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Solutions Test 1B

# Solutions Test 1B - Solutions to Test 1 version B Math 1501...

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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 + 2 = 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 + 2 A = 0 and so, A = 0 or 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 = 0 . 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 = 0 , 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 (
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Solutions Test 1B - Solutions to Test 1 version B Math 1501...

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