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Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 16 pptx

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Applying what you learned in the last section of this chapter to the pocketful of functions you’ve been introduced to the identity, squaring, reciprocal, and absolute value functions, gr

Trang 1

x

f (x) = x2

y

x

y

x

f (x) = x2 –1

y

x

y =

x2

1

y =

x = 1

x = –1

x2 –1

1

–1

–1

Figure 3.13

Answers to Selected Exercises

Answer to Exercise 3.7

h(x) = x ·1x =xx =

 undefined for x = 0

1

x h

Figure 3.14

Your calculator probably won’t indicate the pinhole in the graph

Answer to Exercise 3.8

i

f (3) = 3

3 − 1 + 2(3)

=3

2 + 6 = 7.5

Trang 2

f (y + 1) = y + 1

(y + 1) − 1+ 2(y + 1)

=y + 1y + 2y + 2

= 1 + 1/y + 2y + 2

= 2y + 3 + 1/y iii

f (1/x) = 1/x

(1/x) − 1+ 2(1/x)

=

1 x 1

x −xx

+2 x

=

1 x 1−x x

+x2

= 1x

  x

1 − x

 +2x

= 1

1 − x +

2 x

If you want to add the fractions, get a common denominator

x (1 − x)x +

2(1 − x) x(1 − x)=

x + 2 − 2x (1 − x)x =

2 − x (1 − x)x iv

f (x + h) = x + h

x + h − 1+ 2(x + h) =

x + h

x + h − 1+

(2x + 2h)(x + h − 1)

x + h − 1

=x + h + 2x

2

+ 2xh − 2x + 2hx + 2h2− 2h

x + h − 1

=2x

2+ 4xh − x + 2h2− h

x + h − 1 v

f (2h)

h =

2h 2h−1 + 2(2h) h

=

2h 2h−1 + 4h h

=

2h 2h−1 +4h(2h−1)2h−1 h

=

2h+8h 2

−4h 2h−1

h

Trang 3

 8h2− 2h 2h − 1



 1 h



= (8h − 2)(h) 2h − 1

  1 h



=8h − 2 2h − 1or

2(4h − 1) 2h − 1

Solution to Exercise 3.11

All but the last two pairs of functions give a decomposition

P R O B L E M S F O R S E C T I O N 3 4

1 The zeros of the function f (x) are at x = −4, −1, 2, and 8 What are the zeros of (a) m(x) = 5f (x)?

(b) g(x) = f (x + 2)?

(c) h(x) = f (2x)?

(d) j (x) = f (x − 1)?

Verify your answers analytically

2 The zeros of the function f (x) are at x = −5, −2, 0, and 5 Find the zeros of the following functions If there is not enough information to determine this, say so (a) g(x) = 3|f (x)|

(b) h(x) = w(f (x)), where w(x) = −2x2

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

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

(e) m(x) = 4f (−x)

(f ) n(x) = −f (x)

3 The graph of y = f (x) is symmetric about the y-axis Which of the following functions

is equal to f (x)?

(a) g(x) = −f (x)

(b) h(x) = f (−x)

(c) j (x) = −f (−x)

4 Using what you know about shifting, flipping, and stretching, match the graphs on page

135 with the equations

(a) y =−3x

(b) y =x−31

(c) y =x+11 − 1

(d) y = 2x+11 + 1

(e) y = −2

x+1− 1

Trang 4

y

y = 2

y = –1

y = –1

x y

x

y

x y

x y

x

y

5 Below is the graph of y = 2x

As x → ∞, y → ∞

As x → −∞, y → 0

2 x

1

x y

Using what you know about shifting, flipping, and sliding, match the graphs on the following page with the equations

(a) y = −2x

(b) y = 2−x

(c) y = 2x+ 1 (d) y = 2−x− 1 (e) y = −2−x+ 1 (f ) y = −2x+ 1

Trang 5

y

x

y

x

y

x

y

x

y

x

y

x

y

x

y

x

(iv)

6 Using what you know about shifting, flipping, and stretching, match the graphs below with the equations

(a) y = |x − 2|

(b) y = −3|x + 1|

(c) y = |x + 1| + 2

(d) y = −|x + 1| − 1

(iv) y

y

7 Applying what you learned in the last section of this chapter to the pocketful of functions you’ve been introduced to (the identity, squaring, reciprocal, and absolute value functions), graph the following functions Label any asymptotes and x- and y-intercepts

Trang 6

(a) f (x) = x12

(b) g(x) = |(x − 1)2| (c) h(x) = |x2− 1|

(d) j (x) =x+11 + 2 (e) m(x) =x−2−1 + 1 (f ) p(x) =

1 x

8 Which of the functions in the previous problem are even?

9 f is a function with domain [−3, 4] The graph of f is given below Sketch g(x) =

f (x + 2) What is the domain of g(x)?

1 2

3

(2, 2)

(–3, –2)

(4, –1)

Graph the functions in Problems 10 through 18 by starting with the graph of a famil-iar function and applying appropriate shifts, flips, and stretches Label all x- and

y-intercepts and the coordinates of any vertices and corners Use exact values,

not numerical approximations.

10 (a) y = (x − 1)2 (b) y = −x2− 1

11 (a) y = |x + 2|

(b) y = −|x| + 2

12 (a) y = −(x + 3)2− 1 (b) y = (x − 3)2+ 1

13 (a) y = −2(x + 1)2+ 3 (b) y + 3 = 7(x + 1)2

14 (a) y =x+42 + 1 (b) y =x−π−1

15 (a) y = −x + π (b) y = −(x + π)

Trang 7

16 (a) y − π = (x − 2π)2

(b) y − π = −(x − 2π)2

17 (a) y = x+3x+2(rewrite x + 3 as x + 2 + 1)

(b) y =x+1x−1

18 (a) y = −1 − 2|x + 1|

(b) y = −1 − 2(x + 1)2

For Problems 19 through 21, let f (x) = |x| Graph the functions on the same set of axes.

19 g(x) = (x − 3)2− 4 and f (g(x))

20 g(x) = −(x + 2)2+ 1 and f (g(x))

21 g(x) = |x − 2| − 3 and f (g(x))

22 Let f (x) = 1x and g(x) = x2 Using what you’ve learned in Section 3.4, graph the following equations

(a) y = f (g(x))

(b) y = |g(x − 1) − 4|

(c) y = |f (x)| − 1

23 The graph of y = f (x) is given below

1

–1

f

Trang 8

(a) y = f (x − 3) (b) y = 2f (x) (c) y = −.5f (x) (d) y = f (x/2)

24 A and B are points on the graph of k(x) The x-coordinate of point A is 6 and the

x-coordinate of point B is (6 + h) Write mathematical expressions, using functional notation, for each of the following

(a) The change in value of the function from point A to point B (b) The average rate of change of the function k over the interval [6, 6 + h]

(c) Suppose that the average rate of change of the function k over the interval [6, 6 + h]

is −5 The functions f , g, and h are defined as follows:

f (x) = k(x) + 2, g(x) = k(x + 2), h(x) = 2k(x)

i Which of the following must also be equal to −5?

A The average rate of change of the function f over the interval [6, 6 + h]

B The average rate of change of the function g over the interval [6, 6 + h]

C The average rate of change of the function h over the interval [6, 6 + h]

ii One of the functions f , g, and h has an average value of −10 on the interval [6, 6 + h] Which is it? Explain briefly

Trang 9

II to the Derivative

4

Linearity and Local Linearity

TO LOCAL LINEARITY

Every day we are bombarded with predictions; weather forecasters, sports announcers, fi-nancial consultants, demographers, experts, and self-proclaimed experts alike are constantly letting us know what they think will happen in the future And, based on certain assump-tions, each of us makes his or her own projections and acts accordingly Let’s look at a couple of examples

for herself She likes to have completed her daily afternoon run by sunset, so she has been checking the local newspaper for sunset times She has recorded the following information.1

1 These times (and all that follow in this problem) are given in Greenwich mean time for a north latitude of 30 ◦ Source, 1998

World Almanac.

139

Trang 10

Date Sunset Time

Tuesday, November 4 5:11 p.m

Wednesday, November 5 5:10 p.m

Thursday, November 6 5:09 p.m

She estimates that on Sunday, November 9, the sun will set at 5:06 p.m Upon what assumptions is her projection based? Is it reasonable to use the same assumptions to predict the time of sunset on November 20, 1997? On December 25, 1997? One year later, on November 6, 1998?

SOLUTION Over the past few days the hour of sunset had been getting earlier at a rate of 1 minute/day.2

In other words, the sunset time has been changing at a rate of −1 minute/day Assuming that this rate of change remains constant for the next few days, then three days later the sunset will be three minutes earlier than it was on November 6

Change in sunset time = (rate of change per day) · (days)

= −1minuteday · (3 days) = −3 minutes The predicted sunset time is

5:09 p.m +



−1minute day



· (3 days) = 5:09 p.m − 3 minutes = 5:06 p.m Keep in mind that this prediction is based upon the assumption that the time of sunset is

changing at a constant rate of −1 minute/day throughout the six-day period If this were

correct, then the graph of sunset time plotted versus the date would be a straight line

time

date

Figure 4.1

If we used the same assumptions to predict the hour of sunset on November 20, 1997, we’d get

5:09 p.m +



−1minute day

 (14 days) = 5:09 p.m − 14 minutes = 4:55 p.m

In fact, on Sunday, November 9, sunset was at 5:07 p.m., while on November 20 it was at 5:02 p.m

Common sense tells us that it is unreasonable to assume that the sun will continue to set

a minute earlier every day over a long period of time By December 25 the days are getting longer; sunset is at 5:07 p.m., not at 4:20 p.m., the prediction based on a constant rate of

2 This is the best we can say given the degree of accuracy in the newspaper accounts The newspaper has rounded off to the nearest minute.

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