Solutions Test 1B - Solutions to Test 1, version B Math...

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Solutions to Test 1, version B Math 1501, Fall 02. WG * September 17, 2002 Exercise 1 lim x →- 2 x 2 - 2 x - 8 x +2 = lim x →- 2 ( x +2)( x - 4) x = lim x →- 2 ( x - 4) = - 6 . Exercise 2 Assume that f ( x )= ± A 2 x 2 if x 2 - Ax if x> 2 . Then f is continuous at 2 iff 4 A 2 = lim x 2 + f ( x ) = lim x 2 + - Ax = - 2 A. This means that 4 A 2 A = 0 and so, A =0or A = - 1 / 2 . 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 )=7 c, i.e. c 2 +7 c =0 . Because - 7 6∈ N , we deduce that c . 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 6∈ N . Exercise 4 Recall that f ( x 1 ( x - 1) 2 + 1 x - 7 , and so, f is continuous on the interval (1 , 7) . Set a =1+0 . 01 and b = 7 - 0 . 001 . We first use that 7 - 1 . 01 > 1 to deduce that f ( a ) = 10000 - 1 7 - 1 . 01 > 10000 - 1 > 0 . * School of Mathematics, Georgia Institute of technology. 1
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Then we use that ( 7 - 1 . 001) 2 > 1 to deduce that f ( b )= 1 ( 7 - 1 . 001) 2 - 1 - 0 . 001 = 1 ( 7 -
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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 1B - Solutions to Test 1, version B Math...

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