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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 = Radical Form 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 x1 3 =y Find the inverse. g(x) 1 27 x 3 ANSWER g–1(x) = 33√ x 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 ANSWER Domain : x > 0 ,range : y < 0. Graph the function. Then state the domain and range. g(x) 3 x2 3 ANSWER 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