hws09 - Math 117 F 09 HW#9(sample solutions P.105#15 Let x0...

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Math 117, F 09, HW#9 (sample solutions) P.105, #15. Let x 0 D , and let ± > 0. Since f is continuous at x 0 , there exists δ 1 > 0 such that | f ( x ) - f ( x 0 ) | < ± for all x D with | x - x 0 | < δ 1 . Likewise, there exists δ 2 > 0 such that | g ( x ) - g ( x 0 ) | < ± for all x D with | x - x 0 | < δ 2 . Set δ = min { δ 1 , δ 2 } > 0. If x D with | x - x 0 | < δ , then | x - x 0 | < δ 1 and | x - x 0 | < δ 2 , so | f ( x ) - f ( x 0 ) | < ± and | g ( x ) - g ( x 0 ) | < ± . Then f ( x 0 ) - ± < f ( x ) < f ( x 0 ) + ± and g ( x 0 ) - ± < g ( x ) < g ( x 0 ) + ± , and hence h ( x 0 ) - ± = max { f ( x 0 ) - ±, g ( x 0 ) - ± } < max { f ( x ) , g ( x ) } = h ( x ) h ( x ) = max { f ( x ) , g ( x ) } < max { f ( x 0 ) + ±, g ( x 0 ) + ± } = h ( x 0 ) + ±, which shows that | h ( x ) - h ( x 0 ) | < ± . Therefore h is continuous at x 0 . ± P.105, #17. Define g : [0 , ) R to be the square root function: g ( x ) = x for x [0 , ). By Exercise 6, p. 104, g is continuous. The function f is just the composition g f , and we are given that
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This note was uploaded on 08/27/2011 for the course MATH 117 taught by Professor Akhmedov,a during the Fall '08 term at UCSB.

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