Practice final Exam b solutions

# Practice final Exam b solutions - Solutions for Practice...

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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.

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Practice final Exam b solutions - Solutions for Practice...

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