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Unit 6

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.

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Unit 6

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  1. Unit 6 GA2 Test Review

  2. 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 Find the indicated real nth root(s) of a. a. n = 3, a = –216 b. n = 4, a = 81 SOLUTION

  3. 1 1 23 323/5 64 ( )3 = (161/2)3 = 43 = 43 = 64 = = 16  a. 163/2 a. 163/2 1 1 1 1 1 1 = = = = ( )3 (321/5)3 323/5 32  5 b. 32–3/5 8 23 8 b. 32–3/5 = = = = Evaluate: (a) 163/2(b)32–3/5 SOLUTION Rational Exponent Form Radical Form

  4. 4x5 = 128 x5 = 32 x = 32  5 x 2 = Solve the equation. SOLUTION

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

  6. 31 31/4 1. (51/3 71/4)3 = = 51/33 71/43 = = 2. 23/4 21/2 (51/3)3 (71/4)3 51 73/4 5 73/4 = 2(3/4 + 1/2) = 25/4 3(1 – 1/4) 33/4 = = = 3 1/2 8 = (22)3/2 (41/2)3 = = = 20 5 Simplify the expressions.

  7.    8 8 4 3 5 5 5 5       5 5 5 3 3 3 24 32 2 24 2 2 = 27 3 = = 81 = 3 4 4  3 53 2 = = 4 4 27 3  3 53 = = 5 = 2 Simplify the expressions.

  8. 3  33(q3)3 3q3 = = 5 5 5 3     33 y5 y5 x10 3  (q3)3 2x1/2y1/4 2x(1 – 1/2)y(3/4 –1/2) = = = = 5  (x2)5 x2 = = y Simplify the expression. Assume all variables are positive. a. b. c.

  9. f (x) + g(x) f (x) – g(x) Let f (x) = –2x2/3andg(x) = 7x2/3. Find the following, state the domain. SOLUTION f (x) + g(x) = –2x2/3 + 7x2/3 = 5x2/3 = (–2 + 7)x2/3 SOLUTION f (x) – g(x) = –2x2/3 – 7x2/3 = [–2 + ( –7)]x2/3 = –9x2/3

  10. f (x) g(x) a. f (x) g(x) = 3x x1/5 = 3(x ) 1 + 1/5 =3x6/5 f (x) 3x = = 3(x ) 1 – 1/5 =3x4/5 g(x) x1/5 Let f (x) = 3xandg(x) = x1/5. Find the following, state the domain. SOLUTION b. SOLUTION

  11. f(g(5)) g(f(5)) a b Let f(x) = 3x – 8 and g(x) = 2x2. Find the following. 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 g (5) = 2(5)2 =2(25) =50 = g(7) Theng( f(3)) Thenf( g(5)) = f(50) = 2(7)2 = 3(50) – 8 = 2(49) = 150 – 8 = 98 = 142.

  12. y = –3x + 1 x = –3y +1 x – 1 = –3y x 1 =y 3 Find the inverse. f(x) = –3x – 1 SOLUTION

  13. ANSWER g–1(x) = 33√ x Find the inverse.

  14. ANSWER f –1(x) = 3√ 4 – x Find the inverse. f(x) = –x3 + 4

  15. ANSWER Domain :x > 0 ,range : y < 0. Graph the function. Then state the domain and range.

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

  17. (x + 2)3/4 – 1 = 7 (x + 2)3/4 = 8 4/3 (x + 2)3/4 = 8 4/3 x + 2 = (8 1/3)4 x + 2 = 24 x + 2 = 16 x = 14 Solve (x + 2)3/4 – 1 = 7 SOLUTION

  18. 3x3/2 = 375 x3/2 = 125 (x3/2)2/3= (125)2/3 x = 25 Solve the equation. Check your solution. 3x3/2 = 375 SOLUTION

  19. 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 orx = –2 Solve SOLUTION

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

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