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Is the statement true that f(x)< x- (x^3/3), where f(x)=arctg x?
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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.
Posted by kjcdb8er on June 8, 2009 at 5:19 PM (Answer #1)
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