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Unformatted text preview: Math 113 Exam #1 Solutions 1. What are the domain and range of the function f ( x ) = 1 4 x ? Answer: f ( x ) is welldefined provided p 4 x ) 6 = 0. In order for 4 x to exist, we must have 4 x , meaning that x 4. In order for 4 x 6 = 0 it must also be the case that x 6 = 4, so the domain of f is ( , 4) . Since 4 x > 0 where f is defined, we see that f ( x ) > 0 for all x in the domain, so the range of f is (0 , + ) . 2. Evaluate lim x 4 x 2 8 x + 7 17 x + 12 Answer: Dividing both numerator and denominator by x yields lim x 1 x 4 x 2 8 x + 7 1 x (17 x + 12) = lim x q 1 x 2 (4 x 2 8 x + 7) 17 + 12 x = lim x q 4 8 x + 7 x 2 17 + 12 x = 4 17 = 2 17 . 3. Let f ( x ) = x 2 x 2 4 for x 6 = 2 a for x = 2 If f ( x ) is continuous at x = 2, then find the value of a . Answer: In order for f to be continuous at x = 2, we must have that a = f (2) = lim x 2 f ( x ) = lim x 2 x 2 x 2 4 ....
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This note was uploaded on 01/18/2012 for the course MATH 2250 taught by Professor Chestkofsky during the Fall '08 term at University of Georgia Athens.
 Fall '08
 CHESTKOFSKY
 Math, Calculus

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