−f(a)(a+h)−(a) =f(a+h)−f(a)a+h−a

The slope msecmsec equals the average rate of change between two points (a,f(a))(a,f(a)) and (a+h,f(a+h)).(a+h,f(a+h)).

msec=f(a+h)−f(a)hmsec=f(a+h)−f(a)h

THE AVERAGE RATE OF CHANGE BETWEEN TWO POINTS ON A CURVE

The average rate of change (AROC) between two points (a,f(a))(a,f(a)) and (a+h,f(a+h))(a+h,f(a+h)) on the curve of ff is the slope of the line connecting the two points and is given by

AROC=f(a+h)−f(a)hAROC=f(a+h)−f(a)h

EXAMPLE 1

Finding the Average Rate of Change

Find the average rate of change connecting the points (2,−6)(2,−6) and (−1,5).(−1,5).

Answer

TRY IT #1

Find the average rate of change connecting the points (−5,1.5)(−5,1.5) and (−2.5,9).(−2.5,9).

Understanding the Instantaneous Rate of Change

Now that we can find the average rate of change, suppose we make hh in Figure 2 smaller and smaller. Then a+ha+h will approach aa as hh gets smaller, getting closer and closer to 0. Likewise, the second point (a+h,f(a+h))(a+h,f(a+h)) will approach the first point, (a,f(a)).(a,f(a)). As a consequence, the connecting line between the two points, called the secant line, will get closer and closer to being a tangent to the function at x=a,x=a, and the slope of the secant line will get closer and closer to the slope of the tangent at x=a.x=a. See Figure 3.

Figure 3 The connecting line between two points moves closer to being a tangent line at x=a.x=a.

Because we are looking for the slope of the tangent at x=a,x=a, we can think of the measure of the slope of the curve of a function ff at a given point as the rate of change at a particular instant. We call this slope the instantaneous rate of change, or the derivative of the function at x=a.x=a. Both can be found by finding the limit of the slope of a line connecting the point at x=ax=a with a second point infinitesimally close along the curve. For a function ff both the instantaneous rate of change of the function and the derivative of the function at x=ax=a are written as f'(a),f'(a), and we can define them as a two-sided limit that has the same value whether approached from the left or the right.

f′(a)=limh→0f(a+h)−f(a)hf′(a)=limh→0f(a+h)−f(a)h

The expression by which the limit is found is known as the difference quotient.

DEFINITION OF INSTANTANEOUS RATE OF CHANGE AND DERIVATIVE

The derivative, or instantaneous rate of change, of a function ff at x=a,x=a, is given by

f'(a)=limh→0f(a+h)−f(a)hf'(a)=limh→0f(a+h)−f(a)h

The expression f(a+h)−f(a)hf(a+h)−f(a)h is called the difference quotient.

We use the difference quotient to evaluate the limit of the rate of change of the function as hh approaches 0.

Derivatives: Interpretations and Notation

The derivative of a function can be interpreted in different ways. It can be observed as the behavior of a graph of the function or calculated as a numerical rate of change of the function.

• The derivative of a function f(x)f(x) at a point x=ax=a is the slope of the tangent line to the curve f(x)f(x) at x=a.x=a. The derivative of f(x)f(x) at x=ax=a is written f′(a).f′(a).
• The derivative f′(a)f′(a) measures how the curve changes at the point (a,f(a)).(a,f(a)).
• The derivative f′(a)f′(a) may be thought of as the instantaneous rate of change of the function f(x)f(x) at x=a.x=a.
• If a function measures distance as a function of time, then the derivative measures the instantaneous velocity at time t=a.t=a.

NOTATIONS FOR THE DERIVATIVE

The equation of the derivative of a function f(x)f(x) is written as y′=f′(x),y′=f′(x), where y=f(x).y=f(x). The notation f′(x)f′(x) is read as “ f prime of x.f prime of x. ” Alternate notations for the derivative include the following:

f′(x)=y′=dydx=dfdx=ddxf(x)=Df(x)f′(x)=y′=dydx=dfdx=ddxf(x)=Df(x)

The expression f′(x)f′(x) is now a function of xx; this function gives the slope of the curve y=f(x)y=f(x) at any value of x.x. The derivative of a function f(x)f(x) at a point x=ax=a is denoted f′(a).f′(a).

HOW TO

Given a function f,f, find the derivative by applying the definition of the derivative.

1. Calculate f(a+h).f(a+h).
2. Calculate f(a).f(a).
3. Substitute and simplify f(a+h)−f(a)h.f(a+h)−f(a)h.
4. Evaluate the limit if it exists: f′(a)=limh→0f(a+h)−f(a)h.f′(a)=limh→0f(a+h)−f(a)h.

EXAMPLE 2

Finding the Derivative of a Polynomial Function

Find the derivative of the function f(x)=x2−3x+5f(x)=x2−3x+5 at x=a.x=a.

Answer

TRY IT #2

Find the derivative of the function f(x)=3x2+7xf(x)=3x2+7x at x=a.x=a.

Finding Derivatives of Rational Functions

To find the derivative of a rational function, we will sometimes simplify the expression using algebraic techniques we have already learned.

EXAMPLE 3

Finding the Derivative of a Rational Function

Find the derivative of the function f(x)=3+x2−xf(x)=3+x2−x at x=a.x=a.

Answer

TRY IT #3

Find the derivative of the function f(x)=10x+115x+4f(x)=10x+115x+4 at x=a.x=a.

Finding Derivatives of Functions with Roots

To find derivatives of functions with roots, we use the methods we have learned to find limits of functions with roots, including multiplying by a conjugate.

EXAMPLE 4

Finding the Derivative of a Function with a Root

Find the derivative of the function f(x)=4x−−√f(x)=4x at x=36.x=36.

Answer

TRY IT #4

Find the derivative of the function f(x)=9x−−√f(x)=9x at x=9.x=9.

Finding Instantaneous Rates of Change

Many applications of the derivative involve determining the rate of change at a given instant of a function with the independent variable time—which is why the term instantaneous is used. Consider the height of a ball tossed upward with an initial velocity of 64 feet per second, given by s(t)=−16t2+64t+6,s(t)=−16t2+64t+6, where tt is measured in seconds and s(t)s(t) is measured in feet. We know the path is that of a parabola. The derivative will tell us how the height is changing at any given point in time. The height of the ball is shown in Figure 4 as a function of time. In physics, we call this the “s-t graph.”

Figure 4

EXAMPLE 5

Finding the Instantaneous Rate of Change

Using the function above, s(t)=−16t2+64t+6,s(t)=−16t2+64t+6, what is the instantaneous velocity of the ball at 1 second and 3 seconds into its flight?

Answer

TRY IT #5

The position of the ball is given by s(t)=−16t2+64t+6.s(t)=−16t2+64t+6. What is its velocity 2 seconds into flight?

Using Graphs to Find Instantaneous Rates of Change

We can estimate an instantaneous rate of change at x=ax=a by observing the slope of the curve of the function f(x)f(x) at x=a.x=a. We do this by drawing a line tangent to the function at x=ax=a and finding its slope.

HOW TO

Given a graph of a function f(x),f( x ), find the instantaneous rate of change of the function at x=a.x=a.

1. Locate x=ax=a on the graph of the function f(x).f(x).
2. Draw a tangent line, a line that goes through x=ax=a at aa and at no other point in that section of the curve. Extend the line far enough to calculate its slope as

   change in ychange in x.change in ychange in x.

EXAMPLE 6

Estimating the Derivative at a Point on the Graph of a Function

From the graph of the function y=f(x)y=f(x) presented in Figure 5, estimate each of the following:

1. ⓐ f(0)f(0)
2. ⓑ f(2)f(2)
3. ⓒ f'(0)f'(0)
4. ⓓ f'(2)f'(2)

Figure 5

Answer

TRY IT #6

Using the graph of the function f(x)=x3−3xf(x)=x3−3x shown in Figure 7, estimate: f(1),f(1), f′(1),f′(1), f(0),f(0), and f′(0).f′(0).

Figure 7

Using Instantaneous Rates of Change to Solve Real-World Problems

Another way to interpret an instantaneous rate of change at x=ax=a is to observe the function in a real-world context. The unit for the derivative of a function f(x)f(x) is

output units input unit output units input unit

Such a unit shows by how many units the output changes for each one-unit change of input. The instantaneous rate of change at a given instant shows the same thing: the units of change of output per one-unit change of input.

One example of an instantaneous rate of change is a marginal cost. For example, suppose the production cost for a company to produce xx items is given by C(x),C(x), in thousands of dollars. The derivative function tells us how the cost is changing for any value of xx in the domain of the function. In other words, C′(x)C′(x) is interpreted as a marginal cost, the additional cost in thousands of dollars of producing one more item when xx items have been produced. For example, C′(11)C′(11) is the approximate additional cost in thousands of dollars of producing the 12^th item after 11 items have been produced. C′(11)=2.50C′(11)=2.50 means that when 11 items have been produced, producing the 12^th item would increase the total cost by approximately $2,500.00.

EXAMPLE 7

Finding a Marginal Cost

The cost in dollars of producing xx laptop computers in dollars is f(x)=x2−100x.f(x)=x2−100x. At the point where 200 computers have been produced, what is the approximate cost of producing the 201^st unit?

Answer

EXAMPLE 8

Interpreting a Derivative in Context

A car leaves an intersection. The distance it travels in miles is given by the function f(t),f(t), where tt represents hours. Explain the following notations:

1. ⓐ f(0)=0f(0)=0
2. ⓑ f′(1)=60f′(1)=60
3. ⓒ f(1)=70f(1)=70
4. ⓓ f(2.5)=150f(2.5)=150

Answer

TRY IT #7

A runner runs along a straight east-west road. The function f(t)f(t) gives how many feet eastward of her starting point she is after tt seconds. Interpret each of the following as it relates to the runner.

1. ⓐ f(0)=0f(0)=0
2. ⓑ f(10)=150f(10)=150
3. ⓒ f′(10)=15f′(10)=15
4. ⓓ f′(20)=−10f′(20)=−10
5. ⓔ f(40)=−100f(40)=−100

Finding Points Where a Function’s Derivative Does Not Exist

To understand where a function’s derivative does not exist, we need to recall what normally happens when a function f(x)f(x) has a derivative at x=ax=a. Suppose we use a graphing utility to zoom in on x=ax=a. If the function f(x)f(x) is differentiable, that is, if it is a function that can be differentiated, then the closer one zooms in, the more closely the graph approaches a straight line. This characteristic is called linearity.

Look at the graph in Figure 8. The closer we zoom in on the point, the more linear the curve appears.

Figure 8

We might presume the same thing would happen with any continuous function, but that is not so. The function f(x)=|x|,f(x)=| x |, for example, is continuous at x=0,x=0, but not differentiable at x=0.x=0. As we zoom in close to 0 in Figure 9, the graph does not approach a straight line. No matter how close we zoom in, the graph maintains its sharp corner.

Figure 9 Graph of the function f(x)=|x|,f(x)=| x |, with x-axis from –0.1 to 0.1 and y-axis from –0.1 to 0.1.

We zoom in closer by narrowing the range to produce Figure 10 and continue to observe the same shape. This graph does not appear linear at x=0.x=0.

Figure 10 Graph of the function f(x)=|x|,f(x)=| x |, with x-axis from –0.001 to 0.001 and y-axis from—0.001 to 0.001.

What are the characteristics of a graph that is not differentiable at a point? Here are some examples in which function f(x)f(x) is not differentiable at x=a.x=a.

In Figure 11, we see the graph of

f(x)={x2,8−x,x≤2x>2.f(x)={ x2,x≤28−x,x>2.

Notice that, as xx approaches 2 from the left, the left-hand limit may be observed to be 4, while as xx approaches 2 from the right, the right-hand limit may be observed to be 6. We see that it has a discontinuity at x=2.x=2.

Figure 11 The graph of f(x)f(x) has a discontinuity at x=2.x=2.

In Figure 12, we see the graph of f(x)=|x|.f(x)=| x |. We see that the graph has a corner point at x=0.x=0.

Figure 12 The graph of f(x)=|x|f(x)=| x | has a corner point at x=0x=0.

In Figure 13, we see that the graph of f(x)=x23f(x)=x23 has a cusp at x=0.x=0. A cusp has a unique feature. Moving away from the cusp, both the left-hand and right-hand limits approach either infinity or negative infinity. Notice the tangent lines as xx approaches 0 from both the left and the right appear to get increasingly steeper, but one has a negative slope, the other has a positive slope.

Figure 13 The graph of f(x)=x23f(x)=x23 has a cusp at x=0.x=0.

In Figure 14, we see that the graph of f(x)=x13f(x)=x13 has a vertical tangent at x=0.x=0. Recall that vertical tangents are vertical lines, so where a vertical tangent exists, the slope of the line is undefined. This is why the derivative, which measures the slope, does not exist there.

Figure 14 The graph of f(x)=x13f(x)=x13 has a vertical tangent at x=0.x=0.

DIFFERENTIABILITY

A function f(x)f(x) is differentiable at x=ax=a if the derivative exists at x=a,x=a, which means that f′(a)f′(a) exists.

There are four cases for which a function f(x)f(x) is not differentiable at a point x=a.x=a.

1. When there is a discontinuity at x=a.x=a.
2. When there is a corner point at x=a.x=a.
3. When there is a cusp at x=a.x=a.
4. Any other time when there is a vertical tangent at x=a.x=a.

EXAMPLE 9

Determining Where a Function Is Continuous and Differentiable from a Graph

Using Figure 15, determine where the function is

1. continuous
2. discontinuous
3. differentiable
4. not differentiable

At the points where the graph is discontinuous or not differentiable, state why.

Figure 15

Answer

TRY IT #8

Determine where the function y=f(x)y=f(x) shown in Figure 18 is continuous and differentiable from the graph.

Figure 18

Finding an Equation of a Line Tangent to the Graph of a Function

The equation of a tangent line to a curve of the function f(x)f(x) at x=ax=a is derived from the point-slope form of a line, y=m(x−x1)+y1.y=m(x−x1)+y1. The slope of the line is the slope of the curve at x=ax=a and is therefore equal to f′(a),f′(a), the derivative of f(x)f(x) at x=a.x=a. The coordinate pair of the point on the line at x=ax=a is (a,f(a)).(a,f(a)).

If we substitute into the point-slope form, we have

The equation of the tangent line is

y=f'(a)(x−a)+f(a)y=f'(a)(x−a)+f(a)

THE EQUATION OF A LINE TANGENT TO A CURVE OF THE FUNCTION F

The equation of a line tangent to the curve of a function ff at a point x=ax=a is

y=f'(a)(x−a)+f(a)y=f'(a)(x−a)+f(a)

HOW TO

Given a function f,f, find the equation of a line tangent to the function at x=a.x=a.

1. Find the derivative of f(x)f(x) at x=ax=a using f′(a)=limh→0f(a+h)−f(a)h.f′(a)=limh→0f(a+h)−f(a)h.
2. Evaluate the function at x=a.x=a. This is f(a).f(a).
3. Substitute (a,f(a))(a,f(a)) and f′(a)f′(a) into y=f'(a)(x−a)+f(a).y=f'(a)(x−a)+f(a).
4. Write the equation of the tangent line in the form y=mx+b.y=mx+b.

EXAMPLE 10

Finding the Equation of a Line Tangent to a Function at a Point

Find the equation of a line tangent to the curve f(x)=x2−4xf(x)=x2−4x at x=3.x=3.

Answer

Analysis

We can use a graphing utility to graph the function and the tangent line. In so doing, we can observe the point of tangency at x=3x=3 as shown in Figure 19.

Figure 19 Graph confirms the point of tangency at x=3.x=3.

TRY IT #9

Find the equation of a tangent line to the curve of the function f(x)=5x2−x+4f(x)=5x2−x+4 at x=2.x=2.

Finding the Instantaneous Speed of a Particle

If a function measures position versus time, the derivative measures displacement versus time, or the speed of the object. A change in speed or direction relative to a change in time is known as velocity. The velocity at a given instant is known as instantaneous velocity.

In trying to find the speed or velocity of an object at a given instant, we seem to encounter a contradiction. We normally define speed as the distance traveled divided by the elapsed time. But in an instant, no distance is traveled, and no time elapses. How will we divide zero by zero? The use of a derivative solves this problem. A derivative allows us to say that even while the object’s velocity is constantly changing, it has a certain velocity at a given instant. That means that if the object traveled at that exact velocity for a unit of time, it would travel the specified distance.

INSTANTANEOUS VELOCITY

Let the function s(t)s(t) represent the position of an object at time t.t. The instantaneous velocity or velocity of the object at time t=at=a is given by

s′(a)=limh→0s(a+h)−s(a)hs′(a)=limh→0s(a+h)−s(a)h

EXAMPLE 11

Finding the Instantaneous Velocity

A ball is tossed upward from a height of 200 feet with an initial velocity of 36 ft/sec. If the height of the ball in feet after tt seconds is given by s(t)=−16t2+36t+200,s(t)=−16t2+36t+200, find the instantaneous velocity of the ball at t=2.t=2.

Answer

Analysis

This result means that at time t=2t=2 seconds, the ball is dropping at a rate of 28 ft/sec.

TRY IT #10

A fireworks rocket is shot upward out of a pit 12 ft below the ground at a velocity of 60 ft/sec. Its height in feet after tt seconds is given by s=−16t2+60t−12.s=−16t2+60t−12. What is its instantaneous velocity after 4 seconds?

MEDIA

Access these online resources for additional instruction and practice with derivatives.

• Estimate the Derivative
• Estimate the Derivative Ex. 4

12.4 Section Exercises

Verbal

1.

How is the slope of a linear function similar to the derivative?

2.

What is the difference between the average rate of change of a function on the interval [x,x+h][ x,x+h ] and the derivative of the function at x?x?

3.

A car traveled 110 miles during the time period from 2:00 P.M. to 4:00 P.M. What was the car's average velocity? At exactly 2:30 P.M., the speed of the car registered exactly 62 miles per hour. What is another name for the speed of the car at 2:30 P.M.? Why does this speed differ from the average velocity?

4.

Explain the concept of the slope of a curve at point x.x.

5.

Suppose water is flowing into a tank at an average rate of 45 gallons per minute. Translate this statement into the language of mathematics.

Algebraic

For the following exercises, use the definition of derivative limh→0f(x+h)−f(x)hlimh→0f(x+h)−f(x)h to calculate the derivative of each function.

6.

f(x)=3x−4f(x)=3x−4

7.

f(x)=−2x+1f(x)=−2x+1

8.

f(x)=x2−2x+1f(x)=x2−2x+1

9.

f(x)=2x2+x−3f(x)=2x2+x−3

10.

f(x)=2x2+5f(x)=2x2+5

11.

f(x)=−1x−2f(x)=−1x−2

12.

f(x)=2+x1−xf(x)=2+x1−x

13.

f(x)=5−2x3+2xf(x)=5−2x3+2x

14.

f(x)=1+3x−−−−−√f(x)=1+3x

15.

f(x)=3x3−x2+2x+5f(x)=3x3−x2+2x+5

16.

f(x)=5f(x)=5

17.

f(x)=5πf(x)=5π

For the following exercises, find the average rate of change between the two points.

18.

(−2,0)(−2,0) and (−4,5)(−4,5)

19.

(4,−3)(4,−3) and (−2,−1)(−2,−1)

20.

(0,5)(0,5) and (6,5)(6,5)

21.

(7,−2)(7,−2) and (7,10)(7,10)

For the following polynomial functions, find the derivatives.

22.

f(x)=x3+1f(x)=x3+1

23.

f(x)=−3x2−7x=6f(x)=−3x2−7x=6

24.

f(x)=7x2f(x)=7x2

25.

f(x)=3x3+2x2+x−26f(x)=3x3+2x2+x−26

For the following functions, find the equation of the tangent line to the curve at the given point xx on the curve.

26.

f(x)=2x2−3xx=3f(x)=2x2−3xx=3

27.

f(x)=x3+1x=2f(x)=x3+1x=2

28.

f(x)=x−−√x=9f(x)=xx=9

For the following exercise, find kk such that the given line is tangent to the graph of the function.

29.

f(x)=x2−kx,y=4x−9f(x)=x2−kx,y=4x−9

Graphical

For the following exercises, consider the graph of the function ff and determine where the function is continuous/discontinuous and differentiable/not differentiable.

30.

31.

32.

33.

For the following exercises, use Figure 20 to estimate either the function at a given value of xx or the derivative at a given value of x,x, as indicated.

Figure 20

34.

f(−1)f(−1)

35.

f(0)f(0)

36.

f(1)f(1)

37.

f(2)f(2)

38.

f(3)f(3)

39.

f′(−1)f′(−1)

40.

f′(0)f′(0)

41.

f′(1)f′(1)

42.

f′(2)f′(2)

43.

f′(3)f′(3)

44.

Sketch the function based on the information below:

f′(x)=2xf′(x)=2x, f(2)=4f(2)=4

Technology

45.

Numerically evaluate the derivative. Explore the behavior of the graph of f(x)=x2f(x)=x2 around x=1x=1 by graphing the function on the following domains: [0.9,1.1][ 0.9,1.1 ], [0.99,1.01][ 0.99,1.01 ], [0.999,1.001],[ 0.999,1.001 ], and [0.9999,1.0001][0.9999,1.0001] . We can use the feature on our calculator that automatically sets Ymin and Ymax to the Xmin and Xmax values we preset. (On some of the commonly used graphing calculators, this feature may be called ZOOM FIT or ZOOM AUTO). By examining the corresponding range values for this viewing window, approximate how the curve changes at x=1,x=1, that is, approximate the derivative at x=1.x=1.

Real-World Applications

For the following exercises, explain the notation in words. The volume f(t)f(t) of a tank of gasoline, in gallons, tt minutes after noon.

46.

f(0)=600f(0)=600

47.

f'(30)=−20f'(30)=−20

48.

f(30)=0f(30)=0

49.

f'(200)=30f'(200)=30

50.

f(240)=500f(240)=500

For the following exercises, explain the functions in words. The height, s,s, of a projectile after tt seconds is given by s(t)=−16t2+80t.s(t)=−16t2+80t.

51.

s(2)=96s(2)=96

52.

s'(2)=16s'(2)=16

53.

s(3)=96s(3)=96

54.

s'(3)=−16s'(3)=−16

55.

s(0)=0,s(5)=0.s(0)=0,s(5)=0.

For the following exercises, the volume VV of a sphere with respect to its radius rr is given by V=43πr3.V=43πr3.

56.

Find the average rate of change of VV as rr changes from 1 cm to 2 cm.

57.

Find the instantaneous rate of change of VV when r=3 cm.r=3 cm.

For the following exercises, the revenue generated by selling xx items is given by R(x)=2x2+10x.R(x)=2x2+10x.

58.

Find the average change of the revenue function as xx changes from x=10x=10 to x=20.x=20.

59.

Find R'(10)R'(10) and interpret.

60.

Find R'(15)R'(15) and interpret. Compare R'(15)R'(15) to R'(10),R'(10), and explain the difference.

For the following exercises, the cost of producing xx cellphones is described by the function C(x)=x2−4x+1000.C(x)=x2−4x+1000.

61.

Find the average rate of change in the total cost as xx changes from x=10 to x=15.x=10 to x=15.

62.

Find the approximate marginal cost, when 15 cellphones have been produced, of producing the 16^th cellphone.

63.

Find the approximate marginal cost, when 20 cellphones have been produced, of producing the 21^st cellphone.

Extension

For the following exercises, use the definition for the derivative at a point x=a,x=a, limx→af(x)−f(a)x−a,limx→af(x)−f(a)x−a, to find the derivative of the functions.

64.

f(x)=1x2f(x)=1x2

65.

f(x)=5x2−x+4f(x)=5x2−x+4

66.

f(x)=−x2+4x+7f(x)=−x2+4x+7

67.

f(x)=−43−x2f(x)=−43−x2

Footnotes

• 2http://www.csun.edu/science/health/d...tv&health.html Source provided.

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