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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 < 0 (because the area of the region
−1 below the xaxis for −1 ≤ t ≤ 4 appears much bigger than the area of the region
above the xaxis 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. FREERESPONSE SOLUTIONS ~ 2004 AB (FORM B) Question 5
(a) y
3
2
1
–1 (b)
(c) 0 1 x x ≠ 0, y < 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 FREERESPONSE 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.
 Fall '11
 Mr.Snickles
 Calculus

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