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Is the statement true that f(x)< x- (x^3/3), where f(x)=arctg x?  

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flowerelia | Student, Grade 10 | (Level 2) eNoter

Posted June 8, 2009 at 4:53 PM via web

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Is the statement true that f(x)< x- (x^3/3), where

f(x)=arctg x?

 

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kjcdb8er | Teacher | (Level 1) Associate Educator

Posted June 8, 2009 at 5:19 PM (Answer #1)

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There are a couple of ways to evaluate this inequality. One is to graph arctg (x) and x-x^3/3 and note that the two graphs intersect. Thus it is not always true that f(x) < x - x^3/3

A quick way to graph the two functions is to note what arctg (x) and x-x^3/3 do when x = 0 and when x grows very large. arctg (0) = 0, and 0-0^3/3= 0; so the functions intersect at zero. When x gets very large, arctg (x) = 1 (or -1 for -x). On the other hand for large x, -x^3 dominates, and the function approaches infinity. So for x > 0, f(x) > x - x^3/3.

A second quick approach is to note that both sides of the equation have odd functions. But, thanks to the minus sign infront of the -x^3/4, when they intersect the inequality sign must flip. Thus it can't be true that one side is always less than the other side.

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