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# 101f04fin - B U Department of Mathematics Math 101 Calculus...

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B U Department of Mathematics Math 101 Calculus I Fall 2004 Final Calculus archive is a property of Bo˘gazi¸ci University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. This archive is a non-profit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-profit purpose may result in severe civil and criminal penalties. 1.) Let f ( x ) = - x if x < 0 3 - x if 0 x < 3 ( x - 3) 2 if x > 3 Determine the point(s) at which f ( x ) is discontinuous. Explain in detail. Solution: Check the points x = 0 and x = 3 because these are candidate points of discontinuity. x = 0 : lim x 0 - f ( x ) = lim x 0 - - x = 0 lim x 0 + f ( x ) = lim x 0 + (3 - x ) = 3 lim x 0 f ( x ) doesn’t exist and therefore f is discontinuous at x = 0. x = 3 : lim x 3 - f ( x ) = lim x 3 - (3 - x ) = 0 lim x 3 + f ( x ) = lim x 3 + ( x - 3) 2 = 0 lim x 3 f ( x ) = 0 but f (3) is undefined and therefore f is discontinuous at x = 3, too because for continuity we must have lim x a f ( x ) = f ( a ) . 2.) For what values of r does the function y = e rx satisfy the equation y 00 + 5 y 0 - 6 y = 0? Solution: y = e rx y 0 = re rx and y 00 = r 2 e rx Now substitute in the equation above r 2 e rx + 5 re rx - 6 e rx = 0 e rx ( r 2 + 5 r - 6) = 0 e rx cannot be zero and hence r

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101f04fin - B U Department of Mathematics Math 101 Calculus...

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