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MTH131 Applied Calculus – Spring 2016 Lab 1 – SOLUTIONS 1 13 y = x− 3 3 (b) There are many solutions to each problem. The simplest comes when we set x = 0 and get the corresponding y value. The table below gives that answer along with several others for various values of x; a) b) c) (x1 , y1 ) equation (x1 , y1 ) equation (x1 , y1 ) equation (0, 19) y − 19 = 5x (0, −2) y + 2 = 13 x (0, − 32 ) y + 23 = − 35 x 1. (a) a) y = −3x + 6 b) y = 5x + 19 c) (1, 24) y − 24 = 3(x − 1) (3, −1) y + 1 = 13 (x − 3) (5, − 11 3 ) y+ 11 3 (−3, 4) y − 4 = 5(x + 3) (6, 0) y = 13 (x − 6) (−5, 37 ) y− 7 3 2. (a) 3x1/2 (b) 3x−1/2 (c) 3x−2 (d) 9x−1 (e) 18x3/2 (f) 2x−2 3. a) 31/2 b) 30 c) 3−1/2 3−1 d) 4. (a) 27y 4 (b) u2 v 2 w2 or (uvw)2 5. (a) The two compositions are determined as follows: f (g(x)) = = = f (x + 1) (x + 1) + 1 (x + 1) − 1 x+2 2 and x+1 g x−1 x+1 +1 x−1 x+1 x−1 + x−1 x−1 x + 1 + (x − 1) x−1 2x . x−1 g(f (x)) = = = = = = 3(x − 5) = 5(x + 5) (b) The most obvious set of functions is f (x) = 2x2 + 17 and g(x) = √ x Using these, we have √ f (g(x)) = f ( x) √ = 2( x)2 + 17 = 2x + 17. (c) There are many different answers here (an infinite number in fact). One possibility is √ f (x) = 2x2 and g(x) = x + 17 Using these, we have √ f (g(x)) = f ( x + 17) √ = 2( x + 17)2 = 2(x + 17) = 2x + 34. 6. The first 5 can all be solved by factoring: (a) x2 − 7x + 12 = 0: x2 − 7x + 12 = 0 (x − 3)(x − 4) = 0 ⇒ x = 3, 4 (b) x2 + 7x + 12 = 0: x2 + 7x + 12 = 0 (x + 3)(x + 4) = 0 ⇒ x = −3, −4 (c) x2 − 7x + 12 = 2: x2 − 7x + 12 = 2 x2 − 7x + 10 = 0 (x − 2)(x − 5) = 0 ⇒ x = 2, 5 (d) x2 + 3x − 18 = 0: x2 + 3x − 18 = 0 (x + 6)(x − 3) = 0 ⇒ x = −6, 3 (e) 3x2 + 12x + 11 = 8: 3x2 + 12x + 11 = 2 3x2 + 12x + 9 = 0 x2 + 4x + 3 = 0 (x + 3)(x + 1) = 0 ⇒ x = −3, −1 The last one cannot be factored easily, so we use the quadratic equation. (f) 3x2 + 12x + 11 = 0: 3x2 + 12x + 11 = 0 → a = 3, b = 12, c = 11 p −12 ± 122 − 4(3)(11) x = 2(3) √ −12 ± 144 − 132 = √6 −12 ± 12 = ,or 6√ √ 12 12 = −2 + , −2 − 6 6