Calculus

Solved Math lines If you consider more problems, you can go to http://www.mathtrench.com/ Problem 1: termination: First we write ? x 2 ? 3x ? 1 ? f ( x) ? ? ? ? x?3 ? 5 and straight off we purpose the chain rule to generate: ? x2 ? 3x ? 1 ? d ? x2 ? 3x ? 1 ? ? x 2 ? 3 x ? 1 ? ? (2 x ? 3)( x ? 3) ? ( x 2 ? 3 x ? 1) ? f ( x) ? 5 ? ? ? ? ? 5? ? ? ? ( x ? 3) 2 ? x ? 3 ? dx ? x ? 3 ? ? x?3 ? ? ? 4 4 ? x 2 ? 3x ? 1 ? ? 2 x 2 ? 9 x ? 9 ? x 2 ? 3x ? 1) ? ? x 2 ? 3x ? 1 ? ? x 2 ? 6 x ? 8 ? ? 5? ? 5? ? ? ? ? ? ? 2 ( x ? 3) 2 ? x?3 ? ? ? ? x ? 3 ? ? ( x ? 3) ? 4 4 ? x 2 ? 3x ? 1 ? ( x ? 4)( x ? 2) ? 5? ? ( x ? 3) 2 ? x?3 ? Problem 2: 1) state that the place 4 xn 1 ? xn doesnt converge furnishly on [0, 2]. Solution: Lets compute the pointwise ascertain. We bemuse xn fn ( x) ? 1 ? xn ? For 0 ? x ? 1 : We have x n ? 0 as n ? ? , and then fn ( x) ? as n ? ? . ? xn 0 ? ?0 n 1? x 1 For x ? 1 : We have x n ? 1 , for all n and then fn ( x) ? ? 1n 1 ? n 1?1 2 For1 ? x ? 2 : We have x n ? ? as n ? ? , and then fn ( x) ? xn 1 ? ?1 n 1 1? x ?1 xn as n ? ? . The pointwise learn is ?0 ?1 ? f ( x) ? ? ?2 ?1 ? 0 ? x ?1 x ?1 1? x ? 2 This cultivate is not unceasing and therefore, theres no uniform convergence.

Problem 3: Find the ambit of f ? x ? ? x?5 x ? x?2 2 Solution: The domain is put together by avoiding a division by 0 and grow of negative numbers. In order for the function to be hale defined we need: ? ? x?5? 0 x 2 ? x ? 2 ? 0 ? ( x ? 2)( x ? 1) ? 0 From the precede conditions we find that x ? ?5 and x ? ?2, x ? 1 Theref ore, the domain is dom f ? ( ?5, ?) \ {?2! ,1} Problem 4: Decide if f(x) is continuous at x = 3, if ?x2 ?1 f ( x) ? ? ?2 x ? 6 x?3 x ?3 let off the reasoning tail assembly the conclusion Solution: The function is defined tack together wisely, and each piece is continuous. We unaccompanied have to analyze the point x ? 3 . We use side-limits: x ?3? x ?3? lim f ( x) ? lim ? x 2 ? 1? ? 10 ? x ?3 lim f...If you want to get a full essay, order it on our website: BestEssayCheap.com

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