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Unformatted text preview: Solutions for Practice Final B (I) (a)  f ( x ) f (1)  < 1 / 2 is the same as < x 2 x + 1 < 1 . Clearly x 2 / ( x + 1) > 0 if and only if x > 1. We need only consider such x in analyzing the other inequality. Then, with x > 1 so that x + 1 > 0, x 2 x + 1 < 1 ⇐⇒ x 2 < x + 1 ⇐⇒ x 1 2 2 ≤ 5 4 ⇐⇒ x 1 2 ≤ √ 5 2 ⇐⇒ 1 2 √ 5 2 < x < 1 2 + √ 5 2 . Since this interval is included in the range x > 1, the final answer is 1 2 √ 5 2 < x < 1 2 + √ 5 2 . (I) (b) Writing x = 1 + h , f ( x ) f (1) = 2(1 + h ) 2 (2 + h ) 2(2 + h ) = 3 + 2 h 4 + 2 h h Therefore  f ( x ) f (1)  < is the same as  3 + 2 h   4 + 2 h   h  < . If we have  h  < 1, then  4 + 2 h  > 2 and  3 + 2 h  < 5 so  h  < 1 ⇒  f ( x ) f (1)  < 5 2  h  and the right side is less than in case  h  < 2 / 5. Altogether, δ ( ) = min 1 , 2 5 . 1 (II) (a) √ x √ 3 3 x = 1 √ x + √ 3 so lim x → 3 √ x √ 3 3 x = 1 2 √ 3 . (II) (b) Let y = 2 1 / 3 x . Then as x → 0, y → 0. Since sec 2 ( y ) 1 = tan 2 ( y ), the limit is the same as lim y → tan 2 ( y ) 2 1 / 3 y 2 = 2 2 / 3 . (II) (c) Let y = x 1 / 3 . Then as x → 1, y → 1. Then  x 1  x 1 / 3 1 =  y 3 1  y 1 =  y 1  y 1 ( y 2 + y + 1) . The left limit is 3 and the right limit is 3 so the limit does not exist. (III) Since x and x 2 are themselves continuous functions of x , f is continuous where they agree; i.e., at x = x 2 , which means x = 0 and x = 1. (IV) (a) Differentiating both sides with respect to x , 3 y 2 y y 2 2 xyy = 1 ....
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This note was uploaded on 02/07/2011 for the course MATH 1501 taught by Professor N/a during the Fall '08 term at Georgia Tech.
 Fall '08
 N/A
 Calculus

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