10.07 - For next Wednesday, read section 2.3. Last time we...

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For next Wednesday, read section 2.3. Last time we looked at the function f ( x ) = x | x |: - 2 - 1 1 2 - 4 - 2 2 4 which can also be described by the two-part formula {– x 2 if x < 0, f ( x ) = { { x 2 if x 0. We showed last time that f ( x ) is differentiable at x =0: lim h 0+ ( h | h | – 0)/ h = lim h 0+ h 2 / h = lim h 0+ h = 0 and lim h 0– ( h | h | – 0)/ h = lim h 0– h 2 / h = lim h 0– h = 0. Since the left-hand and right-hand limits agree, the two- sided limit exists. So the derivative of f ( x ) at x =0 is 0. What is the derivative of f ( x ) = x | x | at x =1? . .. ..?. .
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Note that for all h sufficiently close to 0 (specifically, for all h with | h | < 1), f (1+ h ) = (1+ h ) 2 . Hence the two-sided limit lim h 0 ( f (1+ h ) – f (1))/ h equals the two-sided limit lim h 0 ((1+ h ) 2 – (1) 2 )/ h = lim h 0 (2 + h ) = 2. [Explain this with a picture of y = x 2 superimposed on a picture of y = x | x |, and various secant lines.] The general principle at work here is:
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This note was uploaded on 02/13/2012 for the course MATH 141 taught by Professor Staff during the Fall '11 term at UMass Lowell.

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10.07 - For next Wednesday, read section 2.3. Last time we...

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