Unit 6 GA2 Test Review. Find the indicated real n th root ( s ) of a. a. n = 3, a = –216 b. n = 4, a = 81 SOLUTION b. Because n = 4 is even and a = 81.

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Presentation transcript:

Unit 6 GA2 Test Review

Find the indicated real n th 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 3 4 = 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: SOLUTION Rational Exponent FormRadical Form a. 16 3/2 (16 1/2 ) 3 = 4343 = 64 = b. 32 –3/5 = /5 = 1 (32 1/5 ) 3 = = a. 16 3/2 ( ) 3  = = 64= b. 32 –3/ /5 = 1 ( ) 3 5  32 = = = (a) 16 3/2 (b) 32 –3/5

Solve the equation. 4x 5 = 128 x5x5 32= x=  5 x2 = SOLUTION

( x + 5 ) 4 = 16 SOLUTION ( x + 5 ) 4 = 16 ( x + 5 ) = x = – 5 x = 2 – 5 or x= – 2 – 5 x = – 3 or x = –7 Solve the equation.

1. (5 1/3 7 1/4 ) 3 = (5 1/3 ) 3 (7 1/4 ) /4 2 1/2 = 3 (1 – 1/4) Simplify the expressions. = /2 3 (4 1/2 ) 3 = 5 1/ /4 3 = /4 =5 7 3/4 = 2 5/4 =2 (3/4 + 1/2) = /4 = 3 3/4 = (2 2 ) 3/2 = 8 =

Simplify the expressions =81 4 = =3 = 2   3 2  3 = 5 = 2    5 2 = 3  5 4  5 8  5 8  5 = 32  5 24  5 =

Simplify the expression. Assume all variables are positive. =33(q3)333(q3)3  33  3 (q3)3(q3)3  3 == 3q33q3 = x 10  5 y5y5  5 (x2)5(x2)5 5  = y5y5  5 = x2x2 y = 2x (1 – 1/2) y (3/4 –1/2) = 2x 1/2 y 1/4 a. b. c.

Let f (x) = –2x 2/3 and g(x) = 7x 2/3. Find the following, state the domain. f (x) + g(x) 1. SOLUTION f (x) + g(x) = –2x 2/3 + 7x 2/3 = (–2 + 7)x 2/3 = 5x 2/3 f (x) – g(x) 2. SOLUTION f (x) – g(x) = –2x 2/3 – 7x 2/3 = [–2 + ( –7)]x 2/3 = –9x 2/3

Let f (x) = 3x and g(x) = x 1/5. Find the following, state the domain. SOLUTION f (x) g(x) a. f (x) g(x) = 3x x 1/5 = 3(x ) 1 + 1/5 = 3x 6/5 f (x) g(x)g(x) = 3x3x x 1/5 = 3(x ) 1 – 1/5 = 3x 4/5 b.

Let f(x) = 3x – 8 and g(x) = 2x 2. 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  1 33 = y= y SOLUTION

ANSWER g –1 (x) = 3 3 √ x Find the inverse.

f(x) = –x 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.

Graph the function. Then state the domain and range. ANSWER Domain :all real numbers, range: all real numbers.

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 = 2 4 x + 2 = 16 x = 14 SOLUTION

Solve the equation. Check your solution. 3x 3/2 = 375 x 3/2 = 125 (x 3/2 ) 2/3 = (125) 2/3 x = 25 SOLUTION

 x + 1 = 7x + 15 (x + 1) 2  = ( 7x + 15) 2 x 2 + 2x + 1 = 7x + 15 x 2 – 5x – 14 = 0 (x – 7)(x + 2) = 0 x – 7 = 0 or x + 2 = 0 x = 7 or x = –2 SOLUTION Solve

x + 6 – 4 x = x – 2 – 4 x + 6 = – 12 x + 6 = 3 Solve the equation. Check for extraneous solutions. x + 6 = 9 x = 3 SOLUTION