chapter 1 functions and mathematical models - wikispacesman+ch+01.pdf · 20. a. b. x-dilation by 2,...

Problem Set 1-11. a. 20 m; D17.5 m; it is below the top of the cliff.

b. s; s; s

c. 5 m

d. There is only one altitude for any given time; somealtitudes correspond to more than one time.

e. Domain: range:

2. a.

This graph also shows the answer for part b below.

b. Answers will vary. and when T M D273. Absolute zero is C.

c. Extrapolation: V (30) and T such that ;interpolation: V (400).

d. There is only one volume for a given temperature; yes,because there is only one temperature for a given volume.

e. Domain: or whatever number is found in part b;range: .

3. a.

n B

0 150,000

12 145,995

24 141,744

36 137,230

48 132,438

60 127,350

72 121,948

84 116,213

96 110,125

108 103,661

120 96,798

b. Changing to 1, we see that the balance becomesnegative at the end of month 241, so the balance willbecome 0 during month 241. In reality, the balance wouldbe paid off at the end of month 241, but with a smallerpayment, $3.04 rather than $1074.64. (After studyinglogarithms in Chapter 7, students will also be able to solvethis equation algebraically.)

Tbl

y 0x D273,

V (T ) = 0

D273V (T ) = 0V (400) M 23, v (30) M 11,

300 425

10

5

20

35V (liters)

T (C)

D30 y 25.0 x M 5.3;

M 5.3M 3.8M 0.3

n B

235 5296.5

236 4248.3

237 3194.9

238 2136.3

239 1072.3

240 3.0438

241 D1072c.

d. False

e. Domain: x an integer; range: The values are calculated only at whole-month intervals.(The range values also jump from one to the next, but ingeneral they are not integers.)

4. a.

b. if you stay within the speed limit.

c. According to the Texas Drivers Handbook, the distancewould be about 240 ft.

d. Police consider the length of the skid marks theindependent variable.

Speed

Distance

0 x 65

Speed

Distance

0 y 150,000.0 x 241,

100 200

100,000

x

y

100 200

100,000

n

B

Precalculus with Trigonometry: Solutions Manual Problem Set 1-1 1 2007 Key Curriculum Press

Chapter 1 Functions and Mathematical Models

5. This graph assumes that the element heats from a roomtemperature of 72F to nearly a maximum temperature of350F in one minute.

Domain: s; range: .

6. a. 1: graphically (and verbally); 2: numerically; 3: algebraically; 4: verbally; 5: verbally

b. 1: graphical to numerical; 2: numerical to graphical, then graphical to numerical for the extrapolation andinterpolation; 3: algebraic to numerical and algebraic tographical; 4: verbal to graphical; 5: verbal to graphical

Problem Set 1-21. a.

b.

c. Linear

d. Answers will vary; e.g., the cost (in thousands of dollars)of manufacturing x items if each item costs $2000 tomanufacture and there is a $3000 start-up.

2. a.

b.

c. Power

d. Answers will vary; e.g., the weight of an animal based onone of its linear dimensions.

3. a.

b.

c. Inverse variation

d. Answers will vary; e.g., the time it takes to go 12 mi at x mi/h.

g(x) 1.2

4 8

20

40

y

x

0 f (x) 12.8

2 4

10

y

x

3 f (x) 23

4 8

10

20

x

y

72F y < 350Fx 0

5 10

200

y

x

4. a.

b.

c. Exponential

d. Answers will vary; e.g., the number of bacteria (in millions)left after x days if 5 days ago there were approximately64.3 million and the death rate from a drug treatment is40% per day.

5. a.

b. y-intercept at the domain-restricted function hasno x-intercepts (the unrestricted function has intercepts at

and ); no asymptotes

c.

6. a.

b. y-intercept at no x-intercepts; no asymptotes

c.

7. a.

b. y-intercept at x-intercepts at andno asymptotes

c.

8. a.

40

y

x

3 3

D20.7453 y 40x = 6;

x = 2,x = D1,y = 12;

4

y

x

(4.36, 20.74)

20

20

40 y 76

y = 40;

4

20

40 (3, 31)

y

x

7 y 16

x = 6x = D2

y = 12;

4

16

y

x

(2, 16)

0.3888 x M 64.3

20

40

y

x

4 4

2 Problem Set 1-2 Precalculus with Trigonometry: Solutions Manual 2007 Key Curriculum Press

b. y-intercept at the domain-restricted function has x-intercepts at and (the unrestrictedfunction has an additional intercept at ); noasymptotes

c.

9. a.

b. y-intercept at x-intercept at no asymptotes

c.

10. a.

b. y-intercept at both the domain-restricted functionand the unrestricted function have an x-intercept at no asymptotes

c.

11. a.


c.

12. a.


c.

13. a.

4

8

y

x

4 4

D9 y 21x = D2;y = 6;

10

20y

x

4 4

D3 y 6.1x = 557;y = 4;

4 8

3

y

x

0 y 8.1

x = 0;y = 0;

4 8

4

8

y

x

0 y 12

x = 0;y = 0;

4 8

4

8

12

y

x

D20 y 70

x = D4x = 2x = 1,x = D2,

y = 16; b. y-intercept at no x-intercepts; asymptote (the x-axis)

c.

14. a.

b. y-intercept at no x-intercepts; asymptote (the x-axis)

c.

15. a.

b. No y-intercept; no x-intercept; asymptote (the y-axis)

c.

16. a.


c.

17. a.

b. y-intercept at x-intercept at asymptotes and

c. Range: all real numbers

18. a.

2 4 6

y

x

8

4

4

x = 4x = D1x = 2;y = 12;

2 4

y

x

4

4

y 0

x = 0;y = 0;

4 8

20

40

y

x

y > 0

x = 0

4

10

20

y

x

(3.3614) y 118.9980

y = 0y = 20;

50

100

y

x

4 4

0.8079 y 11.1387

y = 0y = 3;


b. y-intercept at x-intercept at or asymptote

c. Range: all real numbers

19. Exponential

20. Linear

21. Linear

22. Exponential

23. Quadratic (polynomial)

24. Cubic (polynomial)

25. Power

26. Inverse variation

27. Rational

28. Direct variation

29. a. 30. a.

b. Power (cubic) b. Exponential

31. a. 32. a.

b. Linear b. Quadratic

33. Function; no x-value has more than one corresponding y-value.

34. Not a function; some x-values on the left have twocorresponding y-values.

35. Not a function; there is at least one x-value with more thanone corresponding y-value.

36. Function; no x-value has more than one corresponding y-value.

37. Not a function; there is at least one x-value with more thanone corresponding y-value.

38. Not a function; the x-value in the middle has infinitely manycorresponding y-values.

39. a. A vertical line through a given x-value crosses the graph at the y-values that correspond to that x-value. So, if avertical line crosses the graph more than once, it means that that x-value has more than one y-value.

b. (Sketch not shown.) In Problem 33, any vertical line crossesthe graph at most once, but in Problem 35, any verticalline between the two endpoints crosses the graph twice.

40. It is all right in a function for different xs to produce thesame y, but a relation is not a function if the same xproduces different ys.

s

ftCost

sq ft

Time

Temp

Weight

Length

x = 32.7320;= D0.7320 x = 1 3y = 23; 41. that is, the number (or the variable representing it) that

is being substituted into f.

42. Student research problem

Problem Set 1-3Q1. Quadratic Q2.

Q3. Q4.

Q5. Q6.

Q7. Q8.

Q9. 900 Q10. D

1. a.

b.

c. y-dilation by 2

2. a.

b.

c. y-translation by D3

3. a.

b.

c. x-translation by 4

y

x

5

5

5

g(x) = 9 D (x D 4)2

y

x

5 5

5

g(x) = D3 + 9 D x2

y

x

5 5

5

g(x) = 29 D x2

4

6

y

x

y

x

9x2 D 30x + 25

x2 + x D 56y = abxy = axb

x D 2,


4. a.

b.

c. x-dilation by 3

5. a.

b.

c. x-dilation by 2, y-translation by 1

6. a.

b.

c. x-translation by D3, y-dilation by

7. a. y-translation by 7

b.

8. a. x-translation by D5b.

9. a. x-dilation by 3

b.

10. a. y-dilation by 4

b.

11. a. x-translation by 6, y-dilation by 3

b.

12. a. x-dilation by 3, y-translation by D4

b.

13. No. The domain of is but the domain of the graphis That restriction must be added to the definition of

14. No. The domain of is but the domain of the graphis That restriction must be added to the definition of f (x).

D3 x 1.x 1,f (x)

f (x).D3 x 1.

x 1,f (x)

g(x) = D4 + f(x

3

)

g(x) = 3f (x D 6)

g(x) = 4f (x)

g(x) = f

(x

3

)

g(x) = f (x + 5)

g(x) = 7 + f (x)

12

y

x

5

5

5

g(x) =1

29 D (x + 3)2

y

x

5

5

5

g(x) = 1 + 9 D(x

2

)2

y

x

5 5

5

g(x) = 9 D(x

3

)2 15. a.

b. x-translation by D6

16. a.

b. x-dilation by 2

17. a.

b. y-dilation by 5

18. a.

b. y-translation by 4

19. a.

b. y-dilation by 5, x-translation by D6

4 4

2 x

y

4

2

y

x

4

2

y

x

4

2

y

x

10 4 4 10

4

2 x

y


20. a.

b. x-dilation by 2, y-translation by 4

21. Answers will vary.

Problem Set 1-4Q1. y-dilation by 3 Q2. y-translation by 5

Q3. Q4.

Q5. x is the base, not the exponent.

Q6.

Q7. Q8.

Q9. 120 Q10. C

1. a. 5 C 7(4) H 33 cm;5 C 7(7) H 54 cm

b. 332 H 33421.1943 cm2;752 H 9160.8841 cm2

c. The area depends on the radius, which in turn depends onthe time. Area is the outside function and radius is theinside function.

d. r(t) H 5 X 7t; a r(t) H r(t) 2; a(t) H (5 X 7t)2; a(4) H (33)2 H 1089 H 3421.1973; a(4) H (54)2 H 2916 H 9160.8841

2. a.

b.

c. The radius depends on the area (essentially the number ofbacteria), which in turn depends on the time. Radius is theoutside function and area is the inside function.

d.

Only the positive value makes sense in the context, so

R(t) = 9(1.1)t

= 9(1.1)t R (t) = A(t)A(t) = A(R (t)) = (R (t))2 (R (t))2 =

A(t)

= 2.7258 mm R (10) = 23.3436(R(10))2 = 23.3436 R(5) = 14.4945 = 2.1479 mm;(R(5))2 = 14.4945(R(0))2 = 9 R(0) = 9 = 1.6925 mm;

= 23.3436 mm2A(10) = 9(1.1)10= 14.4945 mm2;A(5) = 9(1.1)5= 9 mm2;A(0) = 9(1.1)0

)()(

x8x = 5

f (x) = ax2 + bx + c

f (5x)f (x + 4)

4

2

y

x

3. a. Answers will vary. Note that shoe size is a discrete graph, because shoe sizes come only in half units.

Sample answer:

b. In S(x), x represents foot length (in inches, for the graphabove). In L(x), x represents age (in years, above). Thecomposite function S L(x) gives shoe size as a functionof age (x represents age). L S(x) would be meaninglesswith the given functions L and S. Because x is substitutedinto S, x must represent foot length. S then gives shoe size.But this is substituted into L, which expects to have anage, not a shoe size, substituted into it. (If we had twocompletely different functions, S giving shoe size as afunction of age and L giving foot length as a function ofshoe size, then L S(x) would give us foot length as afunction of age.)

c. Answers will vary, but should be the composite of thegraphs in part a. Again, shoe size is a discrete graph.

Sample answer:

10

10

5

20 30 40 50 60 70 80

x (yr)

S (L(x)) (size)

)(

)()(

10 20 30 40 50 60 70 80

x (yr)

L(x) (in.)

5 10 15

5

10

x (in.)

S(x) (size)


4. a. The graph of T(x) should be similar to the one below. Thegraphs of S(x) may vary. A sample graph is shown.

b. In T(x), x represents miles per hour. In S(x), x representsthe number of cars per mile. The composite function T S(x) gives the time to travel 1 mi as a function of thenumber of cars per mile (x represents cars per mile). S T(x) would be meaningless with the given functions S and T, because T(x) represents time in minutes and theinput to S must be number of cars per mile.

c. Answers will vary, but should be the composite of the two functions in part a. Sample answer:

5. a.

b.

1 2 3 4 5 6 7

1

2

3

4

5

6

7

x

p (x)

p(h(3)) = p (5) = 3.51 2 3 4 5 6 7

1

2

3

4

5

6

7

x

h (x)

h(3) = 5

20 40 60 80 100

5

10

x (cars per mile)

T (x) (min)

)(

)(

50 100

50

x (cars per mile)

S (x) (mi/h)

20 40 60 80

5

10

x (mi/h)

T (x) (min)

c. ;

d. , which is different from ,found in part c.

e. , which is undefined, because 6 is not in thedomain of h.

6. a.

b.

c. ;

d. is undefined because 6 is not in the domain of g

f (g(6))f (g(2)) M f (28) M 150f (g(3)) M f (39) M 75

20 40x = 48

60

50

100

150

200

250

300

x

f (x)

f (g(4)) M f (48) M 511 2 3 4 5 6

10

20

30

40

50

60

x

g (x)

g(4) M 48

h(p(0)) = h(6)1 2 3 4 5 6 7

1

2

3

4

5

6

7

x

h(x)

1 2 3 4 5 6 7

1

2

3

4

5

6

7

x

p (x)

p (h(2)) = 4.5h(p(2)) = h(5) = 4p (h(5)) = p (4) = 4p (h(2)) = p (3) = 4.5


e. , which is undefined because 55 is not in thedomain of f.

7. a.

b.

c. , which is undefined

d.

e.

f.

g. , which is undefined

h.

8. a.

b.

c. , which is undefined because 5 is notin the domain of u

d.

e.

f.

g.

h.

9. a.

x g(x) f g(x)

1 3 none

2 4 5

3 5 4

4 6 3

5 7 2

b. Domain of :the intersection of this with the domain of g,gives The domain of appears to be

c. 6 is not in the domain of g, so is undefined.but 3 is not in the domain of f.g(1) = (1) + 2 = 3,

g(x)

2 x 5.f g2 x 5.

1 x 5, 2 x 6; 4 x + 2 84 g(x) 8f g

)(

v (v (v(8))) = v (v(2)) = v(6) = 4u(u(6)) = u(2) = 3v (v(10)) = v(8) = 2v (u(10)) = v(6) = 4u(4) = 8; v (u(4)) = v(8) = 2

v(4) = 5; u(v(4)) = u(5)v(6) = 4; u(v(6)) = u(4) = 8v(2) = 6; u(v(2)) = u(6) = 2f (f (f (1))) = f (f (3)) = f (4) = 2g(g(3)) = g(7)f (f (5)) = f (1) = 3g(f (3)) = g(4) = 5f (4) = 2; g(f (4)) = g(2) = 3g(3) = 7; f (g(3)) = f (7)g(2) = 3; f (g(2)) = f (3) = 4g(1) = 2; f (g(1)) = f (2) = 5

20 40 60

50

100

150

200

250

300

x

f (x)

1 2 3 4 5 6

10

20

30

40

50

60

x

g (x)

f (g(5)) M f (55) d.x f (x) g f (x)

1 5 7

2 4 6

3 3 5

4 2 4

5 1 3

Domain of :

; the intersection of this with the domain of f,coincidentally also gives The domainappears to be .

e.

The domains of the composite functions match thecalculations in parts b and d.

f.and 7 is not in the domain of g.

10. a.

x g(x) f g(x)

0 5 11

1 4 12

2 3 11

3 2 8

4 1 3

5 0 none

6 D1 none7 D2 none

b.the intersection of this with the domain of g, is

which agrees with the table.

c.

but 11 is not in the

domain of g, so is undefined.

d.

e. ,with the domain found in part b. The graphcoincides with the graph in part d.

11. a.

Conjecture: For all values of x, . g(f (x)) = xf (g(x)) == 64 = 8g(f (8)) = g((8)2) = g(64)

g(f (5)) = g((5)2) = g(25) = 25 = 5f (g(7)) = f (7) = (7)2 = 7f (g(3)) = f (3) = (3)2 = 3

0 x 4= Dx2 + 2x + 11= D(5 D x)2 + 8(5 D x) D 4f (g(x)) = f (5 D x)

2 4

5

10f

g

f g

y

x

g(f (3))= g(11),g(f (3)) = g(D(3)2 + 8(3) D 4)

= D(2)2 + 8(2) D 4 = 8;f (g(3)) = f (5 D 3)0 x 4,

0 x 7, D1 x 4; D4 Dx 1 1 5 D x 61 g(x) 6

)(

g(5) = 5 + 2 = 7,

= 9 D 4 = 5;= f (4)f (f (5)) = f (9 D 5)

4 8

4

y

x

g

f

g f

f g

4 x 84 x 8.4 x 8,

4 x 8 D8 Dx D4 1 9 D x 51 f (x) 5g f

)(


b. is not in the domain of g, so is undefined, sois undefined.

. No.

c.

but g is defined only fornonnegative x, so is defined only for nonnegative x.

d.

e.

12. a. Translation 3 units to the right.

b. Horizontal dilation by a factor of 2.

c.

Yes

13. If the dotted graph is then the solid graph is. In terms of composition of functions,

the solid graph is where . , where.

14. If the dotted graph is then the solid graph is. In terms of composition of functions,

the solid graph is , where .

15. a.

= xf (g(x)) = g(f (x))= 1.5(D71

3

)+ 3 = D8.

g(f (D8)) = g(2

3(D8) D 2

)= g

(D71

3

)= 1.5

(4

2

3

)+ 3 = 10;

g(f (10)) = g(2

3(10) D 2

)= g

(4

2

3

)= D15;= 2

3 (D19.5) D 2

= f (D19.5)f (g(D15)) = f (1.5(D15) + 3)

f (g(6)) = f (1.5(6) + 3) = f (12) =2

3(12) D 2 = 6;

h(x) = |x|g(x) = h(f (x))1 x 5g(x) = |f (x)|,

1 x 5,f (x),

h(x) = Dxg(x) = f (h(x))g(x) = Dxg(x),

D5 x D1g(x) = f (Dx),1 x 5,f (x),

3f

f gf h

3 3

y

x

= e xDxif x 0

if x < 0= |x|g(f (x)) = g(x2) = x2

2

4

y

x

g

f

g fg f

2 2

f gf (g(x)) = f (x) = (x)2 = x,

2

4

y

x

f

g

f g

2 2

= 9 D9g(f (D9)) = g((D9)2) = g(81) = 81f (g(D9))

g(D9)D9 b.

and coincide with each other, and with theline . and are each others reflections acrossthat line.

c.

d. We want to find such that

.

Check:

and

.

Problem Set 1-5Q1. Inside Q2. Outside

Q3. Q4. 8

Q5. 5 Q6. 4

Q7. 2 Q8. Exponential

Q9. 1 Q10.

1. a. ; ;

b. The air leaks out of the tire as time passes, so thepressure is constantly getting lower. Thus, f is adecreasing function and hence is invertible.

c. Somewhere between and , all the airgoes out of the tire, and the pressure remains 0. So it isnot possible to give a unique time corresponding to apressure of 0 psi; cannot be defined.

d. The graph of is dotted. The two graphs are reflectionsof each other over the line . (They coincidentallyhappen to be very close over most of their length.)

10 20 30 40

10

20

30

40

x

y

y = f (x)

y = x

y = f 1(x)

y = xf D1(x)

f D1(0)

x = 30 minx = 25 min

f (15) = 10.7 psif (10) = 16 psif (5) = 24 psi

g(x) = 2x+3

(m d)(x)

= x= x D 75

+7

5=

1

5(5x D 7) + 7

5j (h(x)) = j(5x D 7)

= x,= x + 7 D 7= 5(1

5x +

7

5

)D 7

h(j(x)) = h(1

5x +

7

5

)=

1

5x +

7

5 j(x) = x + 7

5

5 j(x) = x + 7 5 j(x) D 7 = xh(j(x)) = xj(x)

= x= x D 3 + 3= 32

2

3x +

3

2 (D2) + 3

g(f (x)) = g(2

3x D 2

)= 1.5

(2

3x D 2

)+ 3

= x + 2 D 2 = x;= 23

3

2x +

2

33 D 2

f (g(x)) = f (1.5x + 3) =2

3(1.5x + 3) D 2

g(x)f (x)y = xg(f (x))f (g(x))

f

gf g = g f

4

4

4

y

x


e. As an input for f, x represents time in minutes. As aninput for , it represents pressure in psi.

2. a.

b. Any one-to-one function is invertible. ;; these give the temperature corresponding

to 30 c/min and 80 c/min. By contrast, and give the number of chirps/min corresponding to 30F and80F (0 c/min and 105 c/min).

c. The cricket does not begin chirping until the temperatureis at least 30F. At least for , the number ofchirps/min remains 0, so cannot be defined.

d.

e. As the input to c, x represents temperature in F. As theinput to , it represents number of chirps/min.

3.

Throughout most of its domain, the inverse relation hastwo y-values for every x-value.

4. No y-value comes from more than one x-value. Also, nohorizontal line passes through more than one point of thefunction.

5. Function

y

x

10 20

10

20

x

y

cD1

50 100

50

x

y

y = c (x) y = x

y = c1(x)

cD1(0)20 x 30

c (80)c (30)cD1(80) = 70F

cD1(30) = 50Fc (40) = 5 c/min; c (50) = 30 c/min; c (60) = 55 c/min

f D16. Not a function

7. Not a function

8. Not a function

9. Function

10. Function

4 4

4

4

y

x

4

4

4 4

y

x

y

x

y

x

y

x


11. Not a function

12. Not a function

13. Function

14. Function

15. Function

16. Function

y

x

4

4

4 8

4 4

4

4

y

x

4 4

4

4

y

x

4 4

4

4

y

x

4 4

4

4

y

x

4 4

4

4

y

x

17. Function

18. Function

19. Function

20. Not a function

21.

The inverse relation is a function.

22.

The inverse relation is a function.

8

x

y

f

f 14 8

=x D 4D0.4 = D2.5x + 10

x = D0.4y + 4 D0.4y = x D 4 y = f D1(x)

y

xf 1

f4

4

4

4

y = f D1(x) = 12

x + 3

x = 2y D 6 2y = x + 6

5 10

4

6

8

10

2

4

25

y

x

4

4

y

x

f

f

f 1 f 14 4

4

4

y

x

4 4

4

4

y

x


23.

The inverse relation is not a function.

24.

The inverse relation is not a function.

25.

26. for all x.

27. a. If you drive 1000 mi in a month, yourmonthly cost is $900.

b. is a function becauseno input produces more than one output.

You would havea monthly cost of $758 if you drove 645 mi in a month.

c.

28. a. .Deer that weigh 50, 100, and 150 lb have hides of areasapproximately 5.43, 8.62, and 11.29 ft2, respectively.

b. False. .

c.

d.

100 200

100

200

y

x

A1(x)

A(x)

= (2.5y)1.5 AD1(x) = (2.5x)1.5 y = 0.4x2/3 x =

(y

0.4

)3/2

A(100) 2A(50)

A(150) = 11.2924A(100) = 8.6177;A(50) = 5.4288;

y

x

c (x)

c1(x)

200

200 600 1000

400

600

800

1000

1250 = 645.cD1(758) = 2.5(758) D

cD1(x)cD1(x) = 2.5x D 1250.

c (1000) = 900.

f (f (x)) = Df (x) = D (Dx) = x

f (f (x)) =1

f (x)=

1

(1/x)= x, x 0

5

5

5

5

y

x

f

f 1

y = 2.5x D 7.5x = 0.4y2 + 3 y2 = x D 3

0.4

5

5

5

5

y

x

1f

f

x = D0.5y2 D 2 y2 = x + 2D0.5 y = D2x D 429. a.

Because the domain of d is ,

the range of is .

b. . This means that a 200-ft skid mark iscaused by a car moving at a speed of about 59 mi/h.

c.

d. Because the domain of d now contains negative numbers,the range of contains negative numbers. Now, becausethe range of contains negative numbers,

which is not a function.

30.

Invertible Not invertible

Problem Set 1-6

Q1. Q2.

Q3. Q4.

Q5. There are two y-values for every x-value.

Q6. 3 Q7. Invertible

Q8. A function for which each y-value corresponds to only one x-value in the range.

Q9. 5 Q10. D5

1. a.

b.

y

x

5 5

5

5

y

x

5 5

5

5

y = xx + 3

2

x + 31

2x

x

y

x

y

dD1 = x0.057,dD1dD1

50

50

d(x)

x

dD1(200) = 59.234

dD1(x) 0dD1

x 0 y = dD1(x) = x0.057.x = 0.057y2 y2 = x

0.057


c.

d.

2. a.

b.

c.

d.

3. a.

y

x

6 4

50

50

y

x

5

5

5

5

y

x

5 5

5

5

y

x

5 5

5

5

y

x

5 5

5

5

y

x

5 5

5

5

y

x

5 5

5

5

b.

c.

d.

4. a.

b.

c.

d.

y

x

5 5

5

5

y

x

5 5

5

5

y

x

5 5

5

5

y

x

5 5

5

5

y

x

6 4

50

50

y

x

6 4

50

50

y

x

6 4

50

50


5. The graphs match.

6. The graphs match.

7. a.

This transformation reflects as in 7b all the points on thegraph below the x-axis across the x-axis.

b.

This transformation reflects f (x), for positive values of x,across the y-axis.

c.

d.

8. a.

b. . At time he is 140 meters before (behind or below) the gas station.

c. 140 m and 70 m, respectively. The answers are positivebecause distance is always positive.

x = 10,f (10) = D140 m; f (40) = 70 m

40 80 120

y

x100

200

y

x

5 5

5

5

y

x

5 5

5

5

= 0.5(3 D 2) D 4.5 = D4f (|D3|) = 0.5(|D3| D 2) D 4.5 |f (3)| = |0.5(3 D 2) D 4.5| = |D4| = 4;

y

x

5 5

5

5

y

x

5 5

5

5

d.

e.

9. a.

A negative number raised to an even power is equal to theabsolute value of that number raised to the same power.So, for Jx, the same corresponding y-value occurs, andtherefore .

b.

A negative number raised to an odd power is equal to theopposite of the absolute value of that number raised tothe same power. Because each term in g(x) has x raised toan odd power, g(Dx) has the same effect on g(x) as Dg(x).

c. Function h is odd; function j is even.

d.

The function e(x) is neither odd nor even.

10.

The function f is an even function. f (Dx) = |Dx| = |x| = f (x)

y

x

e(Dx) De(x)

y

x

5

5

5

5

4 4

8

4

4

8

x

y

f (x) = f (Dx)

4 4

8

4

4

8

x

y

x = 93.1662 s

100

200

40 80 120

y

x


11. a. The graphs match.

b.

c.

The graphs match.

12. a.

b.

c. Dilated by a factor of 23; translated up by 37 cents;

The graphs match.

d.

13. a.

b.

c.

d.

14. Answers will vary.

= 6928.2032 mi

x = 2,400,000,00050 y2(x) = 2,400,000,000x2 = 50= 1333.3 mi; x = 50

0.0375y1(x) = 0.0375x = 50

y(5000) =2,400,000,000

50002= 96 lb

y(3000) = 0.0375(3000) = 112.5 lb;

4000 8000

50

100

150y

x

y2 = 2,400,000,000/x2/(x 4000)

y1 = 0.0375x/(0 x and x 4000);

= 2,400,000,000

b = 15040002b40002

= 150

a = 1504000

= 0.0375;a4000 = 150

So 0 x 13

D23Dx + 1 + 34 313 D23Dx + 1 276 Dx + 1 D12 Dx + 1 D12 Dx D13 x 13

y = e0, x = 0D23Dx + 1 + 37, x > 0

1 2 3 4 5

40

80

120

Weight (oz)

Price (cents)

f (2.9) = 2, f (3) = 3, f (3.1) = 3

3

3

y

x

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

g(x) = 2|x D 1|x D 1 + 3

Problem Set 1-71. Answers will vary.

Problem Set 1-8

Review Problems

R1. a.

b.

x y

0 35

1 24.5

2 17.15

3 12.005

4 8.4035

5 5.8825

6 4.1177

c. The graph intersects the line at approximately

Domain: ; range:

d. Asymptote

e.

R2. a. Linear

b. Polynomial (cubic)

c. Exponential

d. Power

e. Rational

f. Answers will vary; e.g., number of items manufacturedand total manufacturing cost.

g.

h.

A quadratic (with a negative -coefficient) fits thispattern.

i. 1-8b: exponential; 1-8c: polynomial (probably quadratic);1-8d: power

j. Figure 1-8e passes the vertical line test: no vertical lineintersects the graph more than once, so no x-valuecorresponds to more than one y-value. Figure 1-8f fails the vertical line test: there is at least one vertical line thatintersects the graph more than once, so more than one y-value corresponds to the same x-value.

x2

y

x

13 f (x) 37

Stress

Time

5 y 35.0 x M 5.5x = 5.5 min.

y = 5 psi

17.15 psi; 5.4 min


R3. a. Horizontal dilation by a factor of 3, vertical translation by D5;

b. Horizontal translation by C4, vertical dilation by a factor of 3.

R4. a.

b.

c.

d. No; the graph is curved.

e. Answers will vary. Possible answer:

f. ; ; , which is

undefined; ; ;

, which is undefined;

g.

h.

i. which is undefined, because 3 isnot in the domain of f.

j. . The intersectionof this with the domain of g, is which agrees with the graph.

R5. a. The inverse does not pass the vertical line test.

y

x

4

4

5 5

72 x

112 ,2 x 6,

72 x 1124 g(x) 8 4 2x D 3 8

f (g(3)) = f (2(3) D 3) = f (3),f (g(4)) = f (2(4) D 3) = f (5) = (5) D 2 = 3

5

5

y

x

f

g

f g

g(g(3)) = g(4) = 5f (f (3)) = f (2)

g(f (6)) = g(5) = 8f (g(6)) = f (3) = 2f (g(5)) = f (8)f (g(4)) = f (5) = 3f (g(3)) = f (4) = 60 t 13

8

40

t

y

W (h(5)) = 0.004(35)2.5 M 29 lbh(5) = 3(5) + 20 = 35 in.

h(t) = 3t + 20

5

5

x

y

f

g

g(x) = 4 D(x

3

)2

D 5

b.

The domain of f corresponds to the range of Therange of f corresponds to the domain of

c. The J reveals that there are two different y-values for some x-values.

d.

It passes the vertical line test; asymptotes.

e.

The curve is invertible because it is increasing. As theinput to v, x represents radius in meters. As the input to

, it represents volume in cubic meters. If is aparticular input to v, then is a point on thegraph of . Plugging the output, , into gives thepoint on the graph of . But thegraph of is just the graph of with all the x- andy-values exchanged, so this point is actually .Thus, .

f. Since no y corresponds to more than one x in the originalfunction, no x corresponds to more than one y in theinverse relation, so the inverse relation is a function.

y

x

vD1(v(x0)) = x0(v(x0), x0)

v(x)vD1(x)vD1(x)(v(x0), vD1(v(x0)))

vD1v(x0)v(x)(x0, v(x0))

x0vD1

4

3

2

1

4321

55

5

5

y

x

x = y2 + 1 y = x D 1.f D1.

f D1.

5 5

5

5

x

y


R6. a.

b. The graph agrees with Figure 1-8k; each of the graphsagrees with those in part a.

c. Because power functions with odd powers satisfy theproperty and power functions with evenpowers satisfy the property .

d.

Discontinuity

Concept Problems

C1. Horizontal dilation by 3 (width from 4 units to 12 units),vertical dilation by 2 (height from 4 units to 8 units),horizontal translation by X3, vertical translation by D5;

C2. a. Answers will vary. The function repeats itself periodically.

b.

c. Odd. It is its own reflection through the origin, so

.f (Dx) = Df (x)

2

= 2

(x D 3

3

)2

D 5g(x) = 2f(1

3(x D 3)

)D 5

6

4 4

y

x

f (Dx) = f (x)f (Dx) = Df (x)

y = f (|x|)

y

x

5 5

5

5

y = |f (x)|

y

x

5 5

5

5

y = f (x)

y

x

5 5

5

5

y = f (x)

y

x

5 5

5

5

d.

e. Horizontal translation C2, vertical translation X3 [(0, 0)moves to (2, 3)];

f. Horizontal dilation by 2

Chapter Test

T1. Exponential

T2. Linear

T3. Polynomial (quadratic)

T4. Power

T5. All except T3. They are invertible; that is, their inverses arealso functions.

T6. Answers will vary.

or

T7. Odd

T8. Neither

T9. Horizontal dilation by 2;

T10. Horizontal translation by , vertical translation by C5;

T11. Horizontal translation by C6, vertical dilation by 2;

T12. Domain: range: 4 y 9D2 x 7;

g(x) = 2f (x D 6)

g(x) = f(x + 1) + 5D1

g(x) = f

(x

2

)

Time

Temperature

Time

Temperature

y

x

f g

1

y = sin(x D 2) + 3

y = 5 sin(x)

1010

5

5

y

x


T13. Vertical dilation by

T14. Horizontal dilation by

T15. Horizontal translation by vertical translation by

T16. Reflection through the line

5

5

y

x

y = x

5

5

y

x

D4D3,

5

5

y

x

32

5

5

y

x

12 T17. The graph fails the vertical line test. (The pre-image graph

fails the horizontal line testit is not one-to-one.)

T18.

which is not defined, because is not in the domain of f.

T19. Horizontal translation by C4, vertical translation by C5, and

vertical dilation by 3 of

T20. varies proportionately to the 0.52 power of x. Powerfunction.

T21. If there are 150 wild oatplants per square meter of land, the percentage loss to thewheat crop will be about 43%.

T22. 60% of the crop means a 40% crop loss. Solve to get

About 129 plants per square meter.

T23.

If you know the percentage loss and want to find thenumber of wild oat plants per square meter.

T24.

If the crop loss is 100% (i.e., the total crop is lost), theremust have been about 750 wild oat plants per square meter.

T25.

T26.

T27. It passes the vertical line test. (The original function passesthe horizontal line testit is one-to-one.)

T28. Answers will vary.

500

500

x

y

L1(x)

L(x)

0 y 1000 x 750,

LD1(100) =

(100

3.2

) 10.52

= 749.3963

x = 3.2y 0.52 y =(

x

3.2

) 10.52

x =

(40

3.2

) 10.52

= 128.6596

40 = 3.2x0.52

L(150) = 3.2(150)0.52 = 43.3288.

L(x)

y = 3x D 4|x D 4| + 5

x

|x|;

D3= f (D3),f (g(1)) = f ((1)2 D 4)g(f (3)) = g(3) = (3)2 D 4 = D1;f (g(3)) = f ((3)2 D 4) = f (5) = 5;


chapter 1 functions and mathematical models - wikispacesman+ch+01.pdf · 20. a. b. x-dilation by 2,...

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