Math 1314 Final Exam Review -- Worked Out Solutions
2
2
1)
(2x + 3) = 5
(2x + 3) 5
2x + 3 5
2x + 3 5 or 2x + 3 5
2x 3 5 or 2x 3 5
3 5 3 5x or x
2 2
2
2
2
2
2
2)
x + 8x + 7 = 0
x + 8x = - 7
x + 8x + 16 = - 7 + 16 (half 8 is 4; 4 4=16; add 16 to both sides)
4 4 9
4 9
4 9
4 3
4 3 or 4 3
1
x x
x
x
x
x x
x
or - 7x
2
2
3)
x + 7x + 7 = 0
7 49 4 7 74 7 21x =
2 2 1 2
b b ac
a
2
2
2
2
2
2
4)
-6x - 2 = (3x + 1)
-6x - 2 = (3x + 1)(3x + 1)
-6x - 2 = 9x + 3x + 3x + 1
-6x - 2 = 9x + 6x + 1
-6x - 2 = 9x + 6x + 1
0 = 9x + 12x + 3
12 144 4 9 34 12 36 12 6x =
2 2 3 6 6
12
+ 6x + 2 + 6x + 2
6
6
b b ac
a
x
6 12 6 181 or 3
6 6 6x
2
2
5)
3x + 10x + 4 = 0
10 100 4 3 44 10 52x =
2 2 3 2 3
Note: 52 4 13 4 13 2 13
10 52 10 2 13 5 1 13 5 13x = = = divide top and bottom by 2 to simplify
2 3 6 3 3
b b ac
a
2 2
2
2
2
2
6)
26x - 39 = x + 5
26x - 39 = x + 5
26 39 = x + 5 x + 5
26 39 = x 5 5 25
26 39 = x 10 25
26 39 = x 10- 26x +39 - 2625
0 =
x +39
x 16 64
0 = x - 8 8
8
x
x x x
x x
x x
x
x
x
7)
8x + 6 = 4
8x + 6 = 4 or 8x + 6 = -4
8x = -2 or 8x = -10
-2 1 -10 5x = = - or x = = -
8 4 8 4
2 2
8)
7x - 6 = 6
7x - 6 = 6
7x - 6 = 36
7x = 42
x = 6
4 2
2 2
2 2
2 2
2 2
9)
x - 13x + 36 = 0
x 9 x 4 0
x 9 0 or x 4 0
x 9 or x 4
x 9 or x 4
3 or x x
2
+ 4x + 4x
any value of x
10)
-4x -4(x - 5)
-4x -
would work
4x + 20
-4x -4x + 20
0 20 True statement -4x -4(x - 5)
Thus, solution set is all real numbers
fo
or -
r
, .
11)
7x - 6 6x - 2
x - 6 - 2
x 4
12)
x5 + 1 - 8
2
x5 + 1 - 8
2
x1 - 3
2
x x1 - 3 or 1 - -3
2 2
x x- 2 or - -4
- 5
- 5
Sub2 2
tract 1 from both sides
x x-2 - 2 -2 or -2 - -4 -2 Multiply both sides by -2
2 2
x -4 or x 8 Reve
rse inequality sign since both sides
were multiplied by a negative number
14) domain = {5, 12, 6}; range = {8, -9, -3, 1}
15) From graph, when x = -4, y = 0. Thus, f(-4) = 0
1 1
1 1
16)
Slope = 3, passing through (2, 7)
Slope-intercept form: y = mx + b
Point-Slope form: (y - y ) ( )
(y - y ) ( )
7 3( 2)
7 3 6 Point-Slope form
77 3 6
m x x
m x x
y x
y x
y x
3 1 Slope-intercept fo m
7
ry x
1 1
2 1
2 1
1 1
1 1
43
17) Passing through (6, 8) and (3, 4)
Point-Slope form: (y - y ) ( )
y 4 - 8 -4 4m = slope = = = =
3 - 6 -3 3
Using (6, 8) as (x , y ) :
(y - y ) ( )
8 ( 6)
Using (
m x x
y
x x
m x x
y x
1 1
1 1
43
3, 4) as (x , y )
(y - y ) ( )
4 ( 3)
m x x
y x
1 2
2 1
2 1
18)
( ) 2 from 2 to 8
( ) ( ) (8) (2) 16 4 4 2 2 1average rate of change =
8 2 6 6 6 3
f x x x x
f x f x f f
x x
3
19)
h(x) = (x + 2)
Points on Graph
x f(x)
-4 -8
-3 -1
-2 0
-1 1
0 8
20)
4x 5f(x) = , g(x) =
x - 5 x+3
4x 5f(x) + g(x) = Note: x 5; x -3
x - 5 x+3
Thus, domain is - ,-3 3,5 5,
21) Find the domain of (f - g)(x) when f(x) = 5x - 2 and g(x) = 3x - 9.
(f - g)(x) ( ) ( ) 5x - 2 3x - 9 5 2 3 9 2 7
Thus, (f - g)(x) 2 7 and its domain is (- , )
f x g x x x x
x
22)
f(x) = -4x + 8, g(x) = 6x + 5
(g f)(x) = g f(x) 4 8
=6 4 8 5
= -24x + 48 5
= -24x + 53
g x
x
-1
23)
f(x) = 4x - 4 y = 4x - 4
x = 4y - 4 Exchange x and y
x + 4 = 4y Solve for y
x + 4 = y
4
x + 4
4
x + 4f ( ) inverse function
4
Note that graph of
y
x
-1
-1
-1
x + 4 f ( ) is a straight line.
4
x + 4Domain of f ( ) is (- , )
4
x + 4Range of f ( ) is (- , )
4
x
x
x
-1
24)
f(x) = 6x - 7 y = 6x - 7
x = 6y - 7 Exchange x and y
x + 7 = 6y Solve for y
x + 7 = y
6
x + 7
6
x + 7f ( ) inverse function
6
y
x
2 2
2
25)
(x + 5) + (y - 4) = 9
h = -5; k = 4 r 9
Thus, center of circle is (-5, 4)
radius = 3
2 2
2 2
2 2
2 2
26)
x + 10x + 25 + y - 4y + 4 = 9
x + 10x y - 4y 9 25 4
x + 10x y - 4y 20
Note: half of 10 is 5; and 5 5 = 25
half of -4 is -2; and 2 2 4
x + 10x + y - 4y 25 + 25 20
5x x
+ 4 + 4
2 2
2
5 2 2 9
5 2 9
5 k = 2 r 9
Center (-5, 2) radius = 3
y y
x y
h
2
2 2 2
2 2
2 2
27)
(-10, -4); 5
h = -10; k = -4; r = 5; r 25
Standard form is:
Equation of circle is: ( 10) ( 4) 25
10 4 25
x h y k r
x y
x y
2 2
2 2
2
2 2
2 2
28)
x + y = 25
x - 0 + y - 0 = 25
h = 0 k = 0 r 25
Center (0, 0) radius = 5
Domain of graph of x + y = 25 is -5, 5
Range of graph of x + y = 25 is -5, 5
2 2
2
2
29)
f(x) = 2 + 3x + x y = 2 + 3x + x
To find x-intercept(s), set y = 0
y = 2 + 3x + x
0 = x +3x + 2
0 = 2 1
Set x + 2 = 0; Set x + 1 = 0
x = -2
x x
x = -1
Thus, x-intercepts are ( -2, 0) and (-1, 0)
2
2
30)
y + 4 = (x + 2)
y = (x + 2) 4
Thus, vertex of parabola is (-2, -4)
3 2
3
3
31)
f(x) = -3x + 5x + 3x + 2
The leading term is -3x .
Since coefficient of -3x is negative and exponent is odd, graph rises to the left
and falls to the right.
4 3 2
32)
x - 3x + x + 7x - 7
x - 1
1 1 -3 1 7 -7
1 -2 -1 6
-----------------------------------------
4 3 23 2
----------------
1 -2 -1 6 -1
x - 3x + x + 7x - 7 1Thus, 2 6
x - 1 x - 1x x x
3 22
2
3 2
3 2
2
33)
9x - 30x + 13x + 33 133 5 4
-3x + 5 -3x + 5
3 5 4
-3x + 5 9x - 30x + 13x + 33
9x - 15x
-----------------------------------
-15x 13
x x
x x
x
2 -15x 25
-----------------------------------
-12x + 33
-12x + 20
-----------------------------------
x
13
4 3 2
34)
f(x) = 2x - 15x + 45x - 45x + 13 has four zeros.
Last coefficient = 13
Leading coefficient = 2
Factors of 13 are 1, 13
Factors of 2 are 1, 2
1 1 13List of possible rational zeros are , ,
1 2 1
4 3 2
13 1 13, 1, , 13,
2 2 2
Note f(1) = 0 and f(1/2) = 0
Thus, two of the zeros are x = 1 and x = 1/2
Corresponding factors are (x - 1) and (x - 1/2)
f(x) = 2x - 15x + 45x - 45x + 1
4 3 23 2
3 22
3
= (x - 1) (x - 1/2) ( ) ( )
To find the other two zeros:
2x - 15x + 45x - 45x + 132 13 32 13 use synthetic division
(x - 1)
and
2 13 32 132 12
(x - 1/2)
x x x
x x xx
2
2
26 use synthetic division
Now set 2 12 26 0 and then use quadratic formula to solve this equation.
12 144 4(2)(26)4 12 64 12 8x = 3 2
2 2(2) 4 4
Thus, th
x
x x
b b ac ii
a
e four zeros are 1, 1/ 2, 3 2 , 3 - 2i i
35)
3xf(x) =
(x - 7)(x - 9)
Note: f(7) = undefined.
f(9) = undefined.
Thus, x 7 and x 9
3xDomain of f(x) = is , 7 7, 9 9,
(x - 7)(x - 9)
36)
xg(x) =
x(x - 2)
Note: g(0) = undefined. g(2) = undefined. Thus, x 0 and x 2.
Vertical asymptotes are vertical lines x = 0 and x = 2.
37)
(3z + 10)(5z - 1) > 0
Finding boundary values: Set (3z + 10) = 0 and Set (5z - 1) = 0
10 1 z = - z =
3 5
Bou10 1
ndary Values are - and 3 5
10region I interval: - , -
3
10 1region II interval: - ,
3 5
1region III interval: ,
5
Checking to see values in which region works by choosing a value from each interval:
(3z + 10)(5z - 1) > 0
If z = - 5 130 > 0 True
If z = 0 -10 > 0 False
If z = 5 600 > 0 True
If z = -10/3 0 > 0 false
If z = 1/5 0 > 0 false
Values that work are (- , -10/3) (1/5, ).
I II III
-1/10 1/5
x38) > 0
x + 1
Finding boundary values: Set x = 0 and Set x + 1 = 0
x = -1
Boundary Values are 0 and -1
re
gion I interval: - , -1
region II interval: -1, 0
region III interval: 0,
Checking to see values in which region works by choosing a value from each interval:
x > 0
x + 1
If x = - 5 1.25 > 0 True
If x = -0.5 -1 > 0 False
If x = 5 0.833 > 0 True
If x = 0 0 > 0 false
If x = -1 undefined > 0 false
Values that work are (- , -1) (0, ).
I II III
-1 0
3 2
2
2
39)
x + 7x - x - 7 > 0
x ( 7) 1( 7) 0
( 7)( 1) 0
( 7)( 1)( 1) 0
Finding boundary values: Set ( 7) 0; Set ( 1) 0; Set ( 1) 0
x =
x x
x x
x x x
x x x
-7 x = 1 x = -1
Boundary Values are -7, -1, and 1
region I interval: - , -7
region II interval: -7, -1
region III interval: -1, 1
region IV interval: 1,
Checkin
3 2
g to see values in which region works (also check boundary values):
x + 7x - x - 7 > 0
If x = -8 -63 > 0 False
If x = -5 48 > 0 True
If x = 0 -7 > 0 False
If x = 2 27 > 0 True
If x = -7 0 > 0 False
If x = -1 0 > 0 False
If x = 1 0 > 0 False
Values that work are (-7, -1) (1, ).
I II III IV
-7 -1 1
40)
The pressure of a gas varies jointly as the amount of the gas (measured in moles) and the
temperature and inversely as the volume of the gas. If the pressure is 1530 kPa
(kiloPascals) when the number of moles is 8, the temperature is 340° Kelvin, and the
volume is 640 cc, find the pressure when the number of moles is 5, the temperature is
280° K, and the volume is 300 cc.
kA T
P =
1530
8
340
640
k(8)(340)Hence, 1530 = 360
640
360A TThus, P =
Find the pressure when the number of moles is 5, the temperature is
280° K, and the volume is 300 cc.
360(5)(280)P =
30
V
P
A
T
V
k
V
16800
x
-2
-1
0
1
2
141) f(x) =
3
Points on graph.
x f(x)
-2 (1/3) 9
-1 (1/3) 3
0 (1/3) 1
1 (1/3) 1/ 3
2 (1/3) 1/ 9
42)
Suppose that you have $10,000 to invest. Which investment yields the greater return over
7 years: 8.75% compounded continuously or 8.9% compounded semiannually?
For the case of 8.75% compounded cont rt
rt (0.089)(7)
nt
nt (2
inuously, we use the formula A = Pe
A = Pe 10000 18,645.13
rFor the case of 8.9% compounded semiannually (n =2), we use the formula A = P 1+
n
r 0.089A = P 1+ 10000 1+
n 2
e
)(7)
18,395.78
Thus, the option of 8.75% compounded continuously has a greater yield.
106
10
log 21643) log 216 = 3
log 6
6 6
3 5
6 6
3 5
6
44)
3 log x + 5 log (x - 6)
log x + log (x - 6)
log x (x - 6)
108
10
log 645) log 6 = 0.861654
log 8
2x - 4
2x - 4
2x - 4
2x - 4
46)
e - 4 = 1350
e - 4 = 1350
e 1354
ln e ln 1354
2 4 ln ln 1354
ln 13542 4 7.210818
ln
2 4 7.210818
2
+ 4 +
11.210818
5.0605409
4
x e
xe
x
x
x
10 10 10
10 10
47)
log (3 + x) - log (x - 2) = log 2
log (3 + x) - log (x - 2) = log 2
(3 + x)log = log 2
(x - 2)
(3 + x)2
(x - 2)
(3 + x)2
(x - 2)
(3 ) 2 4
3 2 4
7 2
7
( 2) ( 2)x x
x x
x x
x x
x
48) -7x - 6y = -16 1
-4x - 2y = -12 2
Multiply 2 by (-3) to elimate y (-3) -4x - 2y = -12 = 12x + 6y = 36
Hence,
-7x - 6y = -16 1
12x + 6y =
36 2
5 20
5 20
5 5
4
To find y, use 2 :
-4x - 2y = -12
-4(4) - 2y = -12
-16 - 2y = -12
-16 - 2y = -12
-2y = 4
y = -2
T
+16 +1
hus, solution is (4, -
6
2)
x
x
x
2 2
2 2
2
22
2
2 2
2
2
49)
5x + 9y = 81 1
y = x + 3 2
Substitute 2 into 1 :
5x + 9 = 81
5x + 9 = 81
5x + 9 = 81
5x + 9x 27 27 81 = 81
14x
y
x +3
x +3x +
3x
54 81 = 81
14x
+9
x x
x
54 = 0
2x 7 27 = 0
Set 2x = 0; Set 7x + 27 = 0
-27 x = 0 x =
7
x
x
Now use 2 to find y.
For x = 0,
y = x + 3
y = 0 + 3= 3
-27For x = ,
7
y = x + 3
-27 6y = + 3 = -
7 7
-27 6Thus, solutions are (0, 3) and ,
3 7
x y
50)
x - y + 3z = -7
5x + z = -1
x + 3y + z = 11
Using Cramer's Rule:
1 1 3 -7 -1 3 1 7 3
M = 5 0 1 D = -1 0 1 D = 5 1 1
1 3 1 11 3 1 1 11
z
x y z
x
1 1 7
D = 5 0 1
1 1 3 11
( ) 46 det (D ) 0 det (D ) 184 det (D ) 46
D 0Thus, x = 0
M 46
Det M
y zD D184 46
y = 4 z = 1M 46 M 46
Solution is (0, 4, -1)
51) Remember that in multiplying matrices, rows come from first matrix and columns from second matrix
R1 C1 R1 C2 R1 C3
-3 -7 -9 -3 -7 -9 -3 -7 -9
-9 3 4 -1 -7 9 3 3 -9
---------------- ------------------ -------------------
27+-21+-36 = -30 3 + 49 + -81 = -29 -9 + -21 + 81 = 51
R2 C1 R2 C2 R2 C3
2 -4 -1 2 -4 -1 2 -4 -1
-9 3 4 -1 -7 9 3 3 -9
---------------- ------------------ -------------------
-18 + -12 + -4 = -34 -2 + 28 + -9 = 17 6 + -12 + 9 = 3
R3 C1 R3 C2 R3 C3
4 2 -1 4 2 -1 4 2 -1
-9 3 4 -1 -7 9 3 3 -9
---------------- ------------------ -------------------
-36 + 6 + -4 = -34 -4 + -14 + -9 = -27 12 + 6 + 9 = 27
30 29 51
A 34 17 3
34 27 27
B
52)
1 2 5
2 1 3 0
1 2 5
n
1
2
3
4
53)
a = 8n
a = 8(1) = 8
a = 8(2) = 16
a = 8(3) = 24
a = 8(4) = 32
1 n n-1
2 1
3 2
4 3
5 4
54)
a = 5 and a = a - 1 for n 2
a = a - 1 = 5 - 1 = 4
a = a - 1 = 4 - 1 = 3
a = a - 1 = 3 - 1 = 2
a = a - 1 = 2 - 1 = 1
3
n
3
1
3
2
3
3
3
4
55)
na =
(n - 1)!
1 1 1a = = = 1
(1 - 1)! (0)! 1
2 8 8a = = = 8
(2 - 1)! (1)! 1
3 27 27a = = =
(3 - 1)! (2)! 2
4 64 64 32a = = = =
(4 - 1)! (3)! 6 3
5
i = 1
56)
(i + 9) (1 + 9) + (2 + 9) + (3 + 9) + (4 + 9) + (5 + 9)
= 10 + 11 + 12 + 13 + 14 = 60
2 1
57) 10, 15, 20, 25, . . .
common difference = a - a = 15 - 10 = 5
2
1
58) 3, -9, 27, -81, 243
-9common ratio = 3
3
a
a
