Math 1005
Exam 3 Review
1. Determine which of the following defines a function.
(a) {(a, 3), (b, 5), (c, 9), (d, 9)}
Yes
(b) {(a, −3), (b, 6), (c, 1), (b, 9)}
No, b corresponds to both 6 and 9.
(c) y
x
No, fails the vertical line test.
(d) y
x
Yes, passes the vertical line test.
2. What is the domain of the following functions?
(a) f (x) = 3 −
3
5−x
5 − x 6= 0
5 6= x
Domain: (−∞, 5) ∪ (5, ∞)
√
(b) g(x) = x − 1
x−1 ≥ 0
x≥1
Domain: [1, ∞)
(c) h(x) = |x − 9| − 4
Domain: (−∞, ∞)
3. Let f (x) = −2x2 − 5. Find
(a) f (3) = −2(3)2 − 5 = −2(9) − 5 = −18 − 5 = −23
(b) f (a) = −2a2 − 5
(c) f (a + 3) = −2(a + 3)2 − 5 = −2(a2 + 6a + 9) − 5 = −2a2 − 12a − 18 − 5 = −2a2 − 12a − 23
4. Let f (x) =
x2 + 2
. Find
2−x
(a) f (4) =
42 + 2
18
=
= −9
2−4
−2
(b) f (x2 ) =
(x2 )2 + 2 x4 + 2
=
2 − x2
2 − x2
(c) f (a + 2) =
(a + 2)2 + 2 a2 + 4a + 4 + 2
a2 + 4a + 6
=
=−
2 − (a + 2)
2−a−2
a
5. Find the domain, x-intercepts, and y-intercept of the following functions.
2x − 5
3x + 2
Domain: 3x + 2 6= 0
2
x 6= −
3
(−∞, − 23 ) ∪ (− 32 , ∞)
(a) f (x) =
x-intercept: 2x − 5 = 0
5
x=
2
y-intercept: f (0) =
2(0) − 5
5
=−
3(0) + 2
2
5
2
(b) g(x) = −5x − 12
Domain: (−∞, ∞)
y=−
x-intercept: −5x − 12 = 0
12
x=−
5
y-intercept: f (0) = −5(0) − 12 = −12
y = −12
6. Use the graph of the function below to find the following:
(a) Domain
[−6, 6]
(b) Range
[−2, 2]
(c) Intervals where f is increasing
[−4, −2] ∪ [1, 2]
(d) Intervals where f is decreasing
[−6, −4] ∪ [−2, 1]
(e) Intervals where f is constant
[2, 6]
7. Use the graph of the function below to find the following:
(a) Domain
[−4, 4)
(b) Range
[−2, 5)
8.
f (x) =
(a) Find f (−2), f (−1), f (0).
f (−2) = −2 + 2 = 0
f (−1) = −1 − 2 = −3
f (0) = 0 − 2 = −2
(b) Graph f
3
2
1
-4
-2
2
4
-1
-2
-3
(c) Find the domain and range of f .
Domain: (−∞, ∞)
Range: (−∞, ∞)
x + 2 : x < −1
x − 2 : x ≥ −1
9. Use transformations to sketch a graph of the function f (x) = 3 + |x − 2|
8
6
4
2
-2
2
4
6
√
10. Use transformations to sketch a graph of the function g(x) = − x + 1 − 2
1
-2
-1
1
2
3
-1
-2
-3
-4
-5
11. Test whether the following functions are even, odd, or neither.
(a) f (x) = −2x4 − 5x2
f (−x) = 2(−x)4 − 5(−x)2 = 2x4 − 5x2 = f (x)
f (x) is even.
(b) g(x) = 7x3 − 2x5
g(−x) = 7(−x)3 − 2(−x)5 = −7x3 + 2x5 = −(7x3 − 2x5 ) = −g(x)
g(x) is odd.
12. Graph f (x) = −x2 + 6x + 5
−b
−6
Vertex:
=
= −3
2a
2(1)
f (−3) = (−3)2 + 6(−3) + 5 = −4
(−3, −4)
x-intercepts: x2 + 6x + 5 = 0
(x + 5)(x + 1) = 0
x = −5, x = −1
10
5
-6
-4
-2
13.
(a) Find the vertex form of f (x) = 2x2 − 4x + 8
f (x) = (2x2 − 4x) + 8
f (x) = 2(x − 2x) + 8
2
2
f (x) = 2(x − 2x + −2
) + 8 − 2( −2
)
2
2
2
f (x) = 2(x − 1) + 6
(b) What is the vertex of f ?
(1, 6)
(c) The graph of f is a parabola. Does the parabola open up or down?
up
14. Solve the quadratic inequality x2 + x < 12.
x2 + x − 12 < 0
x-intercepts: x2 + x − 12 = 0
(x + 4)(x − 3) = 0
x = −4, x = 3
5
-4
-2
2
4
-5
-10
(−4, 3)
15. The annual U.S. box office revenue in billions of dollars can be modeled by the function
B(x) = −0.19x2 + 1.2x + 7.6 where x is years after 2002. What is the maximum box office revenue
achieved?
−b
−1.2
Vertex: x =
=
= 3.16
2a
2(−0.19)
B(3.16) = 0.19(3.16)2 + 1.2(3.16) + 7.6 = 9.49
Since the parabola opens down (3.16, 9.49) is a maximum
Therefore, the maximum box office revenue achieved is 9.49 billion dollars.
16. Given f (x) = 3x + 2 and g(x) = x2 − 9 find the following:
(a) ( f + g)(x) and its domain.
( f + g)(x) = (3x + 2) + (x2 − 9) = x2 + 3x − 7
Domain: (−∞, ∞)
(b) ( f · g)(x) and its domain.
( f · g)(x) = (3x + 2)(x2 − 9) = 3x3 + 2x2 − 27x − 18
Domain: (−∞, ∞)
(c) ( f − g)(x) and its domain. ( f − g)(x) = (3x + 2) − (x2 − 9) = 3x + 2 − x2 + 9 = −x2 + 3x + 11
Domain: (−∞, ∞)
f
(d) ( )(x) and its domain.
g
f
3x + 2
( )(x) = 2
g
x −9
Domain: x2 − 9 6= 0
(x + 3)(x − 3) 6= 0
x 6= 3, x 6= −3
(−∞, −3) ∪ (−3, 3) ∪ (3, ∞)
√
17. Given f (x) = 4 − x and g(x) = x2 find the following:
Domain of f (x): (−∞, 4]
Domain of g(x): (−∞, ∞)
(a) ( f ◦ g)(x)√and its domain.
f (x2 ) = 4 − x2
Domain: 4 − x2 ≥ 0
x-intercepts: 4 − x2 = 0
−(x − 2)(x + 2) = 0
x = −2, x = 2
4
2
-3
-2
-1
1
2
3
-2
-4
(−2, 2) ∩ (−∞, ∞) = (−2, 2)
(b) (g√
◦ f )(x) and √
its domain.
g( 4 − x) = ( 4 − x)2 = 4 − x
Domain: (−∞, ∞) ∩ (−∞, 4] = (−∞, 4]
x
2x − 4
and g(x) =
find the following:
x−1
x
Domain of f (x): (−∞, 1) ∪ (1, ∞)
Domain of g(x):(−∞, 0) ∪ (0, ∞)
18. Given f (x) =
(a) ( f ◦ g)(x) and its domain.
2x − 4
2x − 4
2x − 4
2x − 4
2x − 4
2x − 4
x
2x − 4
x
x
x
f
=
=
=
= x =
·
=
2x − 4
2x − 4 x
2x − 4 − x
x−4
x
x
x−4
x−4
−1
−
x
x
x
x
x
Domain: All real numbers except x can’t equal 4 or 0.
(−∞, 0) ∪ (0, 4) ∪ (4, ∞)
(b) (g ◦ f )(x) and itsdomain.
x
4(x − 1)
2x
2x − 4x + 4
2
−4
−
x
4 − 2x x − 1 4 − 2x
x−1
x−1
g
=
= x−1 x x−1 =
=
·
=
x
x
x−1
x−1
x
x
x−1
x−1
x−1
Domain: All real numbers except x can’t equal 0 or 1.
(−∞, 0) ∪ (0, 1) ∪ (1, ∞)
19. Determine if the following functions are one-to-one.
(a) 4 + x2 , x ≥ 0
25
20
15
10
5
-2
-1
1
2
3
4
5
Yes, passes the horizontal line test.
(b) g(x)
No, fails the horizontal line test.
20. Find the inverse of the following functions and state the domain and range of the inverse.
(a) f (x) = 4x − 3
Domain: (−∞, ∞)
Range: (−∞, ∞)
y = 4x − 3
x = 4y − 3
4y = x + 3
y = 41 x + 34
f −1 (x) = 41 x + 34
Domain: (−∞, ∞)
Range: (−∞, ∞)
4
2
-4
-2
2
-2
-4
4
√
(b) f (x) = x + 3
Domain: [0, ∞)
Range: [3, ∞)
√
y = x+3
√
x = y+3
√
y = 3−x
y = (3 − x)2
f −1 (x) = (3 − x)2
Domain: [3, ∞)
Range: [0, ∞)
10
8
6
4
2
2
4
6
8
21. List all the zeros and their multiplicities and find the degree of the following polynomial functions:
(a) P(x) = (x2 − 1)(x + 2)2 (x − 5)4 (x2 + 2)
(x2 − 1) = 0
(x + 1)(x − 1) = 0
(x2 + 2) = 0
x2 = −2
√
x = ± √−2
x = ±i 2
Zeros: 1 (with multiplicity
1), −1 (with multiplicity
1), −2 (with multiplicity 2), 5 (with
√
√
multiplicity 4), i 2 (with multiplicity 1), −i 2 (with multiplicity 1)
Degree of P(x): 1 + 1 + 2 + 4 + 1 + 1 = 10
(b) Q(x) = (x2 − 5x + 6)(x2 − 5x + 7)
x2 − 5x + 6 = 0
(x − 3)(x − 2) = 0
x2 − 5x +p7 = 0
√
√
√
5 ± 52 − 4(1)(7) 5 ± 25 − 28 5 ± −3 5 ± i 3
x=
=
=
=
2(1)
2
2
2
√
√
5+i 3 5−i 3
Zeros: 3, 2,
,
(all with multiplicity 1)
2
2
Degree of Q(x): 1 + 1 + 1 + 1 = 4
10
22. Use algebraic long division to find the quotient and the remainder of (11x − 2 + 12x2 ) ÷ (x + 2).
12x − 13
2
x+2
12x + 11x − 2
− 12x2 − 24x
− 13x − 2
13x + 26
24
Quotient: 12x − 13
Remainder: 24
23. Divide (2x2 + 7x − 5) ÷ (x + 4) using synthetic division.
7
−5
−8
4
−1
−1
2
−4
2
Quotient: 2x − 1
Remainder: -1
24. Test whether (x − 1) is a factor of 5x4 − 2x2 − 3.
5
1
5
0
−2
0
−3
5
5
3
3
5
3
3
0
Yes, since we get a remainder of 0 we know (x − 1) is a factor of P(x).
25. Find P(2) given P(x) = 4x6 + 20x5 − 24x4 − 3x2 − 13x + 30 using synthetic division.
4
2
4
20
− 24
0
−3
− 13
30
8
56
64
128
250
474
28
32
64
125
237
504
P(2) = 504
26. For the following question, do NOT graph the function on your calculator.
Does P(x) = x3 − 19x + 30 have at least one real zero between −6 and −4? How do you know?
Show your work.
P(−6) = (−6)3 − 19(−6) + 30 = −72
P(−4) = (−4)3 − 19(−4) + 30 = 42
Since P(−6) is negative and P(−4) is positive we know there is at least one real zero between −6
and −4.
27. Solve the polynomial inequality P(x) = (x − 1)(x − 2)2 (x + 2)2 ≤ 0.
x-intercepts: 1 has odd degree, 2 has even degree, −2 has even degree.
P(−3) = (−)(+)(+) = (−) Therefore, the graph starts below the x-axis.
10
-2
-1
1
2
-10
-20
(−∞, 1]
28. List all the possible rational zeros of x4 + 5x2 + 4.
Factors of 4: ±1, ±4, ±2
Factors of 1: ±1
Possible rational zeros: ±1, ±4, ±2
29. Given 3 is a zero of P(x) = 6x3 − 17x2 − 4x + 3, write P(x) as a product of linear factors.
6
3
6
− 17
−4
3
18
3
−3
1
−1
0
P(x) = (x − 3)(6x2 + x − 1)
P(x) = (x − 3)(2x + 1)(3x − 1)
30. Given
5
2
is a zero of P(x) = 2x3 − 9x2 − 2x + 30, find all the zeros of P(x) exactly.
2
5
2
2
−9
−2
30
5
− 10
− 30
−4
− 12
0
P(x) = (x − 25 )(2x2 − 4x − 12)
2x2 − 4x − 12 = 0
2(x2 − 2x − 6) = 0
x2 − 2x − 6 = 0
x2 − 2x = 6
2
−2 2
−2
2
x − 2x +
= 6+
2
2
(x − 1)2 =√7
x−1 = √
± 7
x = 1± 7
√
√
Zeros: 52 , 1 + 7, 1 − 7
31. Given 6 + i is a zero of P(x) = 2x2 − 24x + 74, find all the zeros of P(x) exactly.
Since 6 + i is a zero, its conjugate 6 − i must also be a zero. A polynomial of degree 2 can have at
most two zeros, so we have
Zeros: 6 + i, 6 − i
32. Find the horizontal and vertical asymptotes of f (x) =
3x2 + 8
2x2 + 6x
Vertical Asymptote:
2x2 + 6x = 0
2x(x + 3) = 0
x = 0, x = −3
Therefore, the vertical lines x = 0, and x = −3 are the vertical asymptotes of f (x)
Horizontal Asymptote:
The degree of the numerator and denominator are both 2, so there is a horizontal asymptote at
y = 23 .