G 5 0 g 4 0 and g 4 1 so the answer is g 4

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Unformatted text preview: te minimum may occur at the end points x = −5 or x = 4 , or at x = −4 where g ′( x) = f ( x) changes from negative to positive. g (−5) = 0, g (4) > 0, and g (−4) = −1 , so the answer is g (−4) = −1 . (d) At a point of inflection, the derivative changes from increasing to decreasing or vice-versa. There are three such points: at x = −3, x = 1, and x = 2 . −3 6 FREE-RESPONSE SOLUTIONS ~ 2004 AB Question 6 (a) y 3 2 1 –1 (b) dy = x 2 ( y − 1) ⇒ dx 1 x y > 1, x ≠ 0 . (c) 0 y − 1 = 2e x3 3 dy ∫ y − 1 = ∫ x dx 2 x3 3 ⇒ y = 1 + 2e . 1 ⇒ ln y − 1 = x 3 + C . x = 0, y = 3 ⇒ C = ln 2 . 3 2004 BC AP Calculus Free-Response Solutions and Notes Question 1 See AB Question 1. Question 2 See AB Question 2. 7 8 FREE-RESPONSE SOLUTIONS ~ 2004 BC Question 3 (a) x(4) = x(2) + ∫ x′(t ) dt = 1 + ∫ 3 + cos ( t 2 ) dt 2 2 (b) y − y (2) = y′(2) −7 ( x − x(2) ) ⇒ y − 8 = ( x − 1) . x′(2) 3 + cos(4) (c) Speed = dx dy + = dt dt (d) x′(t ) = 3 + cos ( t 2 ) ⇒ x′′(4) = 4 4 2 2 ( 3 + cos(4) ) + ( −7 ) 2 d [ x′(t )] dt t =4 ≈ 7.133 . 2 . 1 1 ≈ 2.303 . y′(t ) d = 2t + 1 ⇒ y′(t ) =...
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This note was uploaded on 07/07/2013 for the course MATH AP taught by Professor Mr.snickles during the Fall '11 term at Benjamin N Cardozo High School.

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