Answer Key

Final Exam: Math 140

Sections A - G, J - N, P - R, T, and XW

Fall 2012

Question 1. Solve the following equation by the square root method: (3y + 9)2 = 81.

Solution. This is Problem #17, section 1.2.

3y + 9 = +9 (6 points)

3y = −9+9 (2 points)

3y = 0,−18 (1 point)

y = 0,−6 (1 point)

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Question 2. Solve the following inequality: x2 − 10x > 0. Write your answer in intervalnotation.

Solution. This is problem # 5, Section 4.5.

x(x− 1) > 0 (3 points)

-1-1 0 1 1110

+++++++++ - - - - - - - - - - ++++++++++++++

( 3 points)

x = −1: (−1)(−1− 1) > 0.

x = 1: (1)(1− 10) < 0.

x = 11: (11)(11− 10) > 0. (2 points)

(−∞, 0) ∪ (10,∞). (2 points)

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Question 3. Let R(x) = 7x+23x+4

. What is the equation of the horizontal or oblique asymptoteof R(x)?

Solution. This is problem # 7, Section 5.3.

Since numerator has same degree as denominator, R(x) has a horizontal asymptote. (4points) Since the leading coefficient of the numerator is 7 and the leading coefficient of thedenominator is 3, the equation of this asymptote is y = 7

3. (6 points)

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Question 4. Write the expression as a single logarithm. Express powers as factors.

log7

√x− log7 x

5.

Solution. This is problem # 29 of the homework for Section 6.5.

log7

√x− log7 x

5 = log7

√x

x5(4 points)

= log7 x12−5 (2 points)

= log7 x−9/2 (1 point)

= −9

2log7 x. (3 points)

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Question 5. Find the function that is finally graphed after the following transformationsare applied to the graph of y = x3 in the order listed. Simplify completely.

(a) Shift up 2 units.

(b) Reflect about the x-axis.

(c) Reflect about the y-axis.

Solution. This is problem # 17 of the homework for Section 3.5.

y = x3

(a): y = x3 + 2 (3 points)

(b): y = −(x3 + 2) = −x3 − 2 (3 points)

(c): y = −(−x)3 − 2 = x3 − 2 (4 points)

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Question 6. Construct a polynomial function f with the following characteristics.

• zeros: -1, 1, and 3

• degree 3

• y-intercept: -6

Solution. This is problem # 20 of the homework for Section 5.1.(x+ 1)(x− 1)(x− 3) has zeros at −1, 1, and 3 and its degree is 3 (4 points). However, itsy-intercept is (1)(−1)(−3) = 3. (3 points) To correct for this, we multiply by −2:

−2(x+ 1)(x− 1)(x− 3) (3 points)

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Question 7. For the function f defined by f(x) = 2x2 + 4x + 5, find the following values.Simplify completely.

(a) f(−3)

(b) f(x− 2).

Solution. This is problem # 23 of the homework for Section 3.1.

f(−3) = 18− 12 + 5 = 11. (3 points)

2(x− 2)2 + 4(x− 2) + 5(4 points) = 2(x2 − 4x+ 4) + 4x− 8 + 5

= 2x2 − 8x+ 8 + 4x− 8 + 5

= 2x2 − 4x+ 5 (3 points)

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Question 8. Let

f(x) =

x2 if x < 0−2 if x = 0

3x+ 1 if x > 0

Find:

(a) f(−3)

(b) f(0)

(c) f(4)

Solution. This is problem # 12 of the homework for Section 3.4.

f(−3) = (−3)2 = 9 (3 points)

f(0) = −2 (4 points)

f(4) = 3(4) + 1 = 13 (3 points)

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Question 9. Two vehicles, a car and a truck, leave an intersection at the same time. Thecar heads east at an average speed of 30 miles per hour, while the truck heads south at anaverage speed of 40 miles per hour. How far apart are the car and the truck after 2 hours?

Solution. Compare with Problem # 19 of the homework for Section 2.1.

60

80

(4 points)

√(80)2 + (60)2 (3 points) =

√6400 + 3600

=√

10, 000

= 100 (3 points)

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Question 10. Consider the function f(x) = log2(x− 6).

(a) Sketch a graph of f .

(b) Find the domain of f .

(c) Use the graph to determine the range of f .

(d) What is the domain of f−1?

(e) What is the range of f−1?

Solution. Compare with problem # 31 of the homework for Section 6.4.

(6 points)

domain of f = (6,∞)

range of f = (−∞,∞) (2 points)

domain of f−1 = (−∞,∞)

range of f−1 = (6,∞) (2 points)

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Question 11. Solve the equation 121x = 119x+2.

Solution. Compare with problem #30 of the homework for Section 6.3.

(112)x = 119x+2

112x = 119x+2 (3 points)

2x = 9x+ 2 (4 points)

0 = 7x+ 2

−2

7= x (3 points)

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Question 12. Construct a rational function that has the given graph.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Solution. Compare with problem # 11 of the homework for Section 5.3.

x2 (4 points)

(x− (−1))(x− 1) ( 6 points)

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Question 13. Use the graph to answer the following questions.

(a) What are the coordinates of the center of the circle? (1,−1) (1 point)

(b) What is the radius of the circle? 2 (1 point)

(c) What is the equation of the circle? (x− 1)2 + (y + 1)2 = 4 (3 points)

(d) What are the x-intercepts of the circle? (Note: you cannot find them by looking at thegraph.) (1 +

√3, 0), (1−

√3, 0)

Solution. Set y = 0:

(x− 1)2 + 1 = 4 (3 points)

(x− 1)2 = 3

x− 1 = +√

3

x = 1+√

3 (2 points)

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Question 14. A function g is given by the set of ordered pairs below:

g = {(−1, 2), (0, 1), (1, 4), (2,−1), (3, 5), (7,−3)}.

(a) g(3) =

(b) Solve g(x) = 5

We also want to make a new function h by adding an ordered pair as shown below:

h = {(−1, 2), (0, 1), (1, 4), (2,−1), (3, 5), (7,−3), ( , )}.

(c) Give an example of an ordered pair we could add so that h is still a one-to-one function.

(d) Give an example of an ordered pair we could add that would make h not a one-to-onefunction.

Solution. (a) g(3) = 5. (2 points)

(b) x = 5 (2 points)

(c) (100, 100) (3 points)

(d) (100,−3) (3 points)

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Question 15. The surface area S (in square meters) of a hot-air ballon is given by S(r) =4πr2 where r is the radius of the balloon (in meters). If the radius r is increasing with time t(in seconds) according to the formula r(t) = 2

3t3, t ≥ 0, find the surface area S of the balloon

as a function of the time t.

Solution.

S(r(t)) (4 points) = 4π

(2

3t3)2

(3 points)

= 4π4

9t6

=16

9πt6, t ≥ 0 (3 points)

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Question 16. Find the domain of each function below.

(a) h(x) =√

3x− 1.

(b) g(x) = 2x2−2x−3

.

Solution. (a) 3x− 1 ≥ 0 (3 points), x ≥ 13. (2 points)

(b) g(x) = 2(x−3)(x+1)

(3 points), x 6= 3,−1. (2 points)

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