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E3FreeAns

E3FreeAns - 0 1 toward horiz asymptote W = 2 1d W-1 x =-1 2...

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Answers – Fall 2009 Exam 3 version A 1a. W 0 ( x ) = 2 e - x ; W 00 ( x ) = - 2 e - x ; increasing and concave down 1b. W (0) = 2 hundred wolves now; lim x →∞ W ( x ) = 4 hundred in the long run 1c. graph is increasing, conc down from ( 0 , 2 ) toward horiz asymptote W = 4 1d. W - 1 ( x ) = - ln(2 - 1 2 x ) 3a. number line reveals max is f (2) = 2 + 2 ln(8) ; of two endpoints left to check, the min is f ( - 3) = - 3 + 2 ln(3) by inspection 3b. max area A = 8 2 m 2 , when x = 2 3c. A 00 = - 12 x at x = 2 ; concave down .... max 4. lim x 0 + g ( x ) = -∞ ; lim x →∞ g ( x ) = inc: ( 0 , 1 ) , ( 4 , ) dec: ( 1 , 4 ) HTL: at x = 1 , 4 Local min at: x = 4 point for graph: ( 4 , - 12 . 8 ) Local max at: x = 1 point for graph: ( 1 , - 9 ) conc up: ( 2 , ) conc down: ( 0 , 2 ) inflection at x = 2 point for graph: ( 2 , - 10 . 4 ) 5a. dy dx = x - 1 / 4 e x 2 +1 2 x - 1 4 x 5b. x = p 1 / 8 5c. 7 4 e 2

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Answers – Fall 2009 Exam 3 version B 1a. W 0 ( x ) = 2 e - 2 x ; W 00 ( x ) = - 4 e 2 - x ; increasing and concave down 1b. W (0) = 1 hundred wolves now; lim x →∞ W ( x ) = 2 hundred in the long run 1c. graph is increasing, conc down from
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Unformatted text preview: ( 0 , 1 ) toward horiz asymptote W = 2 1d. W-1 ( x ) =-1 2 ln(2-x ) 3a. number line reveals max is f (1) = 1-2 ln(4) ; of two endpoints left to check, the min is f (-3) =-3-2 ln(12) by inspection 3b. max area A = 24 √ 3 m 2 , when x = √ 3 3c. A 00 =-24 x at x = √ 3 ; concave down. ... max 4. lim x → + g ( x ) = ∞ ; lim x →∞ g ( x ) =-∞ inc: ( 1 , 4 ) dec: ( 0 , 1 ) , ( 4 , ∞ ) HTL: at x = 1 , 4 Local min at: x = 1 point for graph: ( 1 , 4 . 5 ) Local max at: x = 4 point for graph: ( 4 , 6 . 4 ) conc up: ( 0 , 2 ) conc down: ( 2 , ∞ ) inﬂection at x = 2 point for graph: ( 2 , 5 . 2 ) 5a. dy dx = x-1 / 2 e x 2 +6 ± 2 x-1 2 x ² 5b. x = 1 / 2 5c. 3 2 e 7...
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E3FreeAns - 0 1 toward horiz asymptote W = 2 1d W-1 x =-1 2...

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