math 1111 final exam review. 1. identify the type of function. f(x) = 5 constant
TRANSCRIPT
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Math 1111
Final Exam Review
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1. Identify the type of function.
f(x) = 56
4
2
-2
-5 5
Constant
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-2 2
2
1
-1
2. Identify the type of function.
4)(
2 x
xxf
Quotient of two polynomials.
RationalFunction
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3. Identify the type of function.
1572)( 3 xxxf
PolynomialFunction
Highest Exponent 3
-10 -5 5 10
10
8
6
4
2
-2
-4
-6
-8
-10
-12
-14
-16
-18
-20
-22
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4. Identify the type of function.
2847)( 2 xxxf
QuadraticFunction
Highest Exponent 2
-10 10
55
50
45
40
35
30
25
20
15
10
5
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5. Identify the type of function.
23)( 1 xxf
ExponentialFunction
Variable in the Exponent
8
6
4
2
-2
-5 5
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6. Identify the type of function.
6
4
2
-2
-5 5
Linear
3)( xxf
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7. $1500 is invested at a rate of 6¼% compounded continuously. What is the balance at the end of two years?
• A = balance at the end of investment period
• P = Principal (Money invested)
• r = rate = 0.0625• t = time in years = 2
rtPeA)2)(0625.0(1500eA
)125.0(1500eA)13315.1(1500A
725.1699A $1699.73
Continuous Compounding
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8. Evaluate the expression.
• Round your answer to 3 decimal places.
37.4 e
≈ 26.565
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10. Evaluate the expression.
• When t = 15.
• Round your answer to 2 decimal places.
te 076.0300
≈ 95.95
)15(076.0300 e14.1300 e
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11. What transformations are used to create the “child”?
xxf 2)( Parent
52)( xxgChild
6
4
2
-2
-5 5
4
2
-2
-5 5
1. Reflection of x-axis.
2. Vertical shift of 5 units.
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12. What transformations are used to create the “child”?
xxf 2)( Parent
12)( 4 xxgChild
6
4
2
-2
-5 5
6
4
2
51. Reflection of y-axis.
2. Vertical shift up of 1 unit.
3. Horizontal shift right of 4 units.
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13. $2100 is invested at a rate of 7% compounded monthly. What is the balance at the end of 10 years?
• A = balance at the end of investment period
• P = Principal (Money invested)
• r = rate = 0.07• n = number of periods
per year = 12• t = number of years
nt
n
rPA
1
$4220.29
)10(12
12
07.012100
A
12000583.12100A 00966.22100A
n Compoundings per year
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15a. Given f(x) = x3 – 2x2 – 21x – 18answer the following questions.
• What is the degree of the polynomial?
3• According to the Fundamental Theorem of Algebra, how many zeros will
this polynomial have?
3• Use Descartes’ Rule of Signs to determine the number of possible positive
real zeros.
1 possible positive zero, one sign variation• Use Descartes’ Rule of Signs to determine the number of possible negative
real zeros.
f(-x) = (-x)3 – 2(-x)2 -21(-x) – 18 = -x3 – 2x2 + 21x - 18 2 or 0 possible negative zeros, two sign variations
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15b. Given f(x) = x3 – 2x2 – 21x – 18answer the following questions.
• Use the Rational Root Test to list All possible rational zeros of the polynomial.
• Where p is a factor of -18 and q is a factor of 1
p
q
1; 2; 3; 6; 9; 18p
q
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Given all of these candidates, how can you tell which is the actual root?
f(x) = x3 – 2x2 – 21x – 18
1; 2; 3; 6; 9; 18p
q
Synthetic Division
Graphing calculator
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15c. Given f(x) = x3 – 2x2 – 21x – 18answer the following questions.
• Use the synthetic division to find all zeros of the polynomial.
1 -2 -21 -18-1
1
-1
-3
3
-18
18
0
)183))(1(( 2 xxx
)183)(1( 2 xxx)3)(6)(1( xxx
}6,1,3{ xEach zero has a multiplicity of one.
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16a. Given f(x) = x4 + x3 – 11x2 + x – 12answer the following questions.
• What is the degree of the polynomial?
4• According to the Fundamental Theorem of Algebra, how many zeros will
this polynomial have?
4• Use Descartes’ Rule of Signs to determine the number of possible positive
real zeros.3 or 1 possible positive zero, three sign variations
• Use Descartes’ Rule of Signs to determine the number of possible negative real zeros.
f(-x) = (-x)4 + (-x)3 – 11(-x)2 + (-x) – 12 = x4 – x3 – 11x2 – x – 12
1 possible negative zero, one sign variation
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16b. Given f(x) = x4 + x3 – 11x2 + x – 12answer the following questions.
• Use the Rational Root Test to list All possible rational zeros of the polynomial.
• Where p is a factor of -12 and q is a factor of 1
p
q
12;6;4;3;2;1 q
p
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16c. Given f(x) = x4 + x3 – 11x2 + x – 12answer the following questions.1 1 -11 1 -123
1
3
4
12
1
3
4
0)44)(3( 23 xxxx
{ 4, 3, , }x i i Extract
the root.
12
0
0)]4(1)4()[3( 2 xxxx
0)1)(4)(3( 2 xxx
Factor by grouping
2( 1) 0x 2 1x
1x i
Each zero has a multiplicity of one.
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17a. Find and plot the y-intercept. Write as an ordered pair.
5( )
3
xf x
x
Set x = 0
0 5
0 3y
5
3y
50,3
y-intercept
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17b. Find and plot the zeros. Write as an ordered pair.
5( )
3
xf x
x
Set f(x) = y = 0
50
3
x
x
5x
5, 0
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-5 5 10
8
6
4
2
-2
-4
-6
-8
(4, 9)17. Vertical Asymptote
50,3
5, 0
5( )
3
xf x
x
Set the denominator = 0
3 0x 3x
x = 3
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17. Horizontal Asymptote
50,3
5, 0
-5 5 10
8
6
4
2
-2
-4
-6
-8
(4, 9)5
( )3
xf x
x
Since the degree of the two polynomials is the same find the ratio of the leading coefficient of the numerator divided by the leading coefficient of the denominator.
n
m
ayb
x = 3
1
1
1y y = 1
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-5 5 10
8
6
4
2
-2
-4
-6
-8
(4, 9)17. Find function values to help you graph.
50,3
5, 0
5( )
3
xf x
x
x = 3
1
9( 6)f
y = 1
16,9
( 1)f 1
1, 1
(2)f 7
2, 7
(4)f 9
4,9
(6)f 113
116,3
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-5 5 10
8
6
4
2
-2
-4
-6
-8
(4, 9)
50,3
5, 0
5( )
3
xf x
x
x = 3
y = 1
16,9
1, 1
2, 7
4,9
116,3
17. Find function values to help you graph.
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18a. Find and plot the y-intercept. Write as an ordered pair.
2 2( )
3
x xf x
x
Set x = 0
20 0 2
0 3y
2
3y
2
0,3
y-intercept
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18b. Find and plot the zeros. Write as an ordered pairs.
2 2( )
3
x xf x
x
Set f(x) = y = 0
2 20
3
x x
x
20 2x x
2, 0 (1,0)0 ( 2)( 1)x x
{ 2,1}x
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-5 5 10
8
6
4
2
-2
-4
-6
-8
(4, 9)18. Vertical Asymptote
20,3
2, 0
2 2( )
3
x xf x
x
Set the denominator = 0
3 0x
3x x =
3
1, 0
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18. Horizontal Asymptote 2 2
( )3
x xf x
x
Since the degree of the numerator is greater than the degree of the denominator there is no horizontal asymptote.
Slant Asymptote
y = x + 4
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-20 -15 -10 -5 5 10 15 20
22
20
18
16
14
12
10
8
6
4
2
-2
-4
-6
18. Find function values to help you graph.
( 3)f 2
3
2 2( )
3
x xf x
x
2
0,3
2, 0
x = 3
1, 0
( 1)f 1
2(2)f 4(4)f 18
(6)f 13.3
y =x +
4
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19a. Find and plot the y-intercept. Write as an ordered pair.
Set x = 0
y-intercept
43)(
2
xx
xxf
4)0(3)0(
02
y
4
0
y)0,0(
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19b. Find and plot the zeros. Write as an ordered pairs.
Set f(x) = y = 0
x0
)0,0(
The Only Zero
43)(
2
xx
xxf
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-6 -4 -2 2 4
5
4
3
2
1
-1
-2
-3
-4
-5
19. Vertical Asymptote Set the denominator = 0
x = -4
)0,0(
43)(
2
xx
xxf
430 2 xx)1)(4(0 xx
}1,4{x
x = 1
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19. Horizontal Asymptote
Since the degree of the numerator is less than the degree of the denominator the horizontal asymptote is y = 0.
43)(
2
xx
xxf
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-6 -4 -2 2 4
5
4
3
2
1
-1
-2
-3
-4
-5
19. Find function values to help you graph.
x = -4
)0,0(
)6(fx =
1
43)(
2
xx
xxf
y = 0
)3(f4
37
3
)1(f6
1
)2(f3
1
)4(f6
1
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20. Find the Domain:
1)(
2
x
xxf 01x 1x
}1,|{ xxx
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21. Find the Domain:
10
4)(
2
x
xxf 0102x
102 x
}10,|{ xxx
10x
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22. Match the function with the graph:
A) f(x) = 4x – 5
B) f(x) = 4x + 5
C) f(x) = 4-x + 5
D) f(x) = 4-x – 5 -4 -2 2 4 6
8
7
6
5
4
3
2
1
-1
-2
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23. Match the function with the graph:
A) f(x) = 3x-1
B) f(x) = 3x – 1
C) f(x) = 31- x
D) f(x) = 3-x – 1 -4 -2 2 4 6
8
7
6
5
4
3
2
1
-1
-2
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24. Match the function with the graph:
A) f(x) = 5x+1 – 2
B) f(x) = 5x+2 – 1
C) f(x) = 5x-1+ 2
D) f(x) = 5x-2 + 1 -4 -2 2 4
8
7
6
5
4
3
2
1
-1
-2