review3s11answers - REVIEW 3, SPRING 2011 SOLUTIONS 1. (a)...

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REVIEW 3, SPRING 2011 SOLUTIONS 1. (a) f 0 ( x ) = 2 x ln4 ± 2tan x ln4 (b) dy dx = e tan x x ² x sec 2 x ± tan x x 2 (c) dy dx = 3 x 3 +2 x (ln3)(3 x 2 + 2) (d) dy dx = x sin x ± sin x x + cos x ln x ² (e) f 0 ( x ) = 1 p 25 ± x 2 2. (a) f 0 ( x ) = (6 x ± 2) 3 ( x + 4) 2 ± 9 3 x ± 1 + 2 x + 4 ² (b) f 0 ( x ) = e x ± 3 3 p 6 + 3 x (3 x + 1) 2 ² ± 1 + 1 6 + 3 x ± 6 3 x + 1 ² 3. (a) dy dx = ± 1 p 1 ± x 2 (b) d 2 y dx 2 = ± x p (1 ± x 2 ) 3 4. (a) t = 1 second, t = 4 seconds (b) (0 ; 1) and t > 4 seconds (c) displacement: -16 feet; total distance: 38 feet (d) speeding up: t > 4, 1 < t < 5 2 ; slowing down: 0 < t < 1, 5 2 < t < 4 5. Cost of producing the 51 st item ³ $150 (Marginal cost); Average cost decreases by $1.95 per additional unit (Marginal average cost) 6. Revenue is increasing by $15/unit (Marginal revenue) Production increases by 30 units/ week ( dx dt ) 7. Distance increases by 2 p 5km/min Angle of elevation is decreasing by 1 rad/min 8. Height is increasing by 2 3 ± cm/sec 9. For f ( x ): critical numbers are x = 1, ± 1, 0, p 3, ± p 3 local minima at x = 0, p 3, ± p 3; local maxima at x = ± 1, 1 For g ( x ): critical numbers are x = ± 2, 4
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review3s11answers - REVIEW 3, SPRING 2011 SOLUTIONS 1. (a)...

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