# 12.5: Derivatives

- Page ID
- 114141

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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** (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**-

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.

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.

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).

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

- Calculate f(a+h).f(a+h).
- Calculate f(a).f(a).
- Substitute and simplify f(a+h)−f(a)h.f(a+h)−f(a)h.
- 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**-

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**-

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**-

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**-

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.

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

- Locate x=ax=a on the graph of the function f(x).f(x).
- 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:

- ⓐ f(0)f(0)
- ⓑ f(2)f(2)
- ⓒ f'(0)f'(0)
- ⓓ f'(2)f'(2)

Figure **5**

**Answer**-

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:

- ⓐ f(0)=0f(0)=0
- ⓑ f′(1)=60f′(1)=60
- ⓒ f(1)=70f(1)=70
- ⓓ f(2.5)=150f(2.5)=150

**Answer**-

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.

- ⓐ f(0)=0f(0)=0
- ⓑ f(10)=150f(10)=150
- ⓒ f′(10)=15f′(10)=15
- ⓓ f′(20)=−10f′(20)=−10
- ⓔ 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.

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.

- When there is a discontinuity at x=a.x=a.
- When there is a corner point at x=a.x=a.
- When there is a cusp at x=a.x=a.
- 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

- continuous
- discontinuous
- differentiable
- not differentiable

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

Figure **15**

**Answer**-

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 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)

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

- 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.
- Evaluate the function at x=a.x=a. This is f(a).f(a).
- 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).
- 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.

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.

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.

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?

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

### 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.

### Footnotes

__2__http://www.csun.edu/science/health/d...tv&health.html Source provided.

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Callstack:
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```