Download Algebra II Chapter 3 Exam

Algebra 2
Chapter 7 Review Answer Key
Evaluate the expression.
3
1.
8
2.
2
36
1
1
 125 3
2
36  6
2
3.
4.
16
1
5.
4
9


 125
3

2
1

  5  25
2
16
1

4
1
2
Simplify the expression. Assume all variables are positive.
6.
 2 14  5 14 


2
9.
4
4
5
4
4
4
7.
5
 32 x 5 y 5 z 5
  32 x
5
 21  51  10
5
y
5 5
z
3
48

5
 2 xyz
1 1

16 4
 16  3  3
25  3 20
3
2
25  20
3
2
4 3 3
 3 250
5 3
 3 125  2
48  3
5 5
3
48
8.
3
10.
 53 2
Perform the indicated operation and state the domain. Let f ( x)  x 3 and g ( x)  x  1
11.
f ( x)  g ( x)
12.
13.
f ( x)  g ( x)
x  x  1
 x   ( x  1)
x x  1
 x 3  x 1
 x3  x  1
 x4  x3
3
3
Domain: All Real #s
14.
f ( x)  g ( x)
f ( x)
g ( x)
x 
3
( x  1)
Domain: All real #s except – 1
3
Domain: All Real #s
Domain: All Real #s
15.
16.
f ( g ( x))
 f x  1
 x  1
3
Domain: All real #s
g ( f ( x))
 
 x   1
 g x3
3
Domain: All real #s
4
9
3
4
9
1 3
4 4
9
1
2

1
3
Find the inverse function.
To find an inverse:
1. Change the f(x) to a y
2. Switch the x and y’s.
3. Solve for y.
4. Change y to a f-1(x)
17.
f ( x)  3 x  2
18.
f ( x)  
3
x6
4
3
x6
4
3
x   y6
4
4
4 3 
  x  6     y 
3
3 4 
4
 ( x  6)  y
3
4
 ( x  6)  f 1 ( x)
3
y
y  3x  2
x  3y  2
x  2  3y
x2
y
3
x2
 f 1 ( x)
3
19.
f ( x)  x 4 , x  0
20.
f ( x)  2 x 3
y  2x 3
x  2y3
y  x4
x
 y3
2
x 3 3
3
 y
2
x  y4
4
x  4 y4
4 xy
4 x f
1
( x)
3
x
y
2
3
x
 f
2
1
( x)
Graph the function (not the inverse). Then state the domain and range. Also label the starting
point/vertex.
21.
f ( x)  x  1  4
22.
f ( x)  3 x  1  2
23.
f ( x)  23 x  3  1
Starting Point: (– 1, 4)
Starting Point: (1, 2)
Starting Point: (– 3, 1)
Domain: All real #s ≥ – 1
Domain: All real #s
Domain: All real #s
Range: All real #s ≥ 4
Range: All real #s
Range: All real #s
24.
f ( x)  2 x  2  1
25.
f ( x)  33 x  4  6
Starting Point: (2, – 1)
Starting Point: (– 4, 6)
Domain: All real #s ≥ 2
Domain: All real #s
Range: All real #s ≥ – 1
Range: All real #s
Solve the equation. Check for extraneous solutions.
26.
4
3x  2
 3x 
4
 2
4
4x  x  9
27.
 4x   
2
4
3x  16
16
x
3
x9
2x  3  x  3
28.

2

2x  3

2
  x  3
2
2 x  3  x  3x  3
4x  x  9
2x  3  x 2  6x  9
3x  9
0  x 2  8 x  12
0  x  6 x  2 
x3
x  6, x  2
Answer checks.
Answer checks.
After checking,
only x = 6 works
29.

x2 2  x
x2 2  x
x2

2
 x  2
2
x  2  ( x  2)( x  2)
x  2  x 2  4x  4
0  x 2  3x  2
0  ( x  2)( x  1)
x  2, x  1
After checking,
both work.
Chapter test will be half calculator and half non-calculator. Graphing questions will be on the noncalculator part of the test. Be sure you can graph square roots and cube roots by hand!

Also, remember how to determine the domain of a function. There are two things not allowed.
1. You cannot divide by 0
2. You cannot take an EVEN root of a NEGATIVE number.
Examples:
x3
Domain : All real numbers except x = 5.
x 5
(If x = 5, you would divide by 0)
2. Find the domain of: x 3/ 4 Domain: All non-negative numbers.
1. Find the domain of:
( x 3/ 4 means
 x  , and you can’t take a 4
4
3
th
root of a negative number, but 0 is ok)
3. Find the domain of: x 3/ 4 Domain: All positive real numbers.
1
1
( x 3/ 4 means 3/4 
, and you can’t take a 4th root of a negative number, or
3
4
x
x
 
divide by 0, so the domain is all positive real numbers)

Remember what function notation means. f(g(x)) is read “f of g of x.” It does not mean “f times
g times x.”

Be sure to be able to do a composite function. Remember to determine the domain at the start
of the problem, not at the end. The ending result does NOT determine the domain!

Always check for extraneous solutions. There is a good chance one will appear on your test.

Remember that properties of rational exponents are the same as properties of integer
exponents.