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# 10.12 - For Thursday read section 2.4"Paradox"Solve x2 = 1...

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For Thursday, read section 2.4. “Paradox”: “Solve x 2 = 1. We get x = +1 and x = –1. So 1 = x = –1, implying 1 = –1.” ..?.. An equation is only true (or false) in a context and in a domain of validity . Ignore the context and domain of validity and you may get nonsense. That may seem just silly, but later this week I’ll show you a calculus counterpart of this paradox. Section 2.2: The derivative as a function Key facts (to be proved later): Where f ( a ) > 0, f ( x ) is increasing on some neighborhood of a ; where f ( a ) < 0, f ( x ) is decreasing on some neighborhood of a ; where f ( a ) = 0, no conclusion can be drawn without more information. [Sketch x 2 , – x 2 , x | x |, and – x | x |.] [Do problem 3 on page 92.] Note that in each case, the function f ( x ) is either odd or

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even, and that when f ( x ) is odd f ( x ) is even and vice versa. This is no accident: Claim: The derivative of any even function is an odd function. Proof: Suppose f ( x ) is even, so that f (– x ) = f ( x ) for all x . Then for all x we have f ( x ) = lim h 0 ( f ( x + h ) – f ( x ))/ h and f (– x ) = lim h 0 ( f (– x + h ) – f (– x ))/ h = lim h 0 ( f ( x h ) – f ( x ))/ h (since f ( x ) is even) = lim h 0 ( f ( x + h ) – f ( x ))/(– h ) (replace h by – h )
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10.12 - For Thursday read section 2.4"Paradox"Solve x2 = 1...

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