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Unformatted text preview: ⇒ x = 4 3 √ 4 . 1 If x < 4 3 √ 4, f ( x ) < 0, and if x > 4 3 √ 4, f ( x ) > 0. Thus, f ( x ) has an absolute minimum at x = 4 3 √ 4. In this case the length of the ladder is √ f ( x ) = ± (4 3 √ 4 + 4) 2 + 64(4 3 √ 4)2 (4 3 √ 4 + 4) 2 ≈ 14 . 5ft 2. Find f ( θ ) that satisﬁes f 00 ( θ ) = sin θ + cos θ, f (0) = 3 , f (0) = 4 . Solution: f ( θ ) =cos θ + sin θ + C 1 f ( θ ) =sin θcos θ + C 1 θ + C 2 Since f (0) =1 + C 2 = 3, C 2 = 4. And f (0) =1 + C 1 = 4, C 1 = 5. Therefore, f ( θ ) =sin θcos θ + 3 θ + 4 . 2...
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 Fall '06
 McAdam
 Calculus, 8 FT, 4 ft, ladder, ft tall runs, Guoliang Wu

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