Precalculus - Dalton State College

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Precalculus
Final Exam Review Questions
Determine the interval on which the
function is increasing.
a. (0, 1)
b. (4, 3)
c. (3, 0)
d. (1, 2)
Which of the following is symmetric
with respect to the origin?
a. y  (x  4)
2
b. x  y
2
c. y   x  2
d. y  x  x
3
Write an equation for a function that has the
shape of y  x, but is shifted left 3 units
and down 5 units.
a. f (x)  x  3  5
b. f (x)  x  5  3
c. f (x)  x  3  5
d. f (x)  x  5  3
The graph of f is given.
Which graph below
represents the graph of
g(x) = 2f(x) + 1?
a.
b.
c.
d.
Find the zeros of the polynomial function
and state the multiplicity of each.
f(x) = (x + 3)2(x + 1)
a. –3, multiplicity 2, 1 multiplicity 1
b. 3, multiplicity 2, 1 multiplicity 1
c. –3, multiplicity 2, 1 multiplicity 2
d. 3, multiplicity 3, 1 multiplicity 1
For f(x) = 2x4 + 3, use the intermediate
value theorem to determine which interval
contains a zero of f.
a. between 1 and 0
b. between 0 and 1
c. between 1 and 2
d. between 2 and 3
Find the quotient and remainder when
x4 + 5x2 – 3x + 2 is divided by x – 2.
a.
x3  2 x2  9 x  15, R 32
b.
x  2 x  9 x  21, R 44
c.
x3  2 x 2  9 x  21, R 44
d.
x3  7 x2  14 x  25, R 52
3
2
Find a polynomial function of lowest degree
with rational coefficients and 3 and 4i as
some of its zeros.
a.
f ( x)  x  3x  4xi  12i
b.
f ( x)  x3  3x2  16 x  48
c.
f ( x)  x  3x  16 x  48
d.
f ( x)  x3  3x2  16 x  48
2
3
2
Use the rational zeros theorem to determine
which number cannot be a zero of
P(x) = 10x4 + 6x2 – 5x + 2.
a.
1
5
b. 2
c. 5
d.
2
5
Which graph represents the polynomial
x3
.
function f (x)  2
x  3x  4
a.
b.
c.
d.
Solve 3x2 > x + 10.
5

a.  ,    (2, )
3

5

b.  2, 
3

3 

c.  , 2    ,  
5 
5 

d.  , 2    ,  
3 
Convert to a logarithmic equation:
2x = 20.
a. x = log220
b. x = log10
c. 2 = logx20
d. x = log202
Find log 0.001. Do not use a calculator.
a. –4
b. –3
c. 3
d. 1000
Express in terms of sums and
differences of
a.
x2
logarithms: log 3 .
y
1
1
log x  log y
9
3
b. 6log x  3log y
c.
d.
2
1
log x  log y
3
3
2
log x  log y
3
Solve: log3(3x + 6) – log3(x – 6) = 2.
a. 14
b. 10
c.
3

2
d. Does not exist
Solve: 25+x = 32x-4.
a.
b.
c.
d.
9
4
15
4
69
15
25
4
Find log58 using the change-of-base
formula.
a.  0.2041
b.  0.7740
c. 1.2920
d.  1.6
Suppose $2000 is invested at interest rate, k,
compounded continuously, and grows to
$2,473.53 in 5 years. Find the approximate
interest rate.
a. 3.83%
b. 4.25%
c. 4.75%
d. 7.89%
Given the triangle, find cos.
a.
3 109
109
10
c.
3
10 109
b.
109
d.
109
3
Find the exact function value, if it exists, of
7
sin
.
6
a.
c.
1
2
3
2
3
b. 
2
1
d. 
2
Given right triangle ABC with a = 16.5
and A = 23.5º, find c. Standard lettering
has been used.
a. 18.0
b. 41.4
c. 15.1
d. 6.6
Convert 220º to radian measure in terms of
π.
11
a.
9
c.
9
11
11
b.
18
11
d.
9
8
Given that cos   
and that the
89
terminal side is in quadrant III, find sin.
a.
5
89
5
c. 
89
b.
8
5
5
d. 
8

3
Find cos  
exactly in degrees.

 2 
1
a. 120º
b. –60º
c. 150º
d. 30º


For the function y  2 cos  x    4,

4
find the period.
2
a.
3
b. π
c. 2π
d. 4π
Simplify
a.
b.
sec x cot x  cos x .
1 sin x
sin x
cos x  2
sin x
c.
1 csc x
d.
cot x 1
x
Simplify 2 sin  1.
2
2
a.
 cos2x
b.
cos x
c.
sin x
d.
 cos x
Use a sum or a difference identity to
7
find sin
exactly.
12
a.
2 6
4
c.
2 6
4
 2 6
b.
4
d.
2 6
4
Use a half-angle identity to evaluate
 5 
exactly.
cos 

 8 
2 2
a. 
2
2 2
b. 
2
2 3
c. 
2
2 3
d. 
2
Find all the solutions of
3  4 sin x  0 in 0, 2 .
2
a.
c.
 2
b.
,
3 3
 5 7 11
,
6 6
,
6
,
6
d.
  2 5
, , ,
6 3 3 6
 2 4 5
,
3 3
,
3
,
3
Find all the solutions of
2sin x cos x  2 sin x in 0, 2 .
a.
c.
 3
,
4 4
 3
,
2 2
b.
  3 3
, , ,
4 2 4 2
 3
d. 0, , , 
4 4
Solve triangle ABC to find b.
a = 10 m, B = 14º, C = 28º
a. 3.6 m
b. 13.1 m
c. 5.2 m
d. No solution
Solve triangle ABC to find A.
a = 12.5 in., b = 10.5 in., c = 8.5 in.
a. 8.5º
b. 87.9º
c. 81.5º
d. No solution
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