unit 6
DESCRIPTION
Unit 6. GA2 Test Review. 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. - PowerPoint PPT PresentationTRANSCRIPT
Unit 6GA2Test Review
Find the indicated real nth root(s) of a.
a. n = 3, a = –216 b. n = 4, a = 81
SOLUTION
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
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.
Evaluate:
SOLUTIONRational Exponent Form Radical Form
a. 163/2 (161/2)3 = 43= 64=
b. 32–3/5 = 1323/5 = 1
(321/5)3
= 123
18
=
a. 163/2 ( )3 = 16 43= 64=
b. 32–3/51
323/5 = 1( )35 32
=
= 123
18
=
(a) 163/2 (b) 32–3/5
Solve the equation.
4x5 = 128
x5 32=
x = 325
x 2=
SOLUTION
( x + 5 )4 = 16
SOLUTION
( x + 5 )4 = 16
( x + 5 ) = + 4 16
x = + 4 16 – 5x = 2 – 5 or x = – 2 – 5
x = – 3 or x = –7
Solve the equation.
1. (51/3 71/4)3 = (51/3)3 (71/4)3
2. 23/4 21/2
= 3(1 – 1/4)
Simplify the expressions.
= 205
1/2 3(41/2)3=
51/3 3 71/4 3= 51 73/4= 5 73/4=
25/4=2(3/4 + 1/2)=
31
31/4= 33/4=
(22)3/2= 8=
1 4
33.
331 2
1 2
204.
5
Simplify the expressions.
274 34 27 34= 814= = 3
= 23
533 23= 5=
23533 2
245
2=35
4585
85=
325245
=
4 4. 27 3a
3
3
250.
2b
53
. 4
c
Simplify the expression. Assume all variables are positive.
= 33(q3)33 333 (q3)33= = 3q3
=x105
y55(x2)55
=y55
=x2
y
= 2x(1 – 1/2)y(3/4 –1/2) = 2x1/2y1/4
93 27q
10
55
xy
3 4
1 2 1 2
63xyx y
a.
b.
c.
Let f (x) = –2x2/3 and g(x) = 7x2/3. Find the following, state the domain.
f (x) + g(x)1.
SOLUTION
f (x) + g(x) = –2x2/3 + 7x2/3 = (–2 + 7)x2/3 = 5x2/3
f (x) – g(x)2.
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.
SOLUTION
SOLUTION
f (x) g(x)a.
f (x) g(x) = 3x x1/5 = 3(x ) 1 + 1/5 = 3x6/5
f (x)g(x) =
3xx1/5 = 3(x ) 1 – 1/5 = 3x4/5
( )( )f xg x
b.
Let f(x) = 3x – 8 and g(x) = 2x2. Find the following.
a g(f(5))
SOLUTION
To evaluate g(f(5)), you first must find f(5).
f(5)Then g( f(3))
= 3(5) – 8 = 7= g(7)
= 2(7)2
= 2(49)= 98
b f(g(5))
SOLUTION
To evaluate f(g(5)), you first must find g(5).
g (5)
Then f( g(5))
= 2(5)2 = 2(25)
= f(50)
= 3(50) – 8
= 150 – 8
= 142.
= 50
f(x) = –3x – 1
Find the inverse.
y = –3x + 1
x = –3y +1
x – 1 = –3y
x 13 = y
SOLUTION
ANSWER g–1(x) = 33√ x
Find the inverse.
31( )
27g x x
f(x) = –x3 + 4
ANSWER f –1(x) = 3√ 4 – x
Find the inverse.
Graph the function.
Then state the domain and range.
ANSWER
Domain : x > 0 ,range : y < 0.
4 2y x
Graph the function.
Then state the domain and range.
ANSWER
Domain :all real numbers,
range: all real numbers.
3( ) 2 3g x x
Solve (x + 2)3/4 – 1 = 7
(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
SOLUTION
Solve the equation. Check your solution.
3x3/2 = 375
3x3/2 = 375
x3/2 = 125
(x3/2)2/3 = (125)2/3
x = 25
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
SOLUTION
1 7 15x x Solve
x + 6 – 4 x + 6 + 4 = x – 2
– 4 x + 6 = – 12
x + 6 = 3
Solve the equation. Check for extraneous solutions.
6 2 2x x
x + 6 = 9
x = 3
SOLUTION