Math 1414 Course Review
Solve the problem.
1)
A Ferris wheel has a maximum height of 324 feet with a wheel
diameter of 316 feet. Find an equation for the wheel if the center of
the wheel is on the y-axis and y represents the height above the
ground. Write the answer is standard form.
Find the equation of the circle in standard form and sketch the graph.
2)
3x2 + 3y2 – 24x + 12y + 33 = 0
Find the domain of the following functions. Write your answer in setbuilder notation.
3)
h(x) =
10−3x
2x−3
4)
p(x) = log3(– x2 – 3x + 4)
Find and simplify the difference quotient of f,
following functions.
€
5)
f(x) = 5x2 + 7x – 9
6)
€
f(x+h)− f(x)
h
, for the
f(x) = 3 x + 2
For problems #7 and #8:
A)
Use the graph to find the intervals on which the function is
increasing, decreasing, or constant.€
B)
State the domain and range in interval notation.
C)
Find the x-intercept(s) and the y-intercept(s).
D)
Find f(2).
7)
6
5
4
3
2
1
0
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-2
-3
-4
-5
-6
8)
8
7
6
5
4
3
2
1
0
-5 -4 -3 -2 -1
-1 0
-2
1
2
3
4
5
Find the following.
9)
Write the equation of the graph of y = 3 x that has been reflected
across the x-axis, shifted down 3 and to the left 4.
Determine the function whose graph is given.
€
3
10)
11)
2
1
0
-2 -1-1 0
1
2
3
4
5
- 3 -2
6
-2
-3
-4
-5
Solve the problem.
30
25
20
15
10
-1 5
0
-5
-10
-15
-20
-25
-30
1
2
3
4
5
12)
A manufacturer of TV sets determines that their profits (p(x) is
dollars) depends on the number of TV sets, x, that they produce and
sell and is given by: p(x) = – 0.25x2 + 145x – 1350 where x ≥ 0.
Find the number of TV sets the manufacturer should produce and sell
to maximize the profit. What is the maximum profit?
13)
In 2010, the population of Detroit was 713 thousand. By 2013, the
the population of Detroit was 689 thousand. Find and interpret the
average rate of change.
Form a polynomial of degree 3 with the given zeros. Write your
answer in form f(x) = anxn + an – 1xn – 1 + … + a0:
14)
Zeros: – 4, 3, 2
15)
Zeros: – 2, 1 + 2i, 1 – 2i
Find the domain, asymptotes, & intercepts of the following:
16) f(x) =
€
x2 −2x−8
6x2 − x−12
17) g(x) =
€
4x2 −8x
8x3 − 27
18) h(x) =
€
2x3 − x2 −15x
x2 −3x−4
6
Use the information below to graph the function.
19)
i)
f is a rational expression with domain: { x| x ≠ 3 and x ≠ – 5}.
There are no "holes" in the graph.
ii)
The x-intercept is (– 1, 0) (multiplicity of 1) and the y-intercept is
(0, 0.5).
iii)
Vertical Asymptotes: x = – 5 & x = 3 (both have multiplicity of 1)
iv)
Horizontal Asymptote: None
1
Oblique Asymptote: y = – x + 1
4
v)
f(– 7) = 6
Solve the Inequality.
20)
f(– 4) = – 2
f(2) = 4
f(5) = – 2
€
x3 – 4x2 – 25x > – 100
21)
x2 −9x+18
4x2 − 25
≤0
For problems #22 and #23:
A)
Find all the zeros of the function.
B)
Use the zeros to factor the function.
€
C)
Determine the end behavior of the function.
22)
23)
f(x) = 2x3 + 3x2 – 17x + 12
g(x) = – 2x4 + 7x3 – 14x2 + 63x + 36
Use the given zero to find the remaining zeros of the function:
24)
g(x) = 2x3 + 3x2 – 9x + 44;
Zero: – 4
For f(x) = 3x – 9, g(x) = 2x2 + x, find the following:
25)
(f o g)(x)
26)
(g o f)(x)
The given functions are one-to-one. Find the inverse:
7x+5
x−2
27)
f(x) =
29)
h(x) = (x – 5)2; x ≥ 5
€
28) g(x) = 3ln(2x – 1) + 4
Solve the problem.
30)
The amount of a sample of radioactive substance remaining in grams
after t years is given by Q(t) = 200e – 0.015t.
a)
What is the half-life of the radioactive substance?
b)
How many grams are left after 80 years?
31)
Find the amount in an account if $5500 is invested for 15 years at
a)
8.45% compounded quarterly.
b)
8.45% compounded continuously.
32)
The half-life of Cesium-137 is 30 years. If 200 grams are present
now, how much will be left after
a)
30 years?
b)
120 years?
c)
43 years?
Solve the following algebraically. Give the exact value as the answer:
33)
log2(x) + log2(x + 4) = 5
35)
e2x – 7ex – 8 = 0
34)
log(x) + log(x + 6) = log(16)
Write the expression as a single logarithm:
36)
1
2
3ln(3x + 8) + ln(5x – 1) – 2ln(x + 3)
Write the expression as a sum and/or difference of logarithms.
€
Express powers
as factors.
37)
log
xy 5
3
( (x−6) z )
4
Perform the row operations, R3 = – 5r1 + r3, on the augmented matrix:
€ ⎡
38)
39) ⎡1 2 −5 −11⎤
4 −8 2
5⎤
⎢
⎥
⎢
⎥
1
0
−7
−3
1
−4
2
12
⎢
⎥
⎢
⎥
⎢⎣3 2 0
⎥
⎢
6⎦
⎣5 3 −3 −22⎥⎦
€
€
Solve the system for D, Dx, Dy, and Dz (if applicable) using Cramer's
Rule and find the solution to the system:
40)
3x + 2y = 4
4x – 6y = 7
41)
3x + 6y – 3z = 3
7x + 16y + z = – 7
4x – 2y – 24z = 34
Solve the problem.
42)
A copier manufacturer has factories in Austin, Charlotte, and Seattle.
Each plant produces three different models with the daily production
during March and April given in the following matrix:
Copiers produced per day
Model
Model
Model
6570
6790
7425
Austin
Charlotte
=A
Seattle
Due to the sharp rise in the cost of petroleum products, April's profits were
lower than March's profit. The profit per copier is given in the matrix below:
March
Model 6570
Model 6790
Model 7425
a)
b)
c)
April
=B
Calculate AB
What was the daily profit in March of the Austin factory?
What was the total daily profit from all three factories in April?
Solve the following system by substitution.
43)
x2 + y2 = 40
x+y=8
44)
xy2 = 4
x + y2 = 5
Find the sum of the sequence:
6
45)
∑ 2k
8
3
46)
k=1
€
∑
⎛⎜ 2 ⎞⎟k
−
⎝ 3 ⎠
k=5
€
Solve the problem.
47)
It is determined that a particular lake has 340 tons of pollution and a
company begins the process of cleaning the lake. This process will
remove 25% of the pollution a year, but the company will also
produce 30 tons of pollution per year that will be added to the lake.
The amount of pollution (in tons) in a lake after n years is given by the
recursively defined sequence:
p0 = 340; pn = 0.75pn – 1 + 30
How many years will it take for the pollution to fall below 200 tons?
Use the principle of Mathematical Induction to show that the given
statement is true for all natural numbers n.
48)
49)
50)
€
€
€
1
n
3
n
<
∑k =
k=1
n(n+1)
2
3) {x| x > 1.5}
5) 10x + 5h + 7
3
6)
x+h+2 +
€
7b)
7c)
7d)
+ 3)
2)
(x – 4)2 + (y + 2)2 = 9
1
n
€
Answers:
2
1) x€
+ (y – 166)2 = 24,964
7a)
1
n(n
2
– 2 – 3 – 4 – … – (n + 1) = –
4) {x | – 4 < x < 1}
3
2
1
x+2
Inc: [– 6, – 3) U (3, 5]
Dec: (– 2, 1) U (2, 3)
Const ( – 3, – 2) U (1, 2)
0
-2 -1-1 0
2
3
4
5
6
7
8
-2
-3
Domain: [– 6, 5]
Range: [– 4, 2]
-4
x-int: (– 4, 0), (0, 0), (4, 0)
y-int: (0, 0)
-6
f(2) = – 1
1
-5
-7
8a) Inc: (– 3, – 1) U (1, 3); Dec: (– ∞, – 3) U (– 1, 1) U (3, 4); Const: N/A
8b) Domain (– ∞, 4); Range: [0, ∞) 8c) x-int: (– 3, 0) & (1, 0) y-int: (0, 3)
8d) f(2) = 3
11) f(x) =
9) f(x) = –
1 4
x
2
–
5 3
x
2
3
x+4 – 3
3 2
x
2
–
+
17
2
10) f(x) = (x – 1)2 – 3
x–5
12) It will take 290 TV's to
produce a maximum profit of $19,675. 13) – 8 thousand per year; The
€ decreased by 8000 people per year.
population of Detroit
3
2
14) f(x) = x – x – 14x + 24 15) f(x) = x3 + x + 10
€
€
€
€4
3
3
16) Domain: {x| x ≠ – , x ≠ }
17) Domain: {x| x ≠ }
3
Vertical asymptotes: x = –
2
4
3
1
6
&x=
2
3
2
Vertical asymptote: x =
Horizontal asymptote: y =
€
€
Oblique asymptote: None
x-intercepts: (– 2,€0) & (4, €
0)
2
y-intercepts: (0, €)
Horizontal Asymptote: y = 0
€
Oblique asymptote: None
x-intercepts: (0,€0) & (2, 0)
y-intercepts: (0, 0)
3
18) Domain: {x| x ≠ – 1, x ≠ 4}
Vertical asymptotes: x = – 1 & x = 4
Horizontal asymptote: None
Oblique€asymptote: y = 2x + 5
x-intercepts: (0, 0), (– 2.5, 0) & (3, 0)
y-intercepts: (0, 0)
20)
21) (– 2.5, 2.5) U [3, 6] 22a) {– 4, 1,
22b) f(x) = (2x – 3)(x – 1)(x + 4)
10
19)
8
6
4
2
0
-11 -9 -7 -5 -3 -2-1 1
(– 5, 4) U (5, ∞)
3
2
}
22c)

23c)

24)
1
, 4} 23b) g(x)
2
5−3i 7
5+3i 7
{
,
4
4
26) 18x€2 – 105x + 153
3
-4

-6
-8
-10
€
23a) {– 3i, 3i, –
= (x + 3i )(x – 3i )(2x + 1)(x – 4)
25) 6x2 + 3x – 9
}
27) f – 1(x) =
2x+5
x−7
28) g – 1(x) =
e
x−4
3
+1
2
– 1€
29) h (x) = x €+ 5 30a) ≈ 46.21 years 30b) ≈ 60.24 grams
31a) ≈ $19,279.57 31b) ≈ $19,535.79 32a) 100 grams
32b) 12.5 grams 32c) ≈ 74.05
€ {2} 35) {ln(8)}
€ grams 33) {4} 34)
(
36) ln
€
€
(3x+8)3 5x−1
2
(x+3)
)
3
2
37) log(x) + 5log(y) +
€
1
log(z)
3
– 4log(x – 6)
5
7
9
38)
39)
40) D = – 26; Dx = – 38; Dy = 5; (
19
13
,–
5
)
26
41) D = 120; Dx = – 360; Dy = 120; Dz = – 240; (– 3, 1, – 2)
42a)
March
Austin
Charlotte
Seattle
April
€
€
42b) $13,825
42c) $64,950
43) {(2, 6), (6, 2)} 44) {(1, – 2), (1, 2), (4, – 1), (4, 1)} 45) 882
416
46) –
47) 4 years.
48)
6561
1)
The smallest allowable value for n is n = 1.
1
If n = 1, then – ((1) + 1) = – (1)((1) + 3)
€
– (2) = –
– 2 = – 2 True
2)
2
1
(4)
2
So, condition one holds.
Assume the statement is true for n = k. In other words,
1
assume – 2 – 3 – 4 – … – (k + 1) = – k(k + 3) is true. Now,
2
show the statement to be true for n = k + 1, [i.e.,
1
– 2 – 3 – 4 – … – (k + 1) – (k + 2) = – (k + 1)((k + 1) + 3)]:
2
– 2 – 3 – 4 – … – (k + 1) – (k + 2)
1
= – k(k + 3) – (k + 2)
(simplify)
=–
=–
=–
=–
2
1 2
5
k – k–2
2
2
1 2
(k + 5k + 4)
2
1
(k + 1)(k + 4)
2
1
(k + 1)((k + 1)
2
(use assumption)
(factor out –
1
)
2
(factor)
(rewrite 4 as 1 + 3)
+ 3) So, condition two holds.
Thus, by mathematical induction, – 2 – 3 – 4 – … – (n + 1) = –
is true for all natural numbers n.
1
n(n
2
+ 3)
49)
1)
The smallest allowable value for n is n = 1.
If n = 1, then
1
3
1
3
1
(1)
<
(1)
< 1 True So, condition one holds.
2) € Assume the statement is true for n = k. In other words,
1
1
assume
<
is true. Now, show the statement to be true for
k
€
k
3
n = k + 1, [i.e.,
1
<
€
=
€ €
=
€
3
<
1
]:
k +1
k
(3 •3 = 3k + 1)
€3k +1
=
1
k +1
1
k
•
1
3
€
1
3
(use assumption)
3
1
•
k
1
3k
1
k+2k
1
k +1
(simplify)
(3k = k + 2k)
(k + 2k ≥ k + 1)
≤
So, condition two holds.
€
1
1
Thus, by mathematical induction, n < is true for all natural
n
€
3
numbers n.
50) 1)
The smallest allowable value for n is n = 1.
If n = 1, then
€
1
∑k = 1
and
k=1
2)
1(1+1)
2
=1
So, condition one holds.
Assume the statement is true for n = J. In other words,
€ J
J(J+1)
assume
is true. Now, show the statement is
k =
€
∑
2
k=1
true for n = J + 1. In other words, we want to show that
J+1
∑
€
k=1
€
€
[J+1]([J+1]+1)
2
k =€
€
=
(J+1)(J+2)
2
€
=
J2 +3J+2
2
.
J+1
∑k
k=1
J
=
J+1
∑k + ∑k
(apply assumption and definition of Σ)
k=1
k=J+1
J(J+1)
=
+ (J + 1)
(get a common
2
J(J+1)
2(J+1)
=
+
(expand)
2
€ 2
2J+2
J2 +J
=
+
(combine like terms)
2
2
J2 +3J+2
=€
2
€
€
€
€
denominator)
Thus, by mathematical induction,
€n
n(n+1)
for all natural numbers n.
k=
€
∑
€
€
(apply formula four from section 12.1)
k=1
€
2