### Unit 6

```Unit 6
GA2
Test Review
Find the indicated real nth root(s) of a.
a. n = 3, a = –216
b. n = 4, a = 81
SOLUTION
a.
Because n = 3 is odd and a = –216 < 0, –216 has
one real cube root. Because (–6)3 = –216, you
can write = 3√ –216 = –6 or (–216)1/3 = –6.
b.
Because n = 4 is even and a = 81 > 0, 81 has
two real fourth roots. Because 34 = 81 and
(–3)4 = 81, you can write ±4√ 81 = ±3
Evaluate:
(a) 163/2
(b) 32–3/5
SOLUTION
Rational Exponent Form
a. 163/2 = (161/2)3 = 43 = 64
b.
1
1
=
323/5 (321/5)3
1
= 3 = 1
8
2
32–3/5 =
3
a. 163/2=( 16 ) = 43 = 64
b.
32–3/5
1
1
= 3/5 = 5
32
( 32 )3
= 13 = 1
2
8
Solve the equation.
4x5 = 128
SOLUTION
x5 = 32
5
x = 32
x=2
Solve the equation.
( x + 5 )4 = 16
SOLUTION
( x + 5 )4 = 16
(x+5) =+
4
16
x = + 4 16 – 5
x = 2 – 5 or x = – 2 – 5
x = –3
or x = –7
Simplify the expressions.
1.
(51/3 71/4)3 =
2.
23/4 21/2
3.
3
3
(51/3)3 (71/4)3 = 51/3
3
71/4
3
= 2(3/4 + 1/2) = 25/4
1
3
(1 – 1/4)
3/4
=
3
3
=
=
31/4
1 4
 20 
4.  1 2 
 5

1 2
3
=
20
5
1/2 3
= (41/2)3 = (22)3/2 = 8
= 51 73/4 = 5 73/4
Simplify the expressions.
a.
4
27 
3
b.
c.
3
4
250
3
5
4
2
3
4
27
3
=
 53
4
2
=
 2
3
= 5
4
4
27 3 = 4 81 = 3
3
3
5
=
3
5
8
5
8
3
 53
 2
3
 2
= 5
5
 24 =  24
= 5
2
 32
5
Simplify the expression. Assume all
variables are positive.
a.
b.
c.
3
27q
5
x
3
1 2
3
5
 (x2)5
 x10
x2
= 5
= y
= 5
5
5
y
y
5
5
6 xy
3
=  33(q3)3 =  33  (q3)3 = 3q3
10
y
3x
9
3 4
y
1 2
= 2x(1 – 1/2)y(3/4 –1/2) = 2x1/2y1/4
Let f (x) = –2x2/3 and g(x) = 7x2/3.
Find the following, state the domain.
1.
f (x) + g(x)
SOLUTION
f (x) + g(x) = –2x2/3 + 7x2/3 = (–2 + 7)x2/3 = 5x2/3
2.
f (x) – g(x)
SOLUTION
f (x) – g(x) = –2x2/3 – 7x2/3 = [–2 + ( –7)]x2/3 = –9x2/3
Let f (x) = 3x and g(x) = x1/5.
Find the following, state the domain.
a. f (x)
g(x)
SOLUTION
f (x) g(x) = 3x
x1/5 = 3(x ) 1 + 1/5 = 3x6/5
b. f ( x )
g(x)
SOLUTION
f (x)
3x
= 1/5 = 3(x ) 1 – 1/5 = 3x4/5
g(x)
x
Let f(x) = 3x – 8 and g(x) = 2x2.
Find the following.
a
g(f(5))
b
f(g(5))
SOLUTION
SOLUTION
To evaluate g(f(5)),
you first must find f(5).
To evaluate f(g(5)),
you first must find g(5).
f(5) = 3(5) – 8 = 7
Then g( f(3)) = g(7)
= 2(7)2
= 2(49)
= 98
g (5) = 2(5)2 = 2(25) = 50
Then f( g(5)) = f(50)
= 3(50) – 8
= 150 – 8
= 142.
Find the inverse.
f(x) = –3x – 1
SOLUTION
y = –3x + 1
x = –3y +1
x – 1 = –3y
x1
3
=y
Find the inverse.
g(x) 
1
27
x
3
Find the inverse.
f(x) = –x3 + 4
ANSWER f –1(x) = 3√ 4 – x
Graph the function.
Then state the domain and range.
y  4 x  2
Domain : x > 0 ,range : y < 0.
Graph the function.
Then state the domain and range.
g(x) 
3
x2 3
Domain :all real numbers,
range: all real numbers.
Solve (x + 2)3/4 – 1 = 7
SOLUTION (x + 2)3/4 – 1 = 7
(x + 2)3/4 = 8
(x + 2)3/4
4/3
= 8 4/3
x + 2 = (8 1/3)4
x + 2 = 24
x + 2 = 16
x = 14
Solve the equation. Check your solution.
3x3/2 = 375
SOLUTION
3x3/2 = 375
x3/2 = 125
(x3/2)2/3 = (125)2/3
x = 25
Solve x  1 
7 x  15
SOLUTION
x + 1 =  7x + 15
(x + 1)2 = (7x + 15)2
x2 + 2x + 1 = 7x + 15
x2 – 5x – 14 = 0
(x – 7)(x + 2) = 0
x – 7 = 0 or x + 2 = 0
x = 7 or
x = –2
Solve the equation.
Check for extraneous solutions.
x 6 2  x 2
SOLUTION
x+6–4
–4
x+6+4=x–2
x + 6 = – 12
x+6=3
x+6=9
x=3
```