APcal 2004-Solutions

# Therefore the absolute maximum is at x 1 c g 2 2

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Unformatted text preview: f ′(t )dt + C . Since f ′(t ) ≤ 0 for − 1 ≤ t ≤ 4 and f ′(t ) ≥ 0 for 4 ≤ t ≤ 5 , x −1 f reaches the absolute minimum at x = 4. f could reach the absolute maximum at 5 x = −1 or x = 5 . But f (5) − f (−1) = ∫ f ′(t )dt &lt; 0 (because the area of the region −1 below the x-axis for −1 ≤ t ≤ 4 appears much bigger than the area of the region above the x-axis for 4 ≤ t ≤ 5 ). Therefore, the absolute maximum is at x = −1 . (c) g (2) = 2 f (2) = 12 . g ′(2) = ( f ( x) + xf ′( x) ) x = 2 1 = 6 + 2 ( −1) = 4 . An equation for the tangent line is y − g (2) = g ′(2) ( x − 2 ) ⇒ y − 12 = 4( x − 2) . Notes: 1. Using the Product Rule. FREE-RESPONSE SOLUTIONS ~ 2004 AB (FORM B) Question 5 (a) y 3 2 1 –1 (b) (c) 0 1 x x ≠ 0, y &lt; 2 . ∫ dy x5 = ∫ x 4 dx ⇒ ln y − 2 = + C . y (0) = 0 ⇒ C = ln 2 . y−2 5 y − 2 = 2e x5 5 x5 5 ⇒ y = 2 − 2e . 1 Notes: 1. We reject the solution y = 2 + 2e x5 5 because for this solution y (0) ≠ 0 . 15 16 FREE-RESPONSE SOLUTIONS ~ 2004 AB (FORM B) Question 6 1 (a) x n +1 1 x dx = = . ∫0 n +1 0 n +1 (b) The equation for the tangent line at (1, 1) is 1 n 1 y − 1 = n...
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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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