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PRE-CALCULUS A FINAL REVIEW
NAME_____________________________
CHAPTER ONE
(1) Determine which of the following define y as a function of x:
(a)
x-4 =y
(b) 3x + y2 = 8
(c)
(d)
x2 - 4 = y
(e)
___________________________________________________________________________________
(2) To the right is the graph of f(x)
(a) Give the domain using interval notation.
(b) Give the range using interval notation.
(c) Give the interval(s) on which f is increasing
using interval notation.
(d) Give the interval(s) on which f is decreasing
using interval notation.
(e) At what number does f have a relative
maximum?
(f) What is the relative maximum of f?
(g) Find f(4).
_______________________________________________________________________________________________________________________
(3) Find and simplify the difference quotient
f(x + h) - f(x)
, h  0 , for f(x) = 4x – 3
h
(4) Determine whether each function is even, odd or neither:
(a) f(x) = x3 + 3
(b) g(x) =
(c)
(d)
x2 + 7
Skip#5-8
(9) Find the average rate of change of f(x) = 2x2 + x from x1 = -4 to x2 = 6
(10) Graph f(x) = x2 and g(x) = (x – 4)2– 3
1
(12) Graph f(x) = x3 and g(x) = (-x)3
2
(14) To the right is the graph y = f(x).
(11) Graph f(x) = x and g(x) = (13) Graph f(x) =
3
(a) Graph g(x) = f(2x) + 3
(b) Graph g(x) = -2f(x + 3)
(15) Give the domain of each in interval notation:
(a) f(x) =
(c) f(x) =
x+4
6x - 3
x - 10
5x + 20
(b) f(x) = 2x3 – 4x2 + 10
(d) f(x) =
x - 7 + x + 11
.5x
x and g(x) = f -1(x)
(16) Given f(x) = x2 – 7x – 30 and g(x) = 5x – 8:
g(x)
g
(a) Find   (x) =
and its domain
f(x)
f
(b) Find f(g(x)) and its domain
(c) Find (g ○ f)(4)
(17) Given f(x) =
2x and g(x) = 3x – 5:
(a) Find (f – g)(x) = f(x) – g(x) and its domain
(b) Find (f ○ g)(x) and its domain.
(c) Find (f – g)(8)
(18) Find the inverse function, f -1(x), for each:
(a) f(x) = 7x – 1
(b) f(x) =
6
+4
x
(c) f(x) = x3 + 8
CHAPTER TWO
(19) Simplify (4 - 7i ) - (7 - 3i )
(20) Simplify
(21) Simplify (5 - 2i )(6 + 4i )
(22) Simplify
2
(23) Simplify (8 - 5i )
(25) Simplify
(24) Simplify
-75 + 4 -27
16 +
-32
6
3 -2 5
-22

8i
5+i
(26) Solve, using the quadratic formula: 5x2 – 4x + 2 = 0
(27) Give the vertex, domain and range of the parabola with equation f(x) = 20(x – 9)2 – 15
(28) Give the axis of symmetry, domain and range of the parabola with equation f(x) = -3x2 + 24x + 2
For #29-30, Match the letter of the end behavior with the associated function:
(A)
(B)
(29) f(x) = -16x7 + 3x5 – 2x + 10
(C)
(30) f(x) = 4(x + 4)3(x – 8)
(D)
(31) Sketch the graph of y = -5(x – 4)2(x + 3)(x – 1)2
(32) Sketch the graph of f(x) = 2x4(x + 6)(x – 2)3
(33) Divide (8x3 + 6x2 + x + 3) by (2x – 3) using long division and give the remainder.
(34) Divide (3x3 – 5x2 + 3) by (x – 4) using synthetic division and give the answer.
(35) Use synthetic division to divide (4x2 – 6x + 3) by (x + 5) and give the remainder.
(36) List all the possible rational zeros for f(x) = 3x5 + 4x3 – 5x + 8
(37) Use synthetic division to find f(-4) if f(x) = 5x4 + 30x2 – x
(38) Find the roots of x4 – 3x3 – 15x2 + 99x –130 = 0
(39) Find the roots of x3 – 4x2 –17x + 30 = 0
(40) Find a cubic polynomial with real coefficients satisfying these conditions:
-5 and 2 i are zeros and f(3) = -624. Simplify (multiply it out).
(41) Find the vertical asymptote(s) of the graph of f(x) =
3x + 1
x - 3x 2 - 28x
3
(42) Find the horizontal asymptote(s) of the graph of f(x) =
3x
6x - 11
(43) Find the horizontal asymptote(s) of the graph of f(x) =
9x 2 - 5
15x 2 + 1
2
(44) Solve and graph 4x2 + 5x < 6
(45) Solve and graph
 x - 4  x + 2 
x-2
 0
(46) Solve and graph x3  5x2
(47) t varies directly as the square of m and inversely with p. If t = 300 when m = 10 and p = 5,
find t when m = 6 and p = 3
(48) z varies jointly as x and y. If z = -162 when x = 9 and y = 5, find z when x = -10 and y = 8.
(49) Ed throws a rock straight up from the top of a barn 130 feet high with an initial velocity
of 176 feet per second. During which time period will the rock‘s height exceed that of the barn?
Use the position function f(x) = -16t2 + v0t + s0.
(50) Graph the equations:
6x 2 + 1
(a) y =
2x 2 - 8
(b) y =
-3
x-5
CHAPTER THREE
(51) Graph f(x) = 2 x with a dashed line. Graph g(x) = 2x + 4 - 5 with a solid line.
x
 1
(52) Graph f(x) =   with a dashed line. Graph g(x) = log.5x with a solid line.
2
(53) Write the equation in logarithmic form: 2w = d
(54) Write the equation in exponential form: 42 = log3x
Evaluate:
(55) log28
 1 
(56) log6  
 36 
(57) eln22
(58) log7 7
(59) Find the domain of f(x) = log9(x + 12)
(60) Find the domain of f(x) = log(x – 5)2
(61) Expand as much as possible: log(xy 25)
 x5 
(62) Expand as much as possible: ln  7 
y 
(64) Condense into a single log: log315x – log310y
(63) Condense into a single log: 3logx + 2logy
(65) Find the value of log2 24 to the nearest hundredth.
(66) Find the value of e-.48 to the nearest hundredth.
(67) Find the accumulated value of an investment of $3600 for 6 years
at an interest rate of 5.5% if the money is:
nt
r

(a) compounded quarterly (use A = P  1 + 
n

and round to the nearest cent). Label answer.
(b) compounded continuously (use A = Pert ) and round to the nearest cent). Label answer.
(68) $4000 is invested at an annual rate of 3% and the interest is compounded continuously. Find
the number of years (to the nearest tenth) it will take for the amount to be $12,000. Use A = Pert
(69) The exponential function f(x) = 875(1.04)x models the population of a country, f(x), in millions,
x years after 1974.
(a) Find the country’s population in 1974.
(b) Find the country’s population, to the nearest million, in the year 2015 as predicted by this
function.
(70) Skip
3x-1
(71) Solve for x: 8
3x + 8
= 16
 1
(72) Solve for x:  
e
2x-1
= e x - 14
(73) Solve for x to the nearest hundredth: 2x + 1 = 6
(74) Solve for x to the nearest hundredth: 6e3x = 24
(75) Solve for x: log  2x - 4 = 3
(76) Solve for x to the nearest hundredth: ln(x + 5) =
3
4
(77) Solve for x: log4  x - 4 + log4  x + 8 = log4  x + 38
(78) Solve for x: log3  x + 2  log3  x - 5  2
(79) Solve for x: log5 x2 = log5  2x + 8 
(80) Solve for x to the nearest hundredth: e2x – 4ex – 60 = 0
CHAPTER FOUR
(81) Find the radian measure of the central angle of a circle of radius 24 yards that intercepts
an arc of length 108 yards.
(82) Find the length of the arc on a circle of radius 20 feet intercepted by a central angle of 30o .
Express your answer as an exact multiple of  .
(83) Convert 200o to radians. Express your answer as an exact multiple of  .
(84) Convert
13
 radians to degrees.
20
(85) Convert 2.2 radians to degrees. Answer to the nearest hundredth.
(86) Change 6.5 revolutions to:
(a) degrees
(b) radians (exact answer in terms of  )
(87) The radius of a wheel rolling on the ground is 75 cm. The angular speed of the wheel
is 60o per second. Find the linear speed. Answer to the nearest tenth.
(88) Find the value of each to 4 decimal places:
 5 
(a) cos(2.7)
(b) tan 
(c) csc(7)

 9 
 3 
(d) cot 

 7 
(89) Find a cofunction with the same value as the given expression:
3 
(a) sec 52o
(b) cot   
 11 


(90) Find the 6 trigonometric functions of  in the right triangle shown:
29

21
(91) Find the measure of side “a” of the right triangle.
Answer to 2 decimal places and label.
57o
b =140 cm
(92) Find the measure of side “b” of the right triangle.
Answer to 2 decimal places and label.
145 in.
32o
(93) A tower that is 125 feet tall casts a shadow 188 feet long. Find the angle
of elevation,  , of the sun to the nearest degree and label your answer.
188 ft
(94) Find the values of the acute angle  to the nearest degree.
(a) sin  = .9781
(b) tan  = 12.5544
(95) Find the values of the acute angle  to the nearest hundredth of a radian.
(a) cos  = .1234
(b) tan  = .4567
(96) The point (2, -6) is on the terminal side of angle  . Find the exact values
of the 6 trigonometric functions of  .
(97) If cos  =
8
and  is in quadrant  , find the exact values of the other 5 trig functions of 
17
(98) If tan  =
4
and sin  < 0, find the exact values of the other 5 trig functions of 
9
(99) Find the reference angle for each:
(a) 212o
(b)
13

9
(100) Find 2 exact values of  where 0    2 that satisfies:
(a) sin  =
 3
2
(b) tan  = 1
(101) Find an equation for each graph:
(a)
(b)
For #16 & 17: give the amplitude, period, and phase shift. Graph 1 period marking all important
x and y numbers as discussed in class:


(102) y  5cos  3x - 
2

x

(103) y  4csc  +  
2

Give the amplitude and period. Graph 2 periods marking all important x numbers as discussed in class :

(104) y  tan 
3

x

Find the exact value of each:
(105) sin1 1
(109) cos1
1
2

 3  
(113) cot  sin1   
 8 

 2
(106) cos1 
 2 


 3
(110) tan1 
 3 


4

(114) sin  tan1 
9

(107) tan1  1
 3
(108) sin1 
 2 



 1  
(111) csc  cos1   
 5 

(112) cos cos1 2.4
3 

(115) cos1  cos
4 


 4
(116) tan1  tan 
 3


(117) Use the figures shown to find the bearing from O to A
(a)
(b)
71
53
(118) A police helicopter is flying at 550 feet above the ground. A stolen car is sighted at an angle of
depression of 55o . Find the distance of the stolen car, to the nearest tenth of a foot, from a
point on the ground directly below the helicopter.
(119) A hot-air balloon is rising vertically. From a point on level ground 100 feet from the point
directly under the passenger compartment, the angle of elevation to the balloon changes
from 19.2o to 31.7o . How far, to the nearest tenth of a foot, does the balloon rise during
this period?
100 ft.
(120) A boat leaves the entrance to a harbor and travels 80 miles on a bearing of S 25o W.
How many miles, to the nearest tenth, west from the harbor has the boat traveled?
(121) A ship is 20 miles east and 12 miles north of a harbor. What bearing did he take to sail
from the harbor? Answer to the nearest tenth of a degree.




ANSWERS:
(1) a, c, e
(2)
(9) 5
(a)
 -,  
(b)
 6,  
(c)
 3, 1   4, 
(d)
 , 3   1, 4
(10)
(e) -1
(f) 6
(g) -5
(3) 4
(4)
(11)
(a) neither
(b) even
(c) even
(d) odd
(5)
(6)
(a)
4
3
(b)
11
6
(a) y  0 
3
 x  6
4
y3
3
 x  2
4
or
(b) y 
3
9
x
4
2
(7) y  8 x  39
(8) y  3 
1
 x  4
10
(12)
(13)
2 x  3x  5
(17) (a)
Domain: 0, 
6 x  10
(b)
5 
Domain:  ,  
3 
(b) -15
(c) -15
(14)
(18) (a) f 1  x  
x 1
7
(b) f 1  x  
6
x4
(c) f 1  x   3 x  8
(19) 3  4i
(20) 17i 3
(21) 38  8i
1 1 

(15) (a)  ,    ,  
2 2 

(22)
8 2i 2

3
3
(b)
 ,  
(23) 39  80i
(c)
 4,  
(24) 30 11
(d)
7,
(25)
4 20i

13 13
5x  8
x  7 x  30
(26)
2 i 6

5
5
(16) (a)
2
Domain:  , 3   3,10  10,  
(b) 25 x 2  115 x  90
Domain:
(c) -218
 ,  
(27) Vertex:
Domain:
Range:
 9, 15
 ,  
15, 
(28) Axis of Symmetry: x  4
Domain:
Range:
 ,  
 ,50
(39) 6, 1  6
(40) f(x) = -6x3 – 30x2 – 24x –120
(41) x = 0, 7, -4
(29) D
(42) y = 0
(30) A
(43) y =
(31)
(44) (-2, ¾)
3
5
○
○
-2
3/4
(45) (-  , -2]  (2, 4]

○

-2
2
4
(46) {0}  [5,  )
(32)


0
5
(47) 180
(48) 288
(49) between 0 and 11 seconds
(50) (a)
(33) 45
(34) 3 x 2  7 x  28 
115
x4
(35) 133
1 2 4 8
(36) 1, 2, 4, 8,  ,  ,  , 
3 3 3 3
(37) 1764
(38) 5, 2,3  2i
(50) (b)
(53) log2d= w
(54) 342 = x
(55) 3
(56) -2
(57) 22
(58)
1
2
(59) x > -12
(51)
(60) x  5
(61) logx + 25logy
(62) 5lnx – 7lny
(63) logx3y2
 3x 
(64) log3  
 2y 
(65) 4.58
(66) .62
(67) (a) $4996.24
(52)
(b) $5007.49
(68) 36.6 years
(69) (a) 875 million
(b) 4369 million
(70) 1975
(71) x =
35
3
(72) x = 5
(73) x = 1.58
(74) x = .46
(90) cos  =
21
29
sin  =
sec  =
29
21
csc  =
(75) x = 502
(91) 215.58 cm
(76) x = -2.88
(92) 122.97 in.
(77) x = 7
(93) 34
(78) x = 5.875
(94) (a) 78
(79) x = -2, 4
(b) 85
(80) x = 2.30
10
ft.
3
(83)
10
9
(89) (a) csc 38
10
10
sin  =
3 10
10
sec  =
10
csc  =
 10
3
tan  =
15
8
17
8
csc  =
17
15
9 97
97
sin  =
4 97
97
sec  =
 97
9
csc  =
 97
4
(b) -5.6713
(99) (a) 32
(b)
(d) .2282
(100) (a)
sec  =
(b) 13
(98) cos  =
(87) 78.5 cm/sec
(c) 1.5221
(96) cos  =
(97) sin  =
(85) 126.05
(88) (a) -.9041
20
21
cot  =
21
20
(b) .43
(84) 117
(86) (a) 2340
29
20
tan  =
(95) (a) 1.45
(81) 4.5
(82)
20
29
(b) tan
5
22
(b)
15
17
4 5
,
3 3
 5
,
4 4
4
9
tan  = 3
cot  =
1
3
cot  =
8
15
cot  =
9
4
(101a) y = -2sin(4x)
 
(101b) y = 6cos  x  - 4
3 
(102)
(103)
amplitude: 5
2
period:
3
phase shift:
amplitude: none
period: 4

6
phase shift: 2
(104)
amplitude: none
period: 3
(105)

2
(106)
3
4
(107)

4
(108)
(109)

3
(110)

6
(111)
5 6
12
(112) undefined
(113)
 55
3
(114)
4 97
97
(115)
3
4
(116)
(117a) N 19 E
(120) 33.8 miles
(117b) S 37 W
(121) N 59.0 E
(118) 385.1 ft.

3

3
(119) 26.9 ft.