Solve for - Montville Township Public Schools · x í6 í4 í2 0 2 4 6 f(x) 8.6 7.9 6.9 0 í6.9 í7.9 í8.6 62/87,21 Evaluate the function for several x-values in its domain. Use

Graph and analyze each function. Describe the domain, range, intercepts, end behavior, continuity, and where the function is increasing or decreasing.

1.f (x) = 5x2

SOLUTION:Evaluate the function for several x-values in its domain.

Use these points to construct a graph.

The function is a monomial with an even degree and a positive value for a.

D = ( , ), R = [0, ); intercept: 0; continuousforallrealnumbers

decreasing: ( , 0); increasing: (0, )

x 3 2 1 0 1 2 3 f (x) 45 20 5 0 5 20 45

2.g(x) = 8x5



The function is a monomial with an odd degree and a positive value for a.

D = ( , ), R = ( , ); intercept: 0; continuousforallrealnumbers

increasing: ( , )

x 3 2 1 0 1 2 3 f (x) 1944 256 8 0 8 256 1944

3.h(x) = x3



The function is a monomial with an odd degree and a negative value for a.


decreasing: ( , )

x 3 2 1 0 1 2 3 f (x) 27 8 1 0 1 8 27

4.f (x) = 4x4



The function is a monomial with an even degree and a negative value for a.

D = ( , ), R = ( , 0]; intercept: 0; continuousforallrealnumbers

increasing: ( , 0); decreasing: (0, )

x 3 2 1 0 1 2 3 f (x) 324 64 4 0 4 64 324

5.g(x) = x9





increasing: ( , )

x 3 2 1 0 1 2 3 f (x) 6561 170.7 0.3 0 0.3 170.7 6561

6.f (x) = x8






x 3 2 1 0 1 2 3 f (x) 4100.6 160 0.625 0 0.625 160 4100.6

7.





decreasing: ( , )

x 3 2 1 0 1 2 3 f (x) 1093.5 64 0.5 0 0.5 64 1093.5

8.






x 3 2 1 0 1 2 3 f (x) 182.3 16 0.25 0 0.25 16 182.3

9.f (x) = 2x 4



Since the power is negative, the function will be undefined at x = 0.

D = ( , 0) (0, ), R = (0, ); no intercepts; infinitediscontinuityatx = 0;


x 3 2 1 0 1 2 3 f (x) 0.025 0.125 2 2 0.125 0.025

10.h(x) = 3x 7




D = ( , 0) (0, ), R = ( , 0) (0, ); no intercepts; infinite

discontinuity at x = 0; increasing: ( , 0) and (0, )

x 3 2 1 0 1 2 3 f (x) 0.001 0.023 3 3 0.023 0.001

11.f (x) = 8x 5






x 3 2 1 0 1 2 3 f (x) 0.03 0.25 8 8 0.25 0.03

12.g(x) = 7x 2






x 3 2 1 0 1 2 3 f (x) 0.78 1.75 7 7 1.75 0.78

13.




D = ( , 0) (0, ), R = (, 0) (0, ); no intercepts; infinitediscontinuity

at x = 0; increasing: ( , 0) and (0, )

x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.01 0.4 204.8 204.8 0.4 0.01

14.h(x) = x 6






x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.01 0.17 10.67 10.67 0.17 0.01

15.h(x) = x 3





discontinuity at x = 0; decreasing: ( , 0) and (0, )

x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.22 0.75 6 6 0.75 0.22

16.




D = ( , 0) (0, ), R = ( , 0); no intercepts; infinitediscontinuityatx = 0;


x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.03 0.7 179.2 179.2 0.7 0.03

17.GEOMETRY ThevolumeofasphereisgivenbyV(r) = r3, where r is the radius.

a. State the domain and range of the function. b. Graph the function.

SOLUTION:a. The radius of a sphere cannot have a negative length. The radius also cannot be 0 because then the object wouldfail to be a sphere. Thus, D = (0, ), R = (0, )

b. Evaluate the function for several x-values in its domain.


x 0.5 1 1.5 2 2.5 3 3.5 f (x) 0.5 4.2 14.1 33.5 65.5 113.1 179.6


18.



Since the denominator of the power is even, the domain must be restricted to nonnegative values.

D = [0, ), R = [0, ); intercept: 0; ; continuous on [0, ); increasing: (0, )

x 0 1 2 3 4 5 6 f (x) 0 8 9.5 10.5 11.3 12 12.5

19.




decreasing: ( , )

x 6 4 2 0 2 4 6 f (x) 8.6 7.9 6.9 0 6.9 7.9 8.6

20.






x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.17 0.2 0.25 0.25 0.2 0.17

21.



Since the denominator of the power is even and the power is negative, the domain must be restricted to positive values.

D = (0, ), R = (0, ); no intercepts; ; continuous on (0, ); decreasing: (0, )

x 1 2 3 4 5 6 7 f (x) 10.0 8.9 8.3 7.9 7.6 7.4 7.2

22.




D = [0, ), R = [0, ); intercept: 0; ; continuous on [0, ); decreasing: (0, )

x 0 1 2 3 4 5 6 f (x) 0 3 4.6 6 7.1 8.2 9.2

23.




increasing: ( , )

x 3 2 1 0 1 2 3 f (x) 1.45 1.13 0.75 0 0.75 1.13 1.45

24.




D = (0, ), R = ( , 0); no intercept; ; continuous on (0, ); increasing: (0, )

x 0.5 1 1.5 2 2.5 3 3.5 f (x) 0.84 0.5 0.37 0.3 0.25 0.22 0.2

25.






x 3 2 1 0 1 2 3 f (x) 0.48 0.63 1 1 0.63 0.48

26.




increasing: ( , )

x 1.5 1 0.5 0 0.5 1 1.5 f (x) 13.8 7 2.2 0 2.2 7 13.8

27.




D = [0, ), R = ( , 0]; intercept: 0; ; continuous on [0, ); decreasing: (0, )

x 0 0.5 1 1.5 2 2.5 3 f (x) 0 1.2 4 8.1 13.5 19.9 27.4

28.




D = (0, ), R = ( , 0); no intercepts; ; continuous on (0, ); increasing: (0, )

x 0.5 1 1.5 2 2.5 3 3.5 f (x) 14.1 5 2.7 1.8 1.3 1 0.8

29.






x 3 2 1 0 1 2 3 f (x) 0.11 0.22 0.67 0.67 0.22 0.11

Complete each step. a. Create a scatter plot of the data. b. Determine a power function to model the data.c. Calculate the value of each model at x = 30.

30.

SOLUTION:a. Enter the data into a graphing calculator and create a scatter plot.

b. Use the power regression function on the graphing calculator to find values for a and b.

y = 3.54x2.89

c. Graph the regression equation using a graphing calculator. To calculate x = 30, use the CALC function on the graphing calculator.

The value of the model at x = 30 is about 66,098.82.

31.



y = 0.77x5.75.


The value of the model at x = 30 is about 235,906,039.

32.CLIFF DIVING Inthesportofcliffdiving,competitorsperformthreedivesfromaheightof28meters.Judgesaward divers a score from 0 to 10 points based on degree of difficulty, take-off, positions, and water entrance. The table shows the speed of a diver at various distances in the dive.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the speed at which a diver would enter the water from a cliff dive of 30 meters.



f(x) = 4.42x0.5

.

c. Graph the regression equation using a graphing calculator. To calculate the speed at which a diver would enter the water from a cliff dive of 30 meters, use the CALC function on the graphing calculator. Let x = 30.

The speed at which a diver would enter the water from a cliff dive of 30 meters is about 24.25 meters per second.

33.WEATHER Thewindchilltemperatureistheapparenttemperaturefeltonexposedskin,takingintoaccounttheeffect of the wind. The table shows the wind chill temperature produced at winds of various speeds when the actual temperature is 50F.

a. Create a scatter plot of the data. b. Determine a power function to model the data. c. Use the function to predict the wind chill temperature when the wind speed is 65 miles per hour.



f(x) = 55.14x0.0797

.

c. Graph the regression equation using a graphing calculator. To predict the wind chill temperature when the wind speed is 65 miles per hour, use the CALC function on the graphing calculator. Let x = 65.

Thewindchilltemperaturewhenthewindspeedis65milesperhourisabout39.54F


34.f (x) = 3



Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 6 + 3x. Solve for x when the radicand is 0 to find the restriction on the domain and the x-intercept.

D = [2, ), R = [0, ); x-intercept: 2, y-intercept: 3 ; continuouson[2, ); increasing: (2,

)

x 2 1 0 1 2 3 4 f (x) 0 5.2 7.3 9 10.4 11.6 12.7

35.



Solve for x when g(x) is 0 to find the x-intercept.

D = ( , ), R = ( , ); x-intercept: 128, y-intercept: 8; continuousfor

all real numbers; decreasing: ( , )

x 300 200 100 0 100 200 300 f (x) 8.5 7.1 5.9 8 9 9.7 10.2

36.



Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 16x + 48. Solve for x when the radicand is 0 to find the x-intercept to find the restriction on the domain.

D = [3, ), R = ( , 3]; x-intercept: none, y-intercept: 3.71; continuouson[3, );

decreasing: (3, )

x 3 2 1 0 1 2 3 f (x) 3 3.6 3.67 3.71 3.75 3.78 3.8

37.h(x) = 4 +



Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 7x 12. Solve for x when the radicand is 0 to find the restriction on the domain.

, R = [4, ); no intercepts; continuouson ; increasing:

x 2 3 4 5 6 7

f (x) 4 5.4 7 8 8.8 9.5 10.1

38.g(x) = 16



Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, (1 4x)3. Solve for x when the radicand is 0 to find the restriction on the domain.


D = ( , 0.25], R = [16, ); x-intercept: 1.34, y-intercept: 15; continuouson( , 0.25];

decreasing: ( , 0.25)

x 2.5 2 1.5 1 0.5 0 0.25 f (x) 20.5 11 2.5 4.8 10.8 15 16

39.



x 6 4 2 0 2 4 6 f (x) 78.1 71.54 63.81 52.66 61.27 69.53 76.35

Use the maximum function or the trace function on a graphing calculator to approximate the maximum value of f

(x) at (0.28, 49).

The range is restricted to values less than or equal to 49. Also, there is a turning point at x = 0.28. D = ( , ), R = ( , 49.00]; x-intercept: none, y-intercept: 52.66;

continuousforallrealnumbersincreasing:( , 0.28); decreasing: (0.28, )

40.



Solve for x when h(x) is 0 to find the x-intercept.

D = ( , ), R = ( , ); x-intercept: 2034.5, y-intercept: 6.5; continuous

for all real numbers; decreasing: ( , )

x 8 4 0 4 8 12 16 f (x) 6.2 6.4 6.5 6.7 6.9 7.3 8.9

41.g(x) =



Since g(x) includes two even-degree radicals, the domain is restricted to nonnegative values for each radicand, 22

x and 3x 3. Solve for x when each radicand is greater than or equal to 0 to find the restrictions on the domain.

Thus, xmustbe1x22.Substitutethesevaluesforx to find the restrictions on the range. x = 1

x = 22

Use the zero function or the trace function on a graphing calculator to approximate the x-intercept at (6.25, 0).

D = [1, 22], R = [ , ]; x-intercept: 6.25; continuous on [1, 22]; decreasing: (1, 22)

x 1 3 6 12 15 18 22 f (x) 4.58 1.91 0.13 2.58 3.84 5.14 7.94

42.FLUID MECHANICS ThevelocityofthewaterflowingthroughahosewithanozzlecanbemodeledusingV

(P) = 12.1 , where V is the velocity in feet per second and P is the pressure in pounds per square inch.

a. Graph the velocity through a nozzle as a function of pressure. b. Describe the domain, range, end behavior, and continuity of the function and determine where it is increasing or decreasing.



b. Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, P. Thus, P0. D = [0, ), R = [0, ); ; continuous on [0, ); increasing: (0, )

x 0 1 2 3 4 5 6 f (x) 0 12.1 17.1 21 24.2 27.1 29.7

43.AGRICULTURAL SCIENCE The net energy NEm required to maintain the body weight of beef cattle, in

megacalories (Mcal) per day, is estimated by the formula wherem is the animals mass in kilograms. One megacalorie is equal to one million calories. a. Find the net energy per day required to maintain a 400-kilogram steer. b. If 0.96 megacalorie of energy is provided per pound of whole grain corn, how much corn does a 400-kilogram steer need to consume daily to maintain its body weight?

SOLUTION:a. Substitute m = 400.

The net energy per day required to maintain a 400-kilogram steer is approximately 6.89 Mcal. b. Divide 6.89 Mcal by the 0.96 Mcal found in a pound of whole grain corn to find the total amount of corn necessary to maintain a 400-kilogram steer.

It will take about 7.18 pounds of corn to maintain a 400-kilogram steer.

Solve each equation.

44.4 = +

SOLUTION:

Since the each side of the equation was raised to a power, check the solutions in the original equation.

x = 11

x= 75

Neither value for x is a solution for the original equation. Thus, there is no solution.

45.0.5x = +2

SOLUTION:


x = 0

x= 4


46.3 =

SOLUTION:


x = 13

x= 1.75

One solution checks and the other solution does not. Therefore, the solution is x = 13.

47. 10 = 17

SOLUTION:

Since the each side of the equation was raised to a power, check the solution in the original equation.

The solution is x = 7.

48.

SOLUTION:



49.x = +2

SOLUTION:


x = 2

x= 4

The solutions are x = 2 and x = 4.

50.7 + =250

SOLUTION:



51.x = 5 +

SOLUTION:


x = 8

x= 3


52. + 4 =

SOLUTION:


x = 2

x= 10


53. = 20

SOLUTION:


There is no solution.

54. 1 =

SOLUTION:


x = 1


55.

SOLUTION:



Determine whether each function is a monomial function given that a and b are positive integers. Explain your reasoning.

56.y = x4a

SOLUTION:

Yes; sample answer: The function follows the form f (x) = axn, where n is a positive integer. In this case, a =

andn = 4a.

57.G(x) = 2ax4

SOLUTION:

Yes; sample answer: The function follows the form f (x) = axn, where n is a positive integer. In this case, a = 2a

and n = 4.

58.F(b) = 3ab5x

SOLUTION:No; sample answer: The function is not a power function because the variable is in the exponent.

59.y = tab

SOLUTION:


andn = ab.

60.

SOLUTION:


andn = 2b.

61.y = 4abx2

SOLUTION:No; sample answer: The function is not a monomial function because the exponent for x is negative.

62.CHEMISTRY Thefunction canbeusedtoapproximatethenuclearradiusofanelementbasedon

its molecular mass, where r is length of the radius in meters, R0 is a constant (about 1.2 1015

meter), and A is

the molecular mass.

a. If the nuclear radius of sodium is about 3.412 1015 meter, what is its molecular mass?

b. The approximate nuclear radius of an element is 6.030 1015

meter. Identify the element.

c. The ratio of the molecular masses of two elements is 27:8. What is the ratio of their nuclear radii?

SOLUTION:

a. Substitute r = 3.412 1015 and R0 = 1.2 1015

into the function to solve for A.

b. Substitute r = 6.030 1015 and R0 = 1.2 1015


With a molecular mass of 126.88, the element is iodine. c. Although the molecular masses of the two elements are unknown, the ratio of the two masses will always be 27:8. These numbers can be used for the molecular masses to find the ratio of the radii.

Since R0 is a constant, the ratio of the radii will be 3:2.

Element 1 Element 2

Solve each inequality.

63.

SOLUTION:

Since the each side of the equation was raised to a power, check a solution in the original equation.

x = 1

The solution is x 2.

64.

SOLUTION:


x = 5

The solution is x6.

65.

SOLUTION:


x = 7

Since the denominator of the exponent is even, the function must be checked for restrictions on the domain. The

radicand, 1 4x, must be greater than or equal to 0. Solve 1 4x0forx.

The solution accounts for this restriction. So, the solution is x 6.

66. 9

SOLUTION:


x = 24

Since the denominator of the exponent is even, the function must be checked for restrictions on the domain. The radicand, 6 + 3x, must be greater than or equal to 0. Solve 6 + 3x0forx.

Since the solution does not account for this restriction, it must be added to the solution. The solution is 2x25.

67.

SOLUTION:


x = 6

The solution is x5.

68.

SOLUTION:


x = 291

69.CHEMISTRY Boyles Law states that, at constant temperature, the pressure of a gas is inversely proportional toits volume. The results of an experiment to explore Boyles Law are shown.

a. Create a scatter plot of the data. b. Determine a power function to model the pressure P as a function of volume v. c. Based on the information provided in the problem statement, does the function you determined in part b make sense? Explain. d. Use the model to predict the pressure of the gas if the volume is 3.25 liters. e . Use the model to predict the pressure of the gas if the volume is 6 liters.

SOLUTION:a.

b. Use the power regression function on the graphing calculator to find values for a and n.

P(v) = 3.62v1

c. Sample answer: Yes; the problem states that the volume and pressure are inversely proportional, and in the power function, the exponent of the volume variable is 1. d. Graph the regression equation using a graphing calculator. To predict the pressure of the gas if the volume is 3.25 liters, use the CALC function on the graphing calculator. Let v = 3.25.

The pressure of 3.25 liters of the gas is about 1.12 atmospheres. e . Graph the regression equation using a graphing calculator. To predict the pressure of the gas if the volume is 6 liters, use the CALC function on the graphing calculator. Let v = 6.

The pressure of 6 liters of the gas is about 0.60 atmospheres.

Without using a calculator, match each graph with the appropriate function.

a.

b. g (x) = x6

c. h(x) = 4x3

d.

70.

SOLUTION:The end behavior of the graph indicates that n is positive and even. Also, a is positive. The equation that matches

this description is g(x) = x6.

The answer is b.

71.

SOLUTION:There is an infinite discontinuity at x = 0. This indicates that the power is negative. The equation that matches this

description is h(x) = 4x3

.

The answer is c.

72.

SOLUTION:The domain of the graph is restricted to nonnegative values. This indicates that the function has a rational exponentwith an even denominator or is a radical function with an even value for n. The equation that matches this

description is .

The answer is a.

73.

SOLUTION:The end behavior and the continuity of the graph indicate that it is a radical function with an odd value for n. The

equation that matches this description is .

The answer is d.

74.ELECTRICITY ThevoltageusedbyanelectricaldevicesuchasaDVDplayercanbecalculatedusingV =

, where V is the voltage in volts, P is the power in watts, and R is the resistance in ohms. The function I =

canbeusedtocalculatethecurrent,whereI is the current in amps.

a. If a lamp uses 120 volts and has a resistance of 11 ohms, what is the power consumption of the lamp? b. If a DVD player has a current of 10 amps and consumes 1200 watts of power, what is the resistance of the DVD player? c. Ohms Law expresses voltage in terms of current and resistance. Use the equations given above to write OhmsLaw using voltage, resistance, and amperage.

SOLUTION:

a. Substitute V = 120 and R = 11 into V = tosolveforP.

The power consumption of the lamp is 1309 watts.

b. Substitute I = 10 and P = 1200 into I = .

The resistance of the DVD player is 12 ohms.

c. To write Ohms Law in terms of V, R, and I, solve I = forP.

Substitute P = I2R into V = .

Use the points provided to determine the power function represented by the graph.

75.

SOLUTION:

The function is undefined at x = 0, therefore; it is a power function of the form where

a is a real constant and n is a natural number. Start by assuming n = 1 and solve for a by substituting a set of points (2, 2) for x and y .

When n =1, f (x)=4 or forthepoint(2,2).Ifthispowerfunctionistrueforthesecondpoint(1, 4), then

f(x) can represent the graph.

Since f (x) is true for (1, 4), f (x) = or4x1 is a power function for the graph.

76.

SOLUTION:

The end behavior of the function indicates that it is power function of the form f (x) = axn, where n is even and a is

positive. Start by assuming n = 2 and solve for a by substituting a set of points forx and y .

When n =2, f (x) = x2 for the point . If this power function is true for the second point , then f

(x) can represent the graph.

Since f (x) is not true for the second point, repeat the first step but assume that n = 4.



eSolutions Manual - Powered by Cognero Page 1

2-1 Power and Radical Functions


1.f (x) = 5x2






x 3 2 1 0 1 2 3 f (x) 45 20 5 0 5 20 45

2.g(x) = 8x5





increasing: ( , )

x 3 2 1 0 1 2 3 f (x) 1944 256 8 0 8 256 1944

3.h(x) = x3





decreasing: ( , )

x 3 2 1 0 1 2 3 f (x) 27 8 1 0 1 8 27

4.f (x) = 4x4






x 3 2 1 0 1 2 3 f (x) 324 64 4 0 4 64 324

5.g(x) = x9





increasing: ( , )

x 3 2 1 0 1 2 3 f (x) 6561 170.7 0.3 0 0.3 170.7 6561

6.f (x) = x8






x 3 2 1 0 1 2 3 f (x) 4100.6 160 0.625 0 0.625 160 4100.6

7.





decreasing: ( , )

x 3 2 1 0 1 2 3 f (x) 1093.5 64 0.5 0 0.5 64 1093.5

8.






x 3 2 1 0 1 2 3 f (x) 182.3 16 0.25 0 0.25 16 182.3

9.f (x) = 2x 4






x 3 2 1 0 1 2 3 f (x) 0.025 0.125 2 2 0.125 0.025

10.h(x) = 3x 7






x 3 2 1 0 1 2 3 f (x) 0.001 0.023 3 3 0.023 0.001

11.f (x) = 8x 5






x 3 2 1 0 1 2 3 f (x) 0.03 0.25 8 8 0.25 0.03

12.g(x) = 7x 2






x 3 2 1 0 1 2 3 f (x) 0.78 1.75 7 7 1.75 0.78

13.






x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.01 0.4 204.8 204.8 0.4 0.01

14.h(x) = x 6






x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.01 0.17 10.67 10.67 0.17 0.01

15.h(x) = x 3






x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.22 0.75 6 6 0.75 0.22

16.






x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.03 0.7 179.2 179.2 0.7 0.03






x 0.5 1 1.5 2 2.5 3 3.5 f (x) 0.5 4.2 14.1 33.5 65.5 113.1 179.6


18.





x 0 1 2 3 4 5 6 f (x) 0 8 9.5 10.5 11.3 12 12.5

19.




decreasing: ( , )

x 6 4 2 0 2 4 6 f (x) 8.6 7.9 6.9 0 6.9 7.9 8.6

20.






x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.17 0.2 0.25 0.25 0.2 0.17

21.





x 1 2 3 4 5 6 7 f (x) 10.0 8.9 8.3 7.9 7.6 7.4 7.2

22.





x 0 1 2 3 4 5 6 f (x) 0 3 4.6 6 7.1 8.2 9.2

23.




increasing: ( , )

x 3 2 1 0 1 2 3 f (x) 1.45 1.13 0.75 0 0.75 1.13 1.45

24.





x 0.5 1 1.5 2 2.5 3 3.5 f (x) 0.84 0.5 0.37 0.3 0.25 0.22 0.2

25.






x 3 2 1 0 1 2 3 f (x) 0.48 0.63 1 1 0.63 0.48

26.




increasing: ( , )

x 1.5 1 0.5 0 0.5 1 1.5 f (x) 13.8 7 2.2 0 2.2 7 13.8

27.





x 0 0.5 1 1.5 2 2.5 3 f (x) 0 1.2 4 8.1 13.5 19.9 27.4

28.





x 0.5 1 1.5 2 2.5 3 3.5 f (x) 14.1 5 2.7 1.8 1.3 1 0.8

29.






x 3 2 1 0 1 2 3 f (x) 0.11 0.22 0.67 0.67 0.22 0.11


30.



y = 3.54x2.89



31.



y = 0.77x5.75.







f(x) = 4.42x0.5

.







f(x) = 55.14x0.0797

.




34.f (x) = 3





)

x 2 1 0 1 2 3 4 f (x) 0 5.2 7.3 9 10.4 11.6 12.7

35.






x 300 200 100 0 100 200 300 f (x) 8.5 7.1 5.9 8 9 9.7 10.2

36.





decreasing: (3, )

x 3 2 1 0 1 2 3 f (x) 3 3.6 3.67 3.71 3.75 3.78 3.8

37.h(x) = 4 +





x 2 3 4 5 6 7

f (x) 4 5.4 7 8 8.8 9.5 10.1

38.g(x) = 16







x 2.5 2 1.5 1 0.5 0 0.25 f (x) 20.5 11 2.5 4.8 10.8 15 16

39.



x 6 4 2 0 2 4 6 f (x) 78.1 71.54 63.81 52.66 61.27 69.53 76.35


(x) at (0.28, 49).



40.






x 8 4 0 4 8 12 16 f (x) 6.2 6.4 6.5 6.7 6.9 7.3 8.9

41.g(x) =






x = 22



x 1 3 6 12 15 18 22 f (x) 4.58 1.91 0.13 2.58 3.84 5.14 7.94







x 0 1 2 3 4 5 6 f (x) 0 12.1 17.1 21 24.2 27.1 29.7







44.4 = +

SOLUTION:


x = 11

x= 75


45.0.5x = +2

SOLUTION:


x = 0

x= 4


46.3 =

SOLUTION:


x = 13

x= 1.75


47. 10 = 17

SOLUTION:



48.

SOLUTION:



49.x = +2

SOLUTION:


x = 2

x= 4


50.7 + =250

SOLUTION:



51.x = 5 +

SOLUTION:


x = 8

x= 3


52. + 4 =

SOLUTION:


x = 2

x= 10


53. = 20

SOLUTION:



54. 1 =

SOLUTION:


x = 1


55.

SOLUTION:




56.y = x4a

SOLUTION:


andn = 4a.

57.G(x) = 2ax4

SOLUTION:


and n = 4.

58.F(b) = 3ab5x


59.y = tab

SOLUTION:


andn = ab.

60.

SOLUTION:


andn = 2b.

61.y = 4abx2




meter), and A is

the molecular mass.





SOLUTION:







Element 1 Element 2


63.

SOLUTION:


x = 1


64.

SOLUTION:


x = 5

The solution is x6.

65.

SOLUTION:


x = 7



The solution accounts for this restriction. So, the solution is x 6.

66. 9

SOLUTION:


x = 24

Since the denominator of the exponent is even, the function must be checked for restrictions on the domain. The radicand, 6 + 3x, must be greater than or equal to 0. Solve 6 + 3x0forx.

Since the solution does not account for this restriction, it must be added to the solution. The solution is 2x25.

67.

SOLUTION:


x = 6

The solution is x5.

68.

SOLUTION:


x = 291

69.CHEMISTRY Boyles Law states that, at constant temperature, the pressure of a gas is inversely proportional toits volume. The results of an experiment to explore Boyles Law are shown.

a. Create a scatter plot of the data. b. Determine a power function to model the pressure P as a function of volume v. c. Based on the information provided in the problem statement, does the function you determined in part b make sense? Explain. d. Use the model to predict the pressure of the gas if the volume is 3.25 liters. e . Use the model to predict the pressure of the gas if the volume is 6 liters.

SOLUTION:a.

b. Use the power regression function on the graphing calculator to find values for a and n.

P(v) = 3.62v1

c. Sample answer: Yes; the problem states that the volume and pressure are inversely proportional, and in the power function, the exponent of the volume variable is 1. d. Graph the regression equation using a graphing calculator. To predict the pressure of the gas if the volume is 3.25 liters, use the CALC function on the graphing calculator. Let v = 3.25.

The pressure of 3.25 liters of the gas is about 1.12 atmospheres. e . Graph the regression equation using a graphing calculator. To predict the pressure of the gas if the volume is 6 liters, use the CALC function on the graphing calculator. Let v = 6.

The pressure of 6 liters of the gas is about 0.60 atmospheres.

Without using a calculator, match each graph with the appropriate function.

a.

b. g (x) = x6

c. h(x) = 4x3

d.

70.

SOLUTION:The end behavior of the graph indicates that n is positive and even. Also, a is positive. The equation that matches

this description is g(x) = x6.

The answer is b.

71.

SOLUTION:There is an infinite discontinuity at x = 0. This indicates that the power is negative. The equation that matches this

description is h(x) = 4x3

.

The answer is c.

72.

SOLUTION:The domain of the graph is restricted to nonnegative values. This indicates that the function has a rational exponentwith an even denominator or is a radical function with an even value for n. The equation that matches this

description is .

The answer is a.

73.

SOLUTION:The end behavior and the continuity of the graph indicate that it is a radical function with an odd value for n. The

equation that matches this description is .

The answer is d.

74.ELECTRICITY ThevoltageusedbyanelectricaldevicesuchasaDVDplayercanbecalculatedusingV =

, where V is the voltage in volts, P is the power in watts, and R is the resistance in ohms. The function I =

canbeusedtocalculatethecurrent,whereI is the current in amps.

a. If a lamp uses 120 volts and has a resistance of 11 ohms, what is the power consumption of the lamp? b. If a DVD player has a current of 10 amps and consumes 1200 watts of power, what is the resistance of the DVD player? c. Ohms Law expresses voltage in terms of current and resistance. Use the equations given above to write OhmsLaw using voltage, resistance, and amperage.

SOLUTION:

a. Substitute V = 120 and R = 11 into V = tosolveforP.

The power consumption of the lamp is 1309 watts.

b. Substitute I = 10 and P = 1200 into I = .

The resistance of the DVD player is 12 ohms.

c. To write Ohms Law in terms of V, R, and I, solve I = forP.

Substitute P = I2R into V = .

Use the points provided to determine the power function represented by the graph.

75.

SOLUTION:

The function is undefined at x = 0, therefore; it is a power function of the form where

a is a real constant and n is a natural number. Start by assuming n = 1 and solve for a by substituting a set of points (2, 2) for x and y .

When n =1, f (x)=4 or forthepoint(2,2).Ifthispowerfunctionistrueforthesecondpoint(1, 4), then

f(x) can represent the graph.

Since f (x) is true for (1, 4), f (x) = or4x1 is a power function for the graph.

76.

SOLUTION:

The end behavior of the function indicates that it is power function of the form f (x) = axn, where n is even and a is

positive. Start by assuming n = 2 and solve for a by substituting a set of points forx and y .



Since f (x) is not true for the second point, repeat the first step but assume that n = 4.



eSolutions Manual - Powered by Cognero Page 2

2-1 Power and Radical Functions


1.f (x) = 5x2






x 3 2 1 0 1 2 3 f (x) 45 20 5 0 5 20 45

2.g(x) = 8x5





increasing: ( , )

x 3 2 1 0 1 2 3 f (x) 1944 256 8 0 8 256 1944

3.h(x) = x3





decreasing: ( , )

x 3 2 1 0 1 2 3 f (x) 27 8 1 0 1 8 27

4.f (x) = 4x4






x 3 2 1 0 1 2 3 f (x) 324 64 4 0 4 64 324

5.g(x) = x9





increasing: ( , )

x 3 2 1 0 1 2 3 f (x) 6561 170.7 0.3 0 0.3 170.7 6561

6.f (x) = x8






x 3 2 1 0 1 2 3 f (x) 4100.6 160 0.625 0 0.625 160 4100.6

7.





decreasing: ( , )

x 3 2 1 0 1 2 3 f (x) 1093.5 64 0.5 0 0.5 64 1093.5

8.






x 3 2 1 0 1 2 3 f (x) 182.3 16 0.25 0 0.25 16 182.3

9.f (x) = 2x 4






x 3 2 1 0 1 2 3 f (x) 0.025 0.125 2 2 0.125 0.025

10.h(x) = 3x 7






x 3 2 1 0 1 2 3 f (x) 0.001 0.023 3 3 0.023 0.001

11.f (x) = 8x 5






x 3 2 1 0 1 2 3 f (x) 0.03 0.25 8 8 0.25 0.03

12.g(x) = 7x 2






x 3 2 1 0 1 2 3 f (x) 0.78 1.75 7 7 1.75 0.78

13.






x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.01 0.4 204.8 204.8 0.4 0.01

14.h(x) = x 6






x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.01 0.17 10.67 10.67 0.17 0.01

15.h(x) = x 3






x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.22 0.75 6 6 0.75 0.22

16.






x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.03 0.7 179.2 179.2 0.7 0.03






x 0.5 1 1.5 2 2.5 3 3.5 f (x) 0.5 4.2 14.1 33.5 65.5 113.1 179.6


18.





x 0 1 2 3 4 5 6 f (x) 0 8 9.5 10.5 11.3 12 12.5

19.




decreasing: ( , )

x 6 4 2 0 2 4 6 f (x) 8.6 7.9 6.9 0 6.9 7.9 8.6

20.






x 1.5 1 0.5 0 0.5 1 1.5 f (x) 0.17 0.2 0.25 0.25 0.2 0.17

21.





x 1 2 3 4 5 6 7 f (x) 10.0 8.9 8.3 7.9 7.6 7.4 7.2

22.





x 0 1 2 3 4 5 6 f (x) 0 3 4.6 6 7.1 8.2 9.2

23.




increasing: ( , )

x 3 2 1 0 1 2 3 f (x) 1.45 1.13 0.75 0 0.75 1.13 1.45

24.





x 0.5 1 1.5 2 2.5 3 3.5 f (x) 0.84 0.5 0.37 0.3 0.25 0.22 0.2

25.






x 3 2 1 0 1 2 3 f (x) 0.48 0.63 1 1 0.63 0.48

26.




increasing: ( , )

x 1.5 1 0.5 0 0.5 1 1.5 f (x) 13.8 7 2.2 0 2.2 7 13.8

27.





x 0 0.5 1 1.5 2 2.5 3 f (x) 0 1.2 4 8.1 13.5 19.9 27.4

28.





x 0.5 1 1.5 2 2.5 3 3.5 f (x) 14.1 5 2.7 1.8 1.3 1 0.8

29.






x 3 2 1 0 1 2 3 f (x) 0.11 0.22 0.67 0.67 0.22 0.11


30.



y = 3.54x2.89



31.



y = 0.77x5.75.







f(x) = 4.42x0.5

.







f(x) = 55.14x0.0797

.




34.f (x) = 3





)

x 2 1 0 1 2 3 4 f (x) 0 5.2 7.3 9 10.4 11.6 12.7

35.






x 300 200 100 0 100 200 300 f (x) 8.5 7.1 5.9 8 9 9.7 10.2

36.





decreasing: (3, )

x 3 2 1 0 1 2 3 f (x) 3 3.6 3.67 3.71 3.75 3.78 3.8

37.h(x) = 4 +





x 2 3 4 5 6 7

f (x) 4 5.4 7 8 8.8 9.5 10.1

38.g(x) = 16







x 2.5 2 1.5 1 0.5 0 0.25 f (x) 20.5 11 2.5 4.8 10.8 15 16

39.



x 6 4 2 0 2 4 6 f (x) 78.1 71.54 63.81 52.66 61.27 69.53 76.35


(x) at (0.28, 49).



40.






x 8 4 0 4 8 12 16 f (x) 6.2 6.4 6.5 6.7 6.9 7.3 8.9

41.g(x) =






x = 22



x 1 3 6 12 15 18 22 f (x) 4.58 1.91 0.13 2.58 3.84 5.14 7.94







x 0 1 2 3 4 5 6 f (x) 0 12.1 17.1 21 24.2 27.1 29.7







44.4 = +

SOLUTION:


x = 11

x= 75


45.0.5x = +2

SOLUTION:


x = 0

x= 4


46.3 =

SOLUTION:


x = 13

x= 1.75


47. 10 = 17

SOLUTION:



48.

SOLUTION:



49.x = +2

SOLUTION:


x = 2

x= 4


50.7 + =250

SOLUTION:



51.x = 5 +

SOLUTION:


x = 8

x= 3


52. + 4 =

SOLUTION:


x = 2

x= 10


53. = 20

SOLUTION:



54. 1 =

SOLUTION:


x = 1


55.

SOLUTION:




56.y = x4a

SOLUTION:


andn = 4a.

57.G(x) = 2ax4

SOLUTION:


and n = 4.

58.F(b) = 3ab5x


59.y = tab

SOLUTION:


andn = ab.

60.

SOLUTION:


andn = 2b.

61.y = 4abx2




meter), and A is

the molecular mass.





SOLUTION:







Element 1 Element 2


63.

SOLUTION:


x = 1


64.

SOLUTION:


x = 5

The solution is x6.

65.

SOLUTION:


x = 7



The solution accou

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Solve for - Montville Township Public Schools · x í6 í4 í2 0 2 4 6 f(x) 8.6 7.9 6.9 0 í6.9 í7.9 í8.6 62/87,21 Evaluate the function for several x-values in its domain. Use