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Derivatives

Module by: First Last. E-mail the author

Summary: 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.

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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 1.

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 )= x 3 4x. f( x )= x 3 4x. We can see the slope at various points along the curve.

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

Figure 1: Graph showing tangents to curve at –2, –1, and 2.
Graph of f(x) = x^3 - 4x with tangent lines at x = -2 with a slope of 8, at x = -3 with a slope of -1, and at x=2 with a slope of 8.

Let’s imagine a point on the curve of function f fat x=a x=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 a awith a point just beyond allows us to measure a slope close to that of a tangent line at x=a. x=a.
Graph of an increasing function that demonstrates the rate of change of the function by drawing a line between the two points, (a, f(a)) and (a, f(a+h)).

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 y change in x slope =  change in y change in x

slope =  change in y change in x slope =  change in y change in x

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

m sec = f(a+h)f(a) (a+h)(a)        = f(a+h)f(a) a +h a m sec = f(a+h)f(a) (a+h)(a)        = f(a+h)f(a) a +h a

The slope m sec m sec 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)).

m sec = f( a+h )f( a ) h m sec = f( a+h )f( a ) h

A General Note: 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 f fis the slope of the line connecting the two points and is given by

AROC= f( a+h )f( a ) h AROC= f( a+h )f( a ) h
(4)

Example 1

Problem 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 ).

Solution

We know the average rate of change connecting two points may be given by

AROC= f( a+h )f( a ) h . AROC= f( a+h )f( a ) h .

If one point is ( 2,6 ), ( 2,6 ), or ( 2,f( 2 ) ), ( 2,f( 2 ) ), then f( 2 )=−6. f( 2 )=−6.

The value h his the displacement from 2 2to 1, 1, which equals 12=−3. 12=−3.

For the other point, f( a+h ) f( a+h ) is the y-coordinate at a+h , a+h , which is 2+(−3) 2+(−3) or −1, −1, so f(a+h)=f(−1)=5. f(a+h)=f(−1)=5.

AROC= f(a+h)f(a) h            = 5(6) 3            = 11 3            = 11 3 AROC= f(a+h)f(a) h            = 5(6) 3            = 11 3            = 11 3

Try It:

Exercise 1

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

Solution

3

Understanding the Instantaneous Rate of Change

Now that we can find the average rate of change, suppose we make h hin Figure 2 smaller and smaller. Then a+h a+h will approach a aas h hgets 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.
Graph of an increasing function that contains a point, P, at (a, f(a)). At the point, there is a tangent line and two secant lines where one secant line is connected to Q1 and another secant line is connected to Q2.

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 f fat 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=a x=a with a second point infinitesimally close along the curve. For a function f fboth the instantaneous rate of change of the function and the derivative of the function at x=a x=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)= lim h0 f( a+h )f( a ) h f (a)= lim h0 f( a+h )f( a ) h

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

A General Note: Definition of Instantaneous Rate of Change and Derivative:

The derivative, or instantaneous rate of change, of a function f fat x=a , x=a , is given by

f'(a)= lim h0 f( a+h )f( a ) h f'(a)= lim h0 f( a+h )f( a ) h
(8)

The expression f( a+h )f( a ) h f( 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 h happroaches 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=a x=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=a x=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.

A General Note: 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 = dy dx = df dx = d dx f(x)=Df(x) f (x)= y = dy dx = df dx = d dx f(x)=Df(x)

The expression f (x) f (x) is now a function of x x; 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=a x=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)= lim h0 f( a+h )f( a ) h . f (a)= lim h0 f( a+h )f( a ) h .

Example 2

Problem 1
Finding the Derivative of a Polynomial Function

Find the derivative of the function f(x)= x 2 3x+5 f(x)= x 2 3x+5 at x=a. x=a.

Solution

We have:

f (a)= lim h0 f( a+h )f( a ) h                   Definition of a derivative  f (a)= lim h0 f( a+h )f( a ) h                   Definition of a derivative 

Substitute f(a+h)= (a+h) 2 3(a+h)+5 f(a+h)= (a+h) 2 3(a+h)+5 and f(a)= a 2 3a+5. f(a)= a 2 3a+5.

f (a)= lim h0 (a+h)(a+h)3(a+h)+5( a 2 3a+5) h         = lim h0 a 2 +2ah+ h 2 3a3h+5 a 2 +3a5 h Evaluate to remove parentheses.         = lim h0 a 2 +2ah+ h 2 3a 3h +5 a 2 +3a 5 h Simplify.         = lim h0 2ah+ h 2 3h h         = lim h0 h (2a+h3) h Factor out an h.         =2a+03 Evaluate the limit.         =2a3 f (a)= lim h0 (a+h)(a+h)3(a+h)+5( a 2 3a+5) h         = lim h0 a 2 +2ah+ h 2 3a3h+5 a 2 +3a5 h Evaluate to remove parentheses.         = lim h0 a 2 +2ah+ h 2 3a 3h +5 a 2 +3a 5 h Simplify.         = lim h0 2ah+ h 2 3h h         = lim h0 h (2a+h3) h Factor out an h.         =2a+03 Evaluate the limit.         =2a3

Try It:

Exercise 2

Find the derivative of the function f(x)=3 x 2 +7x f(x)=3 x 2 +7x at x=a. x=a.

Solution

f (a)=6a+7 f (a)=6a+7

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

Problem 1
Finding the Derivative of a Rational Function

Find the derivative of the function f(x)= 3+x 2x f(x)= 3+x 2x at x=a. x=a.

Solution

f (a)= lim h0 f( a+h )f( a ) h         = lim h0 3+( a+h ) 2( a+h ) ( 3+a 2a ) h Substitute f(a+h) and f(a)        = lim h0 (2(a+h))(2a)[ 3+( a+h ) 2( a+h ) ( 3+a 2a ) ] (2(a+h))(2a)(h) Multiply numerator and denominator by (2(a+h))(2a)        = lim h0 ( 2( a+h ) ) (2a)( 3+( a+h ) ( 2( a+h ) ) )(2(a+h)) ( 2a ) ( 3+a 2a ) ( 2( a+h ) )(2a)(h) Distribute        = lim h0 63a+2a a 2 +2hah6+3a+3h2a+ a 2 +ah ( 2( a+h ) )( 2a )(h) Multiply        = lim h0 5 h ( 2( a+h ) )( 2a )( h ) Combine like terms        = lim h0 5 ( 2( a+h ) )( 2a ) Cancel like factors        = 5 ( 2( a+0 ) )( 2a ) = 5 ( 2a )( 2a ) = 5 ( 2a ) 2 Evaluate the limit f (a)= lim h0 f( a+h )f( a ) h         = lim h0 3+( a+h ) 2( a+h ) ( 3+a 2a ) h Substitute f(a+h) and f(a)        = lim h0 (2(a+h))(2a)[ 3+( a+h ) 2( a+h ) ( 3+a 2a ) ] (2(a+h))(2a)(h) Multiply numerator and denominator by (2(a+h))(2a)        = lim h0 ( 2( a+h ) ) (2a)( 3+( a+h ) ( 2( a+h ) ) )(2(a+h)) ( 2a ) ( 3+a 2a ) ( 2( a+h ) )(2a)(h) Distribute        = lim h0 63a+2a a 2 +2hah6+3a+3h2a+ a 2 +ah ( 2( a+h ) )( 2a )(h) Multiply        = lim h0 5 h ( 2( a+h ) )( 2a )( h ) Combine like terms        = lim h0 5 ( 2( a+h ) )( 2a ) Cancel like factors        = 5 ( 2( a+0 ) )( 2a ) = 5 ( 2a )( 2a ) = 5 ( 2a ) 2 Evaluate the limit

Try It:

Exercise 3

Find the derivative of the function f(x)= 10x+11 5x+4 f(x)= 10x+11 5x+4 at x=a. x=a.

Solution

f (a)= 15 ( 5a+4 ) 2 f (a)= 15 ( 5a+4 ) 2

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

Problem 1
Finding the Derivative of a Function with a Root

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

Solution

We have

f (a)= lim h0 f(a+h)f(a) h         = lim h0 4 a+h 4 a h Substitute f(a+h) and f(a) f (a)= lim h0 f(a+h)f(a) h         = lim h0 4 a+h 4 a h Substitute f(a+h) and f(a)

Multiply the numerator and denominator by the conjugate: 4 a+h +4 a 4 a+h +4 a . 4 a+h +4 a 4 a+h +4 a .

    f (a)= lim h0 ( 4 a+h 4 a h )( 4 a+h +4 a 4 a+h +4 a )            = lim h0 ( 16(a+h)16a h4( a+h +4 a ) ) Multiply.            = lim h0 ( 16a +16h 16a h4( a+h +4 a ) ) Distribute and combine like terms.            = lim h0 ( 16 h h ( 4 a+h +4 a ) ) Simplify.            = lim h0 ( 16 4 a+h +4 a ) Evaluate the limit by letting h=0.            = 16 8 a = 2 a   f (36)= 2 36 Evaluate the derivative at x=36.            = 2 6            = 1 3     f (a)= lim h0 ( 4 a+h 4 a h )( 4 a+h +4 a 4 a+h +4 a )            = lim h0 ( 16(a+h)16a h4( a+h +4 a ) ) Multiply.            = lim h0 ( 16a +16h 16a h4( a+h +4 a ) ) Distribute and combine like terms.            = lim h0 ( 16 h h ( 4 a+h +4 a ) ) Simplify.            = lim h0 ( 16 4 a+h +4 a ) Evaluate the limit by letting h=0.            = 16 8 a = 2 a   f (36)= 2 36 Evaluate the derivative at x=36.            = 2 6            = 1 3

Try It:

Exercise 4

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

Solution

3 2 3 2

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)=−16 t 2 +64t+6, s(t)=−16 t 2 +64t+6, where t tis 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
Graph of a negative parabola with a vertex at (2, 70) and two points at (1, 55) and (3, 55).

Example 5

Problem 1

Finding the Instantaneous Rate of Change

Using the function above, s(t)=−16 t 2 +64t+6, s(t)=−16 t 2 +64t+6, what is the instantaneous velocity of the ball at 1 second and 3 seconds into its flight?

Solution

The velocity at t=1 t=1 and t=3 t=3 is the instantaneous rate of change of distance per time, or velocity. Notice that the initial height is 6 feet. To find the instantaneous velocity, we find the derivative and evaluate it at t=1 t=1 and t=3: t=3:

f (a)= lim h0 f(a+h)f(a) h         = lim h0 16 (t+h) 2 +64(t+h)+6(16 t 2 +64t+6) h Substitute s(t+h) and s(t).         = lim h0 16 t 2 32ht h 2 +64t+64h+6+16 t 2 64t6 h Distribute.         = lim h0 32ht h 2 +64h h Simplify.         = lim h0 h (32th+64) h Factor the numerator.         = lim h0 32th+64 Cancel out the common factor h. s (t)=32t+64 Evaluate the limit by letting h=0. f (a)= lim h0 f(a+h)f(a) h         = lim h0 16 (t+h) 2 +64(t+h)+6(16 t 2 +64t+6) h Substitute s(t+h) and s(t).         = lim h0 16 t 2 32ht h 2 +64t+64h+6+16 t 2 64t6 h Distribute.         = lim h0 32ht h 2 +64h h Simplify.         = lim h0 h (32th+64) h Factor the numerator.         = lim h0 32th+64 Cancel out the common factor h. s (t)=32t+64 Evaluate the limit by letting h=0.

For any value of t t, s ( t ) s ( t ) tells us the velocity at that value of t. t.

Evaluate t=1 t=1 and t=3. t=3.

s (1)=−32(1)+64=32 s (3)=−32(3)+64=−32 s (1)=−32(1)+64=32 s (3)=−32(3)+64=−32

The velocity of the ball after 1 second is 32 feet per second, as it is on the way up.

The velocity of the ball after 3 seconds is −32 −32 feet per second, as it is on the way down.

Try It:

Exercise 5

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

Solution

0

Using Graphs to Find Instantaneous Rates of Change

We can estimate an instantaneous rate of change at x=a x=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=a x=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=a x=a on the graph of the function f( x ). f( x ).
  2. Draw a tangent line, a line that goes through x=a x=a at a aand at no other point in that section of the curve. Extend the line far enough to calculate its slope as
    change in y change in x . change in y change in x .

Example 6

Problem 1
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:

a. f(0) f(0) ; b. f(2) f(2) ; c. f'(0) f'(0) ; d. f'(2) f'(2)

Figure 5
Graph of an odd function with multiplicity of two and with two points at (0, 1) and (2, 1).
Solution

To find the functional value, f( a ), f( a ), find the y-coordinate at x=a. x=a.

To find the derivative at x=a, x=a, f ( a ), f ( a ), draw a tangent line at x=a, x=a, and estimate the slope of that tangent line. See Figure 6.

Figure 6
Graph of the previous function with tangent lines at the two points (0, 1) and (2, 1). The graph demonstrates the slopes of the tangent lines. The slope of the tangent line at x = 0 is 0, and the slope of the tangent line at x = 2 is 4.
  1. f(0) f(0) is the y-coordinate at x=0. x=0. The point has coordinates ( 0,1 ), ( 0,1 ), thus f(0)=1. f(0)=1.
  2. f(2) f(2) is the y-coordinate at x=2. x=2. The point has coordinates ( 2,1 ), ( 2,1 ), thus f(2)=1. f(2)=1.
  3. f (0) f (0) is found by estimating the slope of the tangent line to the curve at x=0. x=0. The tangent line to the curve at x=0 x=0 appears horizontal. Horizontal lines have a slope of 0, thus f (0)=0. f (0)=0.
  4. f (2) f (2) is found by estimating the slope of the tangent line to the curve at x=2. x=2. Observe the path of the tangent line to the curve at x=2. x=2. As the x xvalue moves one unit to the right, the y yvalue moves up four units to another point on the line. Thus, the slope is 4, so f (2)=4. f (2)=4.

Try It:

Exercise 6

Using the graph of the function f(x)= x 3 3x f(x)= x 3 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
Graph of the function f(x) = x^3-3x with a viewing window of [-4. 4] by [-5, 7
Solution

2 , 2 ,0, 0, 3 3

Using Instantaneous Rates of Change to Solve Real-World Problems

Another way to interpret an instantaneous rate of change at x=a x=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 x xitems 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 x xin 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 x xitems have been produced. For example, C ( 11 ) C ( 11 ) is the approximate additional cost in thousands of dollars of producing the 12th item after 11 items have been produced. C ( 11 )=2.50 C ( 11 )=2.50 means that when 11 items have been produced, producing the 12th item would increase the total cost by approximately $2,500.00.

Example 7

Problem 1
Finding a Marginal Cost

The cost in dollars of producing x xlaptop computers in dollars is f( x )= x 2 100x. f( x )= x 2 100x. At the point where 200 computers have been produced, what is the approximate cost of producing the 201st unit?

Solution

If f( x )= x 2 100x f( x )= x 2 100x describes the cost of producing x xcomputers, f ( x ) f ( x ) will describe the marginal cost. We need to find the derivative. For purposes of calculating the derivative, we can use the following functions:

f(a+b)= (x+h) 2 100(x+h)       f(a)= a 2 100a f(a+b)= (x+h) 2 100(x+h)       f(a)= a 2 100a

     f (x)= f(a+h)f(a) h Formula for a derivative              = (x+h) 2 100(x+h)( x 2 100x) h Substitute f(a+h) and f(a).              = x 2 +2xh+ h 2 100x100h x 2 +100x h Multiply polynomials, distribute.              = 2xh+ h 2 100h h Collect like terms.              = h (2x+h100) h Factor and cancel like terms.              =2x+h100 Simplify.              =2x100 Evaluate when h=0.       f (x)=2x100 Formula for marginal cost   f (200)=2(200)100=300 Evaluate for 200 units.      f (x)= f(a+h)f(a) h Formula for a derivative              = (x+h) 2 100(x+h)( x 2 100x) h Substitute f(a+h) and f(a).              = x 2 +2xh+ h 2 100x100h x 2 +100x h Multiply polynomials, distribute.              = 2xh+ h 2 100h h Collect like terms.              = h (2x+h100) h Factor and cancel like terms.              =2x+h100 Simplify.              =2x100 Evaluate when h=0.       f (x)=2x100 Formula for marginal cost   f (200)=2(200)100=300 Evaluate for 200 units.

The marginal cost of producing the 201st unit will be approximately $300.

Example 8

Problem 1
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 t trepresents hours. Explain the following notations:

a. f(0)=0 f(0)=0 ; b. f (1)=60 f (1)=60 ; c. f(1)=70 f(1)=70 ; d. f(2.5)=150 f(2.5)=150

Solution

First we need to evaluate the function f(t) f(t) and the derivative of the function f (t), f (t), and distinguish between the two. When we evaluate the function f(t), f(t), we are finding the distance the car has traveled in t thours. When we evaluate the derivative f (t), f (t), we are finding the speed of the car after t thours.

  1. f(0)=0 f(0)=0 means that in zero hours, the car has traveled zero miles.
  2. f (1)=60 f (1)=60 means that one hour into the trip, the car is traveling 60 miles per hour.
  3. f(1)=70 f(1)=70 means that one hour into the trip, the car has traveled 70 miles. At some point during the first hour, then, the car must have been traveling faster than it was at the 1-hour mark.
  4. f(2.5)=150 f(2.5)=150 means that two hours and thirty minutes into the trip, the car has traveled 150 miles.

Try It:

Exercise 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 t t seconds. Interpret each of the following as it relates to the runner.

a. f( 0 )=0 f( 0 )=0 ; b. f( 10 )=150 f( 10 )=150 ; c. f ( 10 )=15 f ( 10 )=15 ; d. f ( 20 )=10 f ( 20 )=10 ; e. f( 40 )=−100 f( 40 )=−100

Solution
  1. After zero seconds, she has traveled 0 feet.
  2. After 10 seconds, she has traveled 150 feet east.
  3. After 10 seconds, she is moving eastward at a rate of 15 ft/sec.
  4. After 20 seconds, she is moving westward at a rate of 10 ft/sec.
  5. After 40 seconds, she is 100 feet westward of her starting point.

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=a x=a . Suppose we use a graphing utility to zoom in on x=a x=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
Graph of a negative parabola that is zoomed in on a point to show that the curve becomes linear the closer it is zoomed in.

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.
Graph of an absolute function.

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.
Graph of an absolute function.

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

Notice that, as x xapproaches 2 from the left, the left-hand limit may be observed to be 4, while as x xapproaches 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.
Graph of a piecewise function where from negative infinity to (2, 4) is a positive parabola and from (2, 6) to positive infinity is a linear line.

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=0 x=0 .
Graph of an absolute function.

In Figure 13, we see that the graph of f(x)= x 2 3 f(x)= x 2 3 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 x xapproaches 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)= x 2 3 f(x)= x 2 3 has a cusp at x=0. x=0.
Graph of f(x) = x^(2/3) with a viewing window of [-3, 3] by [-2, 3].

In Figure 14, we see that the graph of f(x)= x 1 3 f(x)= x 1 3 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)= x 1 3 f(x)= x 1 3 has a vertical tangent at x=0. x=0.
Graph of f(x) = x^(1/3) with a viewing window of [-3, 3] by [-3, 3].

A General Note: Differentiability:

A function f( x ) f( x ) is differentiable at x=a x=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

Problem 1
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
Graph of a piecewise function that has a removable discontinuity at (-2, -1) and is discontinuous when x =1.
Solution

The graph of f( x ) f( x ) is continuous on ( −∞,−2 )( −2, 1 )( 1, ). ( −∞,−2 )( −2, 1 )( 1, ). The graph of f( x ) f( x ) has a removable discontinuity at x=−2 x=−2 and a jump discontinuity at x=1. x=1. See Figure 16.

Figure 16: Three intervals where the function is continuous
Graph of the previous function that shows the intervals of continuity.

The graph of is differentiable on ( −∞,−2 )( −2,−1 )( −1,1 )( 1,2 )( 2, ). ( −∞,−2 )( −2,−1 )( −1,1 )( 1,2 )( 2, ). The graph of f(x) f(x) is not differentiable at x=−2 x=−2 because it is a point of discontinuity, at x=−1 x=−1 because of a sharp corner, at x=1 x=1 because it is a point of discontinuity, and at x=2 x=2 because of a sharp corner. See Figure 17.

Figure 17: Five intervals where the function is differentiable
Graph of the previous function that not only shows the intervals of continuity but also labels the parts of the graph that has sharp corners and discontinuities. The sharp corners are at (-1, -1) and (2, 3), and the discontinuities are at (-2, -1) and (1, 1).

Try It:

Exercise 8

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

Figure 18
Graph of a piecewise function with three pieces.
Solution

The graph of f fis continuous on ( ,1 )( 1,3 )( 3, ). ( ,1 )( 1,3 )( 3, ). The graph of f fis discontinuous at x=1 x=1 and x=3. x=3. The graph of f fis differentiable on ( ,1 )( 1,3 )( 3, ). ( ,1 )( 1,3 )( 3, ). The graph of f fis not differentiable at x=1 x=1 and x=3. x=3.

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=a x=a is derived from the point-slope form of a line, y=m( x x 1 )+ y 1 . y=m( x x 1 )+ y 1 . The slope of the line is the slope of the curve at x=a x=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=a x=a is (a,f(a)). (a,f(a)).

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

The point-slope formula that demonstrates that m = f(a), x1 = a, and y_1 = f(a).

The equation of the tangent line is

y=f'(a)( xa )+f(a) y=f'(a)( xa )+f(a)

A General Note: 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 f fat a point x=a x=a is

y=f'(a)( xa )+f(a) y=f'(a)( xa )+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=a x=a using f (a)= lim h0 f( a+h )f( a ) h . f (a)= lim h0 f( 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)( xa )+f(a). y=f'(a)( xa )+f(a).
  4. Write the equation of the tangent line in the form y=mx+b. y=mx+b.

Example 10

Problem 1

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)= x 2 4x f(x)= x 2 4x at x=3. x=3.

Solution

Using:

f'(a)= lim h0 f( a+h )f( a ) h f'(a)= lim h0 f( a+h )f( a ) h

Substitute f(a+h)= (a+h) 2 4(a+h) f(a+h)= (a+h) 2 4(a+h) and f(a)= a 2 4a. f(a)= a 2 4a.

  f (a)= lim h0 (a+h)(a+h)4(a+h)( a 2 4a) h          = lim h0 a 2 +2ah+ h 2 4a4h a 2 +4a h Remove parentheses.          = lim h0 a 2 +2ah+ h 2 4a 4h a 2 +4a h Combine like terms.          = lim h0 2ah+ h 2 4h h          = lim h0 h (2a+h4) h Factor out h.          =2a+04 f (a)=2a4 Evaluate the limit.   f (3)=2(3)4=2   f (a)= lim h0 (a+h)(a+h)4(a+h)( a 2 4a) h          = lim h0 a 2 +2ah+ h 2 4a4h a 2 +4a h Remove parentheses.          = lim h0 a 2 +2ah+ h 2 4a 4h a 2 +4a h Combine like terms.          = lim h0 2ah+ h 2 4h h          = lim h0 h (2a+h4) h Factor out h.          =2a+04 f (a)=2a4 Evaluate the limit.   f (3)=2(3)4=2

Equation of tangent line at x=3: x=3:

y=f'(a)(xa)+f(a) y=f'(3)(x3)+f(3) y=2(x3)+(3) y=2x9 y=f'(a)(xa)+f(a) y=f'(3)(x3)+f(3) y=2(x3)+(3) y=2x9
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=3 x=3 as shown in Figure 19.

Figure 19: Graph confirms the point of tangency at x=3. x=3.
Graph of f(x) = x^2-4x with a tangent line at x = 3 which has the equation of y = 2x - 9.

Try It:

Exercise 9

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

Solution

y=19x16 y=19x16

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.

A General Note: 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=a t=a is given by

s (a)= lim h0 s( a+h )s( a ) h s (a)= lim h0 s( a+h )s( a ) h

Example 11

Problem 1

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 t tseconds is given by s(t)=−16 t 2 +36t+200, s(t)=−16 t 2 +36t+200, find the instantaneous velocity of the ball at t=2. t=2.

Solution

First, we must find the derivative s ( t ) s ( t ) . Then we evaluate the derivative at t=2, t=2, using s( a+h )=16 ( a+h ) 2 +36(a+h)+200 s( a+h )=16 ( a+h ) 2 +36(a+h)+200 and s( a )=16 a 2 +36a+200. s( a )=16 a 2 +36a+200.

s (a)= lim h0 s(a+h)s(a) h         = lim h0 16 (a+h) 2 +36(a+h)+200(16 a 2 +36a+200) h         = lim h0 16( a 2 +2ah+ h 2 )+36(a+h)+200(16 a 2 +36a+200) h         = lim h0 16 a 2 32ah16 h 2 +36a+36h+200+16 a 2 36a200 h         = lim h0 16 a 2 32ah16 h 2 +36a +36h +200 +16 a 2 36a 200 h         = lim h0 32ah16 h 2 +36h h         = lim h0 h (32a16h+36) h         = lim h0 (32a16h+36)         =32a160+36   s (a)=32a+36   s (2)=32(2)+36         =28 s (a)= lim h0 s(a+h)s(a) h         = lim h0 16 (a+h) 2 +36(a+h)+200(16 a 2 +36a+200) h         = lim h0 16( a 2 +2ah+ h 2 )+36(a+h)+200(16 a 2 +36a+200) h         = lim h0 16 a 2 32ah16 h 2 +36a+36h+200+16 a 2 36a200 h         = lim h0 16 a 2 32ah16 h 2 +36a +36h +200 +16 a 2 36a 200 h         = lim h0 32ah16 h 2 +36h h         = lim h0 h (32a16h+36) h         = lim h0 (32a16h+36)         =32a160+36   s (a)=32a+36   s (2)=32(2)+36         =28
Analysis

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

Try It:

Exercise 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 t tseconds is given by s=16 t 2 +60t12. s=16 t 2 +60t12. What is its instantaneous velocity after 4 seconds?

Solution

–68 ft/sec, it is dropping back to Earth at a rate of 68 ft/s.

Media:

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Key Equations

average rate of change AROC= f( a+h )f( a ) h AROC= f( a+h )f( a ) h
derivative of a function f (a)= lim h0 f( a+h )f( a ) h f (a)= lim h0 f( a+h )f( a ) h

Key Concepts

  • The slope of the secant line connecting two points is the average rate of change of the function between those points. See Example 1.
  • The derivative, or instantaneous rate of change, is a measure of the slope of the curve of a function at a given point, or the slope of the line tangent to the curve at that point. See Example 2, Example 3, and Example 4.
  • The difference quotient is the quotient in the formula for the instantaneous rate of change:
    f( a+h )f( a ) h f( a+h )f( a ) h
  • Instantaneous rates of change can be used to find solutions to many real-world problems. See Example 5.
  • The instantaneous rate of change can be found by observing the slope of a function at a point on a graph by drawing a line tangent to the function at that point. See Example 6.
  • Instantaneous rates of change can be interpreted to describe real-world situations. See Example 7 and Example 8.
  • Some functions are not differentiable at a point or points. See Example 9.
  • The point-slope form of a line can be used to find the equation of a line tangent to the curve of a function. See Example 10.
  • Velocity is a change in position relative to time. Instantaneous velocity describes the velocity of an object at a given instant. Average velocity describes the velocity maintained over an interval of time.
  • Using the derivative makes it possible to calculate instantaneous velocity even though there is no elapsed time. See Example 11.

Section Exercises

Verbal

Exercise 11

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

Solution

The slope of a linear function stays the same. The derivative of a general function varies according to x. x. Both the slope of a line and the derivative at a point measure the rate of change of the function.

Exercise 12

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?

Exercise 13

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?

Solution

Average velocity is 55 miles per hour. The instantaneous velocity at 2:30 p.m. is 62 miles per hour. The instantaneous velocity measures the velocity of the car at an instant of time whereas the average velocity gives the velocity of the car over an interval.

Exercise 14

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

Exercise 15

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

Solution

The average rate of change of the amount of water in the tank is 45 gallons per minute. If f( x ) f( x ) is the function giving the amount of water in the tank at any time t ,t,then the average rate of change of f( x ) f( x ) between t=a t=a and t=b t=b is f(a)+45(ba). f(a)+45(ba).

Algebraic

For the following exercises, use the definition of derivative lim h0 f(x+h)f(x) h lim h0 f(x+h)f(x) h to calculate the derivative of each function.

Exercise 16

f( x )=3x4 f( x )=3x4

Exercise 17

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

Solution

f (x)=2 f (x)=2

Exercise 18

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

Exercise 19

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

Solution

f (x)=4x+1 f (x)=4x+1

Exercise 20

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

Exercise 21

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

Solution

f (x)= 1 (x2) 2 f (x)= 1 (x2) 2

Exercise 22

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

Exercise 23

f( x )= 52x 3+2x f( x )= 52x 3+2x

Solution

16 ( 3+2x ) 2 16 ( 3+2x ) 2

Exercise 24

f( x )= 1+3x f( x )= 1+3x

Exercise 25

f(x)=3 x 3 x 2 +2x+5 f(x)=3 x 3 x 2 +2x+5

Solution

f (x)=9 x 2 2x+2 f (x)=9 x 2 2x+2

Exercise 26

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

Exercise 27

f(x)=5π f(x)=5π

Solution

f (x)=0 f (x)=0

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

Exercise 28

( −2,0 ) ( −2,0 ) and ( −4,5 ) ( −4,5 )

Exercise 29

( 4,−3 ) ( 4,−3 ) and ( −2,−1 ) ( −2,−1 )

Solution

1 3 1 3

Exercise 30

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

Exercise 31

( 7,−2 ) ( 7,−2 ) and ( 7,10 ) ( 7,10 )

Solution

undefined

For the following polynomial functions, find the derivatives.

Exercise 32

f(x)= x 3 +1 f(x)= x 3 +1

Exercise 33

f(x)=3 x 2 7x=6 f(x)=3 x 2 7x=6

Solution

f (x)=6x7 f (x)=6x7

Exercise 34

f(x)=7 x 2 f(x)=7 x 2

Exercise 35

f(x)=3 x 3 +2 x 2 +x26 f(x)=3 x 3 +2 x 2 +x26

Solution

f (x)=9 x 2 +4x+1 f (x)=9 x 2 +4x+1

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

Exercise 36

f(x)=2 x 2 3x x=3 f(x)=2 x 2 3x x=3

Exercise 37

f(x)= x 3 +1 x=2 f(x)= x 3 +1 x=2

Solution

y=12x15 y=12x15

Exercise 38

f(x)= x x=9 f(x)= x x=9

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

Exercise 39

f(x)= x 2 kx, y=4x9 f(x)= x 2 kx, y=4x9

Solution

k=10 k=10 or k=2 k=2

Graphical

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

Exercise 40


Graph of a piecewise function with three segments. The first segment goes from negative infinity to (-3, -2), an open point; the second segment goes from (-3, 1) to (2, 3), which are both open points; the final segment goes from (2, 2), an open point, to positive infinity.

Exercise 41


Graph of a piecewise function with three segments. The first segment goes from negative infinity to (-2, -1), an open point; the second segment goes from (-2, -4), an open point, to (0, 0), a closed point; the final segment goes from (0, 1), an open point, to positive infinity.

Solution

Discontinuous at x=2 x=2 and x=0. x=0. Not differentiable at –2, 0, 2.

Exercise 42

Graph of a piecewise function with two segments and an asymptote at x = 3. The first segment, which has a removable discontinuity at x = -2, goes from negative infinity to the asymptote, and the final segment goes from the asymptote to positive infinity.

Exercise 43

Graph of a piecewise function with two segments. The first segment goes from (-4, 0), an open point to (5, -2), and the final segment goes from (5, 3), an open point, to positive infinity.
Solution

Discontinuous at x=5. x=5. Not differentiable at -4, –2, 0, 1, 3, 4, 5.

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

Figure 20
Graph of an odd function with multiplicity of 2 with a turning point at (0, -2) and (2, -6).

Exercise 44

f( 1 ) f( 1 )

Exercise 45

f( 0 ) f( 0 )

Solution

f( 0 )=2 f( 0 )=2

Exercise 46

f( 1 ) f( 1 )

Exercise 47

f( 2 ) f( 2 )

Solution

f( 2 )=6 f( 2 )=6

Exercise 48

f(3) f(3)

Exercise 49

f ( 1 ) f ( 1 )

Solution

f ( 1 )=9 f ( 1 )=9

Exercise 50

f ( 0 ) f ( 0 )

Exercise 51

f (1) f (1)

Solution

f ( 1 )=3 f ( 1 )=3

Exercise 52

f ( 2 ) f ( 2 )

Exercise 53

f ( 3 ) f ( 3 )

Solution

f ( 3 )=9 f ( 3 )=9

Exercise 54

Sketch the function based on the information below:

f ( x )=2x f ( x )=2x , f( 2 )=4 f( 2 )=4

Technology

Exercise 55

Numerically evaluate the derivative. Explore the behavior of the graph of f(x)= x 2 f(x)= x 2 around x=1 x=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.

Solution

Answers vary. The slope of the tangent line near x=1 x=1 is 2.

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, t tminutes after noon.

Exercise 56

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

Exercise 57

f'(30)=−20 f'(30)=−20

Solution

At 12:30 p.m., the rate of change of the number of gallons in the tank is –20 gallons per minute. That is, the tank is losing 20 gallons per minute.

Exercise 58

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

Exercise 59

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

Solution

At 200 minutes after noon, the volume of gallons in the tank is changing at the rate of 30 gallons per minute.

Exercise 60

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

For the following exercises, explain the functions in words. The height, s ,s, of a projectile after t tseconds is given by s(t)=16 t 2 +80t. s(t)=16 t 2 +80t.

Exercise 61

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

Solution

The height of the projectile after 2 seconds is 96 feet.

Exercise 62

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

Exercise 63

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

Solution

The height of the projectile at t=3 t=3 seconds is 96 feet.

Exercise 64

s'(3)=−16 s'(3)=−16

Exercise 65

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

Solution

The height of the projectile is zero at t=0 t=0 and again at t=5. t=5. In other words, the projectile starts on the ground and falls to earth again after 5 seconds.

For the following exercises, the volume V Vof a sphere with respect to its radius r ris given by V= 4 3 π r 3 . V= 4 3 π r 3 .

Exercise 66

Find the average rate of change of V Vas r rchanges from 1 cm to 2 cm.

Exercise 67

Find the instantaneous rate of change of V Vwhen r=3 cm. r=3 cm.

Solution

36π 36π

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

Exercise 68

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

Exercise 69

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

Solution

$50.00 per unit, which is the instantaneous rate of change of revenue when exactly 10 units are sold.

Exercise 70

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 x xcellphones is described by the function C(x)= x 2 4x+1000. C(x)= x 2 4x+1000.

Exercise 71

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

Solution

$21 per unit

Exercise 72

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

Exercise 73

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

Solution

$36

Extension

For the following exercises, use the definition for the derivative at a point x=a, x=a, lim xa f(x)f(a) xa , lim xa f(x)f(a) xa , to find the derivative of the functions.

Exercise 74

f(x)= 1 x 2 f(x)= 1 x 2

Exercise 75

f(x)=5 x 2 x+4 f(x)=5 x 2 x+4

Solution

f'(x)=10a1 f'(x)=10a1

Exercise 76

f(x)= x 2 +4x+7 f(x)= x 2 +4x+7

Exercise 77

f(x)= 4 3 x 2 f(x)= 4 3 x 2

Solution

4 ( 3x ) 2 4 ( 3x ) 2

Chapter Review Exercises

Finding Limits: A Numerical and Graphical Approach

For the following exercises, use Figure 21.

Figure 21
Graph of a piecewise function with two segments. The first segment goes from (-1, 2), a closed point, to (3, -6), a closed point, and the second segment goes from (3, 5), an open point, to (7, 9), a closed point.

Exercise 78

lim x −1 + f(x) lim x −1 + f(x)

Solution

2

Exercise 79

lim x −1 f(x) lim x −1 f(x)

Exercise 80

lim x1 f(x) lim x1 f(x)

Solution

does not exist

Exercise 81

lim x3 f(x) lim x3 f(x)

Exercise 82

At what values of x x is the function discontinuous? What condition of continuity is violated?

Solution

Discontinuous at x=1( lim xa f(x) does not exist),x=3 (jump discontinuity), and x=7 ( lim xa f(x) does not exist). Discontinuous at x=1( lim xa f(x) does not exist),x=3 (jump discontinuity), and x=7 ( lim xa f(x) does not exist).

Exercise 83

Using Table 2, estimate lim x0 f(x). lim x0 f(x).

Table 2
xxF(x) F(x)
−0.12.875
−0.012.92
−0.0012.998
0Undefined
0.0012.9987
0.012.865
0.12.78145
0.152.678
Solution

3

For the following exercises, with the use of a graphing utility, use numerical or graphical evidence to determine the left- and right-hand limits of the function given as x xapproaches a.  a.  If the function has limit as x xapproaches a, a, state it. If not, discuss why there is no limit.

Exercise 84

f(x)={ | x |1, if x1 x 3 , if x=1   a=1 f(x)={ | x |1, if x1 x 3 , if x=1   a=1

Exercise 85

f(x)={ 1 x+1 , if x=2 (x+1) 2 , if x2   a=2 f(x)={ 1 x+1 , if x=2 (x+1) 2 , if x2   a=2

Solution

lim x2 f(x)=1 lim x2 f(x)=1

Exercise 86

f(x)={ x+3 , if x<1 x 3 , if x>1   a=1 f(x)={ x+3 , if x<1 x 3 , if x>1   a=1

Finding Limits: Properties of Limits

For the following exercises, find the limits if lim xc f( x )=−3 lim xc f( x )=−3 and lim xc g( x )=5. lim xc g( x )=5.

Exercise 87

lim xc ( f(x)+g(x) ) lim xc ( f(x)+g(x) )

Solution

2

Exercise 88

lim xc f(x) g(x) lim xc f(x) g(x)

Exercise 89

lim xc ( f(x)g(x) ) lim xc ( f(x)g(x) )

Solution

−15 −15

Exercise 90

lim x 0 + f(x),f(x)={ 3 x 2 +2x+1 5x+3    x>0 x<0 lim x 0 + f(x),f(x)={ 3 x 2 +2x+1 5x+3    x>0 x<0

Exercise 91

lim x 0 f(x),f(x)={ 3 x 2 +2x+1 5x+3    x>0 x<0 lim x 0 f(x),f(x)={ 3 x 2 +2x+1 5x+3    x>0 x<0

Solution

3

Exercise 92

lim x 3 + ( 3x〚x〛 ) lim x 3 + ( 3x〚x〛 )

For the following exercises, evaluate the limits using algebraic techniques.

Exercise 93

lim h0 ( ( h+6 ) 2 36 h ) lim h0 ( ( h+6 ) 2 36 h )

Solution

12

Exercise 94

lim x25 ( x 2 625 x 5 ) lim x25 ( x 2 625 x 5 )

Exercise 95

lim x1 ( x 2 9x x ) lim x1 ( x 2 9x x )

Solution

10 10

Exercise 96

lim x4 7 12x+1 x4 lim x4 7 12x+1 x4

Exercise 97

lim x3 ( 1 3 + 1 x 3+x ) lim x3 ( 1 3 + 1 x 3+x )

Solution

1 9 1 9

Continuity

For the following exercises, use numerical evidence to determine whether the limit exists at x=a. x=a. If not, describe the behavior of the graph of the function at x=a. x=a.

Exercise 98

f(x)= 2 x4 ; a=4 f(x)= 2 x4 ; a=4

Exercise 99

f(x)= 2 ( x4 ) 2 ; a=4 f(x)= 2 ( x4 ) 2 ; a=4

Solution

At x=4, x=4, the function has a vertical asymptote.

Exercise 100

f(x)= x x 2 x6 ; a=3 f(x)= x x 2 x6 ; a=3

Exercise 101

f(x)= 6 x 2 +23x+20 4 x 2 25 ; a= 5 2 f(x)= 6 x 2 +23x+20 4 x 2 25 ; a= 5 2

Solution

removable discontinuity at a= 5 2 a= 5 2

Exercise 102

f(x)= x 3 9x ; a=9 f(x)= x 3 9x ; a=9

For the following exercises, determine where the given function f(x) f(x) is continuous. Where it is not continuous, state which conditions fail, and classify any discontinuities.

Exercise 103

f(x)= x 2 2x15 f(x)= x 2 2x15

Solution

continuous on (,) (,)

Exercise 104

f(x)= x 2 2x15 x5 f(x)= x 2 2x15 x5

Exercise 105

f(x)= x 2 2x x 2 4x+4 f(x)= x 2 2x x 2 4x+4

Solution

removable discontinuity at x=2. x=2. f(2) f(2) is not defined, but limits exist.

Exercise 106

f(x)= x 3 125 2 x 2 12x+10 f(x)= x 3 125 2 x 2 12x+10

Exercise 107

f(x)= x 2 1 x 2x f(x)= x 2 1 x 2x

Solution

discontinuity at x=0 x=0 and x=2. x=2. Both f(0) f(0) and f(2) f(2) are not defined.

Exercise 108

f(x)= x+2 x 2 3x10 f(x)= x+2 x 2 3x10

Exercise 109

f(x)= x+2 x 3 +8 f(x)= x+2 x 3 +8

Solution

removable discontinuity at x=2. f(2) x=2. f(2) is not defined.

Derivatives

For the following exercises, find the average rate of change f(x+h)f(x) h . f(x+h)f(x) h .

Exercise 110

f(x)=3x+2 f(x)=3x+2

Exercise 111

Exercise 112

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

Exercise 113

f(x)=ln(x) f(x)=ln(x)

Solution

ln(x+h)ln(x) h ln(x+h)ln(x) h

Exercise 114

f(x)= e 2x f(x)= e 2x

For the following exercises, find the derivative of the function.

Exercise 115

f(x)=4x6 f(x)=4x6

Solution

=4 =4

Exercise 116

f(x)=5 x 2 3x f(x)=5 x 2 3x

Exercise 117

Find the equation of the tangent line to the graph of f( x ) f( x ) at the indicated x xvalue.

f(x)= x 3 +4x f(x)= x 3 +4x ; x=2. x=2.

Solution

y=8x+16 y=8x+16

For the following exercises, with the aid of a graphing utility, explain why the function is not differentiable everywhere on its domain. Specify the points where the function is not differentiable.

Exercise 118

f(x)= x | x | f(x)= x | x |

Exercise 119

Given that the volume of a right circular cone is V= 1 3 π r 2 h V= 1 3 π r 2 h and that a given cone has a fixed height of 9 cm and variable radius length, find the instantaneous rate of change of volume with respect to radius length when the radius is 2 cm. Give an exact answer in terms of π π

Solution

12π 12π

Practice Test

For the following exercises, use the graph of f fin Figure 22.

Figure 22
Graph of a piecewise function with two segments. The first segment goes from negative infinity to (-1, 0), an open point, and the second segment goes from (-1, 3), an open point, to positive infinity.

Exercise 120

Exercise 121

lim x −1 + f(x) lim x −1 + f(x)

Exercise 122

lim x −1 f(x) lim x −1 f(x)

Solution

0

Exercise 123

lim x−1 f(x) lim x−1 f(x)

Exercise 124

lim x−2 f(x) lim x−2 f(x)

Solution

−1 −1

Exercise 125

At what values of x xis f fdiscontinuous? What property of continuity is violated?

For the following exercises, with the use of a graphing utility, use numerical or graphical evidence to determine the left- and right-hand limits of the function given as x xapproaches a. a. If the function has a limit as x xapproaches a, a, state it. If not, discuss why there is no limit

Exercise 126

f(x)={ 1 x 3, if x2 x 3 +1,if x>2   a=2 f(x)={ 1 x 3, if x2 x 3 +1,if x>2   a=2

Solution

lim x 2 f(x)= 5 2 a lim x 2 f(x)= 5 2 a and lim x 2 + f(x)=9 lim x 2 + f(x)=9 Thus, the limit of the function as x xapproaches 2 does not exist.

f(x)={ x 3 +1, if x<1 3 x 2 1, if x=1 x+3 +4, if x>1   a=1 f(x)={ x 3 +1, if x<1 3 x 2 1, if x=1 x+3 +4, if x>1   a=1

For the following exercises, evaluate each limit using algebraic techniques.

Exercise 127

lim x−5 ( 1 5 + 1 x 10+2x ) lim x−5 ( 1 5 + 1 x 10+2x )

Solution

1 50 1 50

Exercise 128

lim h0 ( h 2 +25 5 h 2 ) lim h0 ( h 2 +25 5 h 2 )

Exercise 129

lim h0 ( 1 h 1 h 2 +h ) lim h0 ( 1 h 1 h 2 +h )

Solution

1

For the following exercises, determine whether or not the given function f fis continuous. If it is continuous, show why. If it is not continuous, state which conditions fail.

Exercise 130

f(x)= x 2 4 f(x)= x 2 4

Exercise 131

f(x)= x 3 4 x 2 9x+36 x 3 3 x 2 +2x6 f(x)= x 3 4 x 2 9x+36 x 3 3 x 2 +2x6

Solution

removable discontinuity at x=3 x=3

For the following exercises, use the definition of a derivative to find the derivative of the given function at x=a. x=a.

Exercise 132

f(x)= 3 5+2x f(x)= 3 5+2x

Exercise 133

f(x)= 3 x f(x)= 3 x

Solution

f'(x)= 3 2 a 3 2 f'(x)= 3 2 a 3 2

Exercise 134

f(x)=2 x 2 +9x f(x)=2 x 2 +9x

Exercise 135

For the graph in Figure 23, determine where the function is continuous/discontinuous and differentiable/not differentiable.

Figure 23
Graph of a piecewise function with three segments. The first segment goes from negative infinity to (-2, -1), an open point; the second segment goes from (-2, -4), an open point, to (0, 0), a closed point; the final segment goes from (0, 1), an open point, to positive infinity.

Solution

discontinuous at –2,0, not differentiable at –2,0, 2.

For the following exercises, with the aid of a graphing utility, explain why the function is not differentiable everywhere on its domain. Specify the points where the function is not differentiable.

Exercise 136

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

Exercise 137

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

Solution

not differentiable at x=0 x=0 (no limit)

For the following exercises, explain the notation in words when the height of a projectile in feet, s, s, is a function of time t tin seconds after launch and is given by the function s(t). s(t).

Exercise 138

s(0) s(0)

Exercise 139

s(2) s(2)

Solution

the height of the projectile at t=2 t=2 seconds

Exercise 140

s'(2) s'(2)

Exercise 141

s(2)s(1) 21 s(2)s(1) 21

Solution

the average velocity from t=1 to t=2 t=1 to t=2

Exercise 142

s(t)=0 s(t)=0

For the following exercises, use technology to evaluate the limit.

Exercise 143

lim x0 sin(x) 3x lim x0 sin(x) 3x

Solution

1 3 1 3

Exercise 144

lim x0 tan 2 (x) 2x lim x0 tan 2 (x) 2x

Exercise 145

lim x0 sin(x)(1cos(x)) 2 x 2 lim x0 sin(x)(1cos(x)) 2 x 2

Solution

0

Exercise 146

Evaluate the limit by hand.

lim x1 f(x), where  f(x)={ 4x7 x1 x 2 4 x=1 lim x1 f(x), where  f(x)={ 4x7 x1 x 2 4 x=1

At what value(s) of x xis the function below discontinuous?

f(x)={ 4x7x1 x 2 4x=1 f(x)={ 4x7x1 x 2 4x=1

For the following exercises, consider the function whose graph appears in Figure 24.

Figure 24
Graph of a positive parabola.

Exercise 147

Find the average rate of change of the function from x=1 to x=3. x=1 to x=3.

Solution

2

Exercise 148

Find all values of x xat which f'(x)=0. f'(x)=0.

Solution

x=1 x=1

Exercise 149

Find all values of x xat which f'(x) f'(x) does not exist.

Exercise 150

Find an equation of the tangent line to the graph of f fthe indicated point: f(x)=3 x 2 2x6,  x=2 f(x)=3 x 2 2x6,  x=2

Solution

y=14x18 y=14x18

For the following exercises, use the function f(x)=x ( 1x ) 2 5 f(x)=x ( 1x ) 2 5 .

Exercise 151

Graph the function f(x)=x ( 1x ) 2 5 f(x)=x ( 1x ) 2 5 by entering f(x)=x ( ( 1x ) 2 ) 1 5 f(x)=x ( ( 1x ) 2 ) 1 5 and then by entering f(x)=x ( ( 1x ) 1 5 ) 2 f(x)=x ( ( 1x ) 1 5 ) 2 .

Exercise 152

Explore the behavior of the graph of f(x) f(x) around x=1 x=1 by graphing the function on the following domains, [0.9, 1.1], [0.99, 1.01], [0.999, 1.001], and [0.9999, 1.0001]. Use this information to determine whether the function appears to be differentiable at x=1. x=1.

Solution

The graph is not differentiable at x=1 x=1 (cusp).

For the following exercises, find the derivative of each of the functions using the definition: lim h0 f(x+h)f(x) h lim h0 f(x+h)f(x) h

Exercise 153

f(x)=2x8 f(x)=2x8

Exercise 154

f(x)=4 x 2 7 f(x)=4 x 2 7

Solution

f ' (x)=8x f ' (x)=8x

Exercise 155

f(x)=x 1 2 x 2 f(x)=x 1 2 x 2

Exercise 156

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

Solution

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

Exercise 157

f(x)= 3 x1 f(x)= 3 x1

Exercise 158

f(x)= x 3 +1 f(x)= x 3 +1

Solution

f ' (x)=3 x 2 f ' (x)=3 x 2

Exercise 159

f(x)= x 2 + x 3 f(x)= x 2 + x 3

Exercise 160

f(x)= x1 f(x)= x1

Solution

f'(x)= 1 2 x1 f'(x)= 1 2 x1

Footnotes

  1. http://www.csun.edu/science/health/docs/tv&health.html Source provided.

Glossary

average rate of change:
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)) on the curve of f( x ); f( x ); it is given by AROC= f( a+h )f( a ) h . AROC= f( a+h )f( a ) h .
derivative:
the slope of a function at a given point; denoted f (a), f (a), at a point x=a x=a it is f (a)= lim h0 f( a+h )f( a ) h , f (a)= lim h0 f( a+h )f( a ) h , providing the limit exists.
differentiable:
a function f( x ) f( x ) for which the derivative exists at x=a. x=a. In other words, if f ( a ) f ( a ) exists.
instantaneous rate of change:
the slope of a function at a given point; at x=a x=a it is given by f (a)= lim h0 f( a+h )f( a ) h . f (a)= lim h0 f( a+h )f( a ) h .
instantaneous velocity:
the change in speed or direction at a given instant; a function s( t ) s( t ) represents the position of an object at time t ,t,and the instantaneous velocity or velocity of the object at time t=a t=a is given by s (a)= lim h0 s( a+h )s( a ) h . s (a)= lim h0 s( a+h )s( a ) h .
secant line:
a line that intersects two points on a curve
tangent line:
a line that intersects a curve at a single point

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