Solutions Test 1A - Solutions to Test 1, Version A Math...

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Solutions to Test 1, Version A Math 1501, Fall 02. WG * September 10, 2002 Exercise 1 lim x →- 2 ( x 2 - x - 6) 2 x +2 = lim x →- 2 [( x +2)( x - 3)] 2 x = lim x →- 2 ( x x - 3) 2 =0 . Exercise 2 Assume that f ( x )= ± A 2 x 2 if x 2 (1 - A ) x if x> 2 . Then f is continuous at 2 iff 4 A 2 = lim x 2 + f ( x ) = lim x 2 + (1 - A ) x =2(1 - A ) . This means that 4 A 2 A - 2=0 , and so, A =1 / 2or A = - 1 . Exercise 3 For all c R , we have that lim x c f ( x ) = lim x c x 2 = c 2 . Case 1: c N . The function f is continuous a c if and only if c 2 = lim x c f ( x f ( c )=2 c, i.e. c 2 - 2 c . Hence c =0or c =2 . Case 2: c 6∈ N . We have that lim x c f ( x c 2 = f ( c ) . Consequently, f is continuous a c. Summary. f is continuous a c if and only if c , or c , or c 6∈ N . Exercise 4 Recall that f ( x 1 ( x - 1) 2 + 1 x - 10 2 , and so, f is continuous on the interval (1 , 10 2 ) . Set a . 01 ,b = 10 2 - 0 . 001 . We first use that 10 2 - 1 . 01 > 1 to deduce that f ( a ) = 10000 + 1 1 . 01 - 10 2 > 10000 - 1 > 0 . * School of Mathematics, Georgia Institute of technology. 1
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Then we use that 10 2 - 1 . 001 > 1 to deduce that f ( b )= 1 ( 10 2 - 1 . 001) 2 - 1 0 . 001 = 1 ( 10
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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 Institute of Technology.

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Solutions Test 1A - Solutions to Test 1, Version A Math...

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