college algebra fall 2015 final exam prep solutions 3√−8 1 ......12. mixture - amy wants to make...
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College Algebra Fall 2015 Final Exam Prep – Solutions
1. Simplify 3√−8
= 3√−1 ∙ √8
= 3𝑖√4 ∙ √2
= 3𝑖 ∙ 2√2
= 6𝑖√2
1. 6𝑖√2
2. Simplify 𝑖85
85
4= 21
1
4; note the remainder is 1
𝑖85 = 𝑖1
2. 𝑖
3. Simplify 19 − (3 − 14𝑖) + 9𝑖
= 19 − 3 + 14𝑖 + 9𝑖
= 16 + 23𝑖
3. 16 + 23𝑖
4. Simplify (4 − 3𝑖)(4 + 3𝑖)
= 42 + 32
= 25
4. 25
5. Simplify (1 − 2𝑖)3
= [(1 − 2𝑖)(1 − 2𝑖)](1 − 2𝑖)
= (1 − 4𝑖 + 4𝑖2)(1 − 2𝑖)
= (1 − 4𝑖 + 4(−1))(1 − 2𝑖)
= (−3 − 4𝑖)(1 − 2𝑖)
= −3 + 2𝑖 + 8𝑖2
= −3 + 2𝑖 + 8(−1)
= −11 + 2𝑖
5. −11 + 2𝑖
6. Simplify 6+21𝑖
3𝑖
= (6 + 21𝑖
0 + 3𝑖) (
0 − 3𝑖
0 − 3𝑖)
=−18𝑖 − 63𝑖2
−9𝑖2
=−18𝑖 − 63(−1)
−9(−1)
=63 − 18𝑖
9
=63
9−
18
9𝑖
= 7 − 2𝑖
6. 7 − 2𝑖
7. Simplify 3−5𝑖
8−36𝑖
= (3 − 5𝑖
8 − 36𝑖) (
8 + 36𝑖
8 + 36𝑖)
=24 + 68𝑖 − 180𝑖2
82 + 362
=24 + 68𝑖 − 180(−1)
1360
=204 + 68𝑖
1360
=204
1360+
68
1360𝑖
=3 ∙ 68
20 ∙ 68+
1 ∙ 68
20 ∙ 68𝑖
=3
20+
1
20𝑖
7. 3
20+
1
20𝑖
8. Simplify 23−11𝑖
5+𝑖−
3+37𝑖
5+𝑖
=23 − 11𝑖 − 3 − 37𝑖
5 + 𝑖
=20 − 48𝑖
5 + 𝑖
= (20 − 48𝑖
5 + 1𝑖) (
5 − 1𝑖
5 − 1𝑖)
=100 − 260𝑖 + 48𝑖2
52 + 12
=100 − 260𝑖 + 48(−1)
26
=52 − 260𝑖
26
=52
26−
260
26𝑖
= 2 − 10𝑖
8. 2 − 10𝑖
9. Plot −5 + 2𝑖
10. Find |11 + 3𝑖|
= √112 + 32
= √130
10. √130
11. Write the complex number that is plotted below.
11. −5 + 2𝑖
12. Mixture - Amy wants to make 14 gallons of a 54% alcohol
solution by mixing together a 63% alcohol solution and pure water. How many gallons of pure water must she use?
. 63(14 − 𝑥) = .54(14) . 63(14) − .63𝑥 = .54(14) 8.82 − .63𝑥 = 7.56 −.63𝑥 = −1.26 𝑥 = 2 She must use 2 gallons of pure water.
12. ________________________________
13. Uniform Motion - Brenda left the hospital and drove south.
Mandy left two hours later driving 16 mph faster in an effort to catch up to her. After three hours Mandy finally caught up. What was Brenda’s average speed? Let 𝑥 = 𝑅Brenda 𝐷Brenda = 𝐷Mandy
𝑅Brenda ∙ 𝑇Brenda = 𝑅Mandy ∙ 𝑇Mandy
𝑥(5) = (𝑥 + 16)(3)
5𝑥 = 3𝑥 + 48
2𝑥 = 48
𝑥 = 24 mph (Brenda)
13. 𝑥 = 24 mph (Brenda)
14. Work - Working alone, Ted can sweep a porch in 11 minutes. One day his friend Cindy helped him and it only took 5.74 minutes. Find how long it would take Cindy to do it alone. (Round to the nearest hundredth.) Let 𝑥 = length of time for Cindy to sweep the porch alone 1
11+
1
𝑥=
1
5.74
(1
11) (
5.74𝑥
5.74𝑥) + (
1
𝑥) (
11(5.74)
11(5.74)) = (
1
5.74) (
11𝑥
11𝑥)
5.74𝑥 + 11(5.74) = 11𝑥
5.74𝑥 + 63.14 = 11𝑥
63.14 = 5.26𝑥
𝑥 = 12
14. 𝑥 = 12
15. Solve 4𝑥2 + 9 = −5
(Complex plane) 4𝑥2 = −14
𝑥2 =−14
4
𝑥 = ±√−14
4
𝑥 =±𝑖√14
2
15. 𝑥 =
±𝑖√14
2
16. Solve 𝑥2 = −3𝑥 + 4
(Complex plane) 𝑥2 + 3𝑥 − 4 = 0
(𝑥 + 4)(𝑥 − 1) = 0
𝑥 = −4, 𝑥 = 1
16. 𝑥 = −4, 𝑥 = 1
17. Solve 5𝑥2 = −5 + 𝑥 (Complex Plane) 5𝑥2 − 𝑥 + 5 = 0
𝑎 = 5, 𝑏 = −1, 𝑐 = 5
𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
𝑥 =−(−1) ± √(−1)2 − 4(5)(5)
2(5)
𝑥 =1 ± √−99
10
𝑥 =1 ± √−1 ∙ √9 ∙ √11
10
𝑥 =1 ± 3𝑖√11
10
𝑥 =1
10±
3√11
10𝑖
17. ________________________________
18. Solve by completing the square 𝑥2 − 4𝑥 − 42 = 8.
(Complex plane) (You must clearly show each step) 𝑥2 − 4𝑥 = 50
𝑥2 − 4𝑥 + 4 = 50 + 4
(𝑥 − 2)2 = 54
𝑥 − 2 = ±√54
𝑥 = 2 ± √9 ∙ √6
𝑥 = 2 ± 3√6
17. 𝑥 = 2 ± 3√6
19. Find the discriminant and state the number and type of solution. 10𝑥2 + 5𝑥 − 11 = −3 (Complex plane) 10𝑥2 + 5𝑥 − 8 = 0
𝑎 = 10, 𝑏 = 5, 𝑐 = −8
discriminant = 𝑏2 − 4𝑎𝑐
discriminant = 52 − 4(10)(−8)
discriminant = 345
Since the discriminant is positive and not a perfect square,
there are two real irrational solutions
19. discriminant = 345
Since the discriminant is positive and not a perfect square, there are two real irrational solutions
20. Solve
1
𝑥+
1
𝑥2+5𝑥=
4
𝑥2+5𝑥
(Must check for extraneous solutions) 1
𝑥+
1
𝑥(𝑥 + 5)=
4
𝑥(𝑥 + 5)
(1
𝑥) (
𝑥 + 5
𝑥 + 5) +
1
𝑥(𝑥 + 5)=
4
𝑥(𝑥 + 5)
𝑥 + 5 + 1 = 4
𝑥 = −2
Check:
1
−2+
1
−2(−2 + 5) ?
4
−2(−2 + 5)
−1
2+
1
−6 ?
4
−6
−3
6−
1
6 ? −
4
6
−4
6= −
4
6; it checks
𝑥 = −2
20. ________________________________
21. Solve 𝑥−5
𝑥2−3𝑥= 1 −
𝑥+2
𝑥
(Must check for extraneous solutions)
𝑥 − 5
𝑥2 − 3𝑥= 1 −
𝑥 + 2
𝑥
𝑥 − 5
𝑥(𝑥 − 3)=
𝑥(𝑥 − 3)
𝑥(𝑥 − 3)−
(𝑥 + 2)(𝑥 − 3)
𝑥(𝑥 − 3)
𝑥 − 5 = 𝑥2 − 3𝑥 − (𝑥2 − 𝑥 − 6)
𝑥 − 5 = −3𝑥 + 𝑥 + 6
−11 = −3𝑥
𝑥 =11
3
Check:
113 − 5
(113 )
2
− 3 (113 )
? 1 −
113 + 2
113
−6
11=
−6
11
𝑥 =11
3
21. 𝑥 =
11
3
22. Solve
5𝑥+1
𝑥+3−
𝑥+4
𝑥+3= 0
(Must check for extraneous solutions)
5𝑥 + 1 − (𝑥 + 4)
𝑥 + 3= 0
4𝑥 − 3
𝑥 + 3= 0
4𝑥 − 3 = 0
𝑥 =3
4
Check
5 (34) + 1
34 + 3
−
34 + 4
34 + 3
? 0
0 = 0
𝑥 =3
4
22. 𝑥 =
3
4
23. Solve 𝑥 − 8 = √43 − 6𝑥 (Must check for extraneous solutions)
(𝑥 − 8)2 = (√43 − 6𝑥)2
𝑥2 − 16𝑥 + 64 = 43 − 6𝑥
𝑥2 − 10𝑥 + 21 = 0
(𝑥 − 3)(𝑥 − 7) = 0
𝑥 = 3, 𝑥 = 7
Check
3 − 8 ? √43 − 6(3)
−5 ≠ √25
𝑥 = 3 is an extraneous solution
7 − 8 ? √43 − 6(7)
−1 ≠ √1
𝑥 = 7 is an extraneous solution
No Solution
23. No Solution
24. Solve √8 − 𝑥 − √1 − 3𝑥 = −1 (Must check for extraneous solutions)
√8 − 𝑥 = √1 − 3𝑥 − 1
(√8 − 𝑥)2
= (√1 − 3𝑥 − 1)2
8 − 𝑥 = 1 − 3𝑥 − 2√1 − 3𝑥 + 1
8 − 𝑥 = 2 − 3𝑥 − 2√1 − 3𝑥
6 + 2𝑥 = 2√1 − 3𝑥
3 + 𝑥 = √1 − 3𝑥
(3 + 𝑥)2 = (√1 − 3𝑥)2
9 + 6𝑥 + 𝑥2 = 1 − 3𝑥
𝑥2 + 9𝑥 + 8 = 0
(𝑥 + 8)(𝑥 + 1) = 0
𝑥 = −8, 𝑥 = −1
Check
√8 − (−8) − √1 − 3(−8) ? −1
√16 − √25 ? −1
4 − 5 = −1
√8 − (−1) − √1 − 3(−1) ? −1
√9 − √4 ? −1
3 − 2 ≠ −1
𝑥 = −1 is an extraneous solution
𝑥 = −8
24. 𝑥 = −8
25. Solve √2𝑥 − 5 = −2 + √4𝑥 − 3 (Must check for extraneous solutions)
(√2𝑥 − 5)2
= (−2 + √4𝑥 − 3)2
2𝑥 − 5 = 4 − 4√4𝑥 − 3 + 4𝑥 − 3
2𝑥 − 5 = 1 + 4𝑥 − 4√4𝑥 − 3
−2𝑥 − 6 = −4√4𝑥 − 3
𝑥 + 3 = 2√4𝑥 − 3
(𝑥 + 3)2 = (2√4𝑥 − 3)2
𝑥2 + 6𝑥 + 9 = 4(4𝑥 − 3)
𝑥2 + 6𝑥 + 9 = 16𝑥 − 12
𝑥2 − 10𝑥 + 21 = 0
(𝑥 − 3)(𝑥 − 7) = 0
𝑥 = 3, 𝑥 = 7
Check
√2(3) − 5 ? −2 + √4(3) − 3
√6 − 5 ? −2 + √12 − 3
1 ? −2 + 3
1 = 1
√2(7) − 5 ? −2 + √4(7) − 3
√9 ? −2 + √25
3 = 3
𝑥 = 3, 𝑥 = 7
25. 𝑥 = 3, 𝑥 = 7
26. Solve {
−6𝑥 + 2𝑦 − 6𝑧 = −12 −6𝑥 + 3𝑦 − 𝑧 = 14
−5𝑥 − 2𝑦 − 𝑧 = −16
Equation 1Equation 2Equation 3
Eliminate z Eqn 1 − 6 ∙ (Eqn 2) −6𝑥 + 2𝑦 − 6𝑧 = −12 36𝑥 − 18𝑦 + 6𝑧 = −84 30𝑥 − 16𝑦 = −96 15𝑥 − 8𝑦 = −48 Equation 4 Eliminate z Eqn 1 − 6 ∙ (Eqn 3) −6𝑥 + 2𝑦 − 6𝑧 = −12 30𝑥 + 12𝑦 + 6𝑧 = 96 24𝑥 + 14𝑦 = 84 12𝑥 + 7𝑦 = 42 Equation 5 Eliminate y 7 ∙ (Eqn 4) + 8 ∙ (Eqn 5) 105𝑥 − 56𝑦 = −336 96𝑥 + 56𝑦 = 336 201𝑥 = 0 𝑥 = 0 15(0) − 8𝑦 = −48 𝑦 = 6 −6(0) + 2(6) − 6𝑧 = −12 −6𝑧 = −24 𝑧 = 4 (𝑥, 𝑦, 𝑧) = (0, 6, 4)
26. (𝑥, 𝑦, 𝑧) = (0, 6, 4)
28. Determine the domain and range of the relation 𝐴 = {(1, 3), (3, 4), (5, 3), (3, 6)}
28. 𝐷 = {1, 3, 5} 𝑅 = {3, 4, 6}
29. Determine the domain and range of the relation.
(Express your answer using interval notation)
29. 𝐷 = [1, ∞) 𝑅 = [2, ∞)
30. Is the relation {(1, 3), (−3, 4), (5, 3), (3, 6)} a function? 30. Yes
31. Is the relation a function?
31. No
32. Does the equation |𝑥2 − 4| = 5𝑦 define y as a function of x? 32. Yes
33. Does the equation 𝑥 ± 𝑦 = 9 define y as a function of x? 33. No
34. Does the equation 15𝑥 = 𝑦2 + 1 define y as a function of x? 34. No
35. Find the domain of the function, 𝑓(𝑥) = 𝑥2 + 9𝑥 + 20 35. 𝐷 = (−∞, ∞)
36. Find the domain of the function 𝑓(𝑥) = √𝑥 − 10 + 3
𝑥 − 10 ≥ 0
𝑥 ≥ 10
𝐷 = [10, ∞)
36. 𝐷 = [10, ∞)
37. Find the domain of the function 𝑓(𝑥) =𝑥+1
𝑥2−9
𝑥2 − 9 ≠ 0
(𝑥 + 3)(𝑥 − 3) = 0
𝑥 = −3, 𝑥 = 3
𝐷 = (−∞, −3) ∪ (−3, 3) ∪ (3, ∞)
37. 𝐷 = (−∞, −3) ∪ (−3, 3) ∪ (3, ∞)
38. Find the domain of the function 𝑓(𝑥) =
√𝑥+5
𝑥2+5𝑥+6
𝑥 + 5 ≥ 0 and 𝑥2 + 5𝑥 + 6 ≠ 0
𝑥 ≥ −5 (𝑥 + 3)(𝑥 + 2) = 0
𝑥 = −3, 𝑥 = −2
𝐷 = [−5, −3) ∪ (−3, −2) ∪ (−2, ∞)
38. 𝐷 = [−5, −3) ∪ (−3, −2) ∪ (−2, ∞)
39. Find the domain of the function 𝑓(𝑥) =
2𝑥
√1−𝑥
1 − 𝑥 > 0
1 > 𝑥
𝐷 = (−∞, 1)
39. 𝐷 = (−∞, 1)
42. Let 𝑓(𝑥) =
𝑥+2
𝑥−5 and 𝑔(𝑥) =
𝑥−1
𝑥+4, find the domain of
(𝑓𝑔)(𝑥) 𝐷𝑓: 𝑥 ≠ 5
𝐷𝑔: 𝑥 ≠ −4
𝐷𝑓𝑔 = 𝐷𝑓 ∩ 𝐷𝑔
𝐷𝑓𝑔 = (−∞, −4) ∪ (−4, 5) ∪ (5, ∞)
42. 𝐷𝑓𝑔 = (−∞, −4) ∪ (−4, 5) ∪ (5, ∞)
43. Let 𝑓(𝑥) =
𝑥+2
𝑥−5 and 𝑔(𝑥) =
𝑥−1
𝑥+4,
find the domain of (𝑓
𝑔) (𝑥)
𝐷𝑓: 𝑥 ≠ 5
𝐷𝑔: 𝑥 ≠ −4
𝑔(𝑥) ≠ 0 ⇒ 𝑥 ≠ 1
𝐷𝑓𝑔
= (−∞, −4) ∪ (−4, 1) ∪ (1, 5) ∪ (5, ∞)
43. 𝐷𝑓𝑔
= (−∞, −4) ∪ (−4, 1) ∪ (1, 5) ∪ (5, ∞)
44. Let 𝑓(𝑥) =𝑥+2
𝑥−5 and 𝑔(𝑥) =
𝑥−1
𝑥+4,
find the domain of (𝑔
𝑓) (𝑥)
𝐷𝑓: 𝑥 ≠ 5
𝐷𝑔: 𝑥 ≠ −4
𝑓(𝑥) ≠ 0, ⇒ 𝑥 ≠ −2
𝐷𝑔𝑓
= (−∞, −4) ∪ (−4, −2) ∪ (−2, 5) ∪ (5, ∞)
44. 𝐷𝑔𝑓
= (−∞, −4) ∪ (−4, −2) ∪ (−2, 5) ∪ (5, ∞)
45. Let 𝑓(𝑥) =
𝑥+2
𝑥−5 and 𝑔(𝑥) =
𝑥−1
𝑥+4, find the
domain of (𝑓 ∘ 𝑔)(𝑥) 𝑥 ∈ 𝐷𝑔 and 𝑔(𝑥) ∈ 𝐷𝑓
𝐷𝑔: 𝑥 ≠ −4 𝑔(𝑥) ≠ 5
𝑥−1
𝑥+4=
5
1
5𝑥 + 20 = 𝑥 − 1 4𝑥 = −21
𝑥 =−21
4
𝑥 = −5.25 𝑔(−5.25) = 5 𝑥 ≠ −5.25 𝐷𝑓𝑔 = (−∞, −5.25) ∪ (−5.25, −4) ∪ (−4, ∞)
45. 𝐷𝑓𝑔 = (−∞, −5.25) ∪ (−5.25, −4) ∪ (−4, ∞)
46. Let 𝑓(𝑥) =
𝑥+2
𝑥−5 and 𝑔(𝑥) =
𝑥−1
𝑥+4, find the
domain of (𝑔 ∘ 𝑓)(𝑥) 𝑥 ∈ 𝐷𝑓 and 𝑓(𝑥) ∈ 𝐷𝑔
𝑥 ≠ 5 𝑓(𝑥) ≠ −4
𝑥 + 2
𝑥 − 5=
−4
1
𝑥 + 2 = −4𝑥 + 20 5𝑥 = 18
𝑥 =18
5
𝑥 = 3.6 𝑓(3.6) = −4 𝑥 ≠ 3.6 𝐷𝑓∘𝑔 = (−∞, 3.6) ∪ (3.6,5) ∪ (5, ∞)
46. 𝐷𝑓∘𝑔 = (−∞, 3.6) ∪ (3.6,5) ∪ (5, ∞)
47. Let 𝑓(𝑥) = {𝑥2 + 1 𝑥 < 1
2𝑥 − 1 𝑥 ≥ 1
I. 𝑓(−2) = 5
II. 𝑓(1) = 1
III. 𝑓(4) = 7
48. Let 𝑓(𝑥) = 10 − 3𝑥
I. 𝑓(𝑥 + 2) = −3𝑥 + 4
II. 𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ= −3
48.
49. Let 𝑓(𝑥) = 𝑥2 − 11 and 𝑔(𝑥) = 5𝑥 + 4
I. (𝑓 ∘ 𝑔)(𝑥) = 25𝑥2 + 18𝑥 + 5 II. (𝑔 ∘ 𝑓)(𝑥) = 5𝑥2 − 51
49.
50.
50. ℎ(6) = 4 ℎ(5) = 6 ℎ(3) = 3 ℎ(5) = 5
51. Find the inverse of the relation {(−5, −1), (0, 4), (2, 6), (5, 9)} 51. {(−1, −5), (4, 0), (6, 2), (9, 5)}
52. Find the inverse of 𝑓(𝑥) =
1
𝑥−2
𝑦 =1
𝑥 − 2
𝑥 =1
𝑦 − 2
𝑥(𝑦 − 2) = 1
𝑦 − 2 =1
𝑥
𝑦 =1
𝑥+ 2
𝑓−1(𝑥) =1
𝑥+ 2
52. 𝑓−1(𝑥) =
1
𝑥+ 2
53.)
54.)
55.) It passes the vertical line test. It is a function It fails the horizontal line test. It is not a 1-1 function