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Let $f(x) = 30 - 2x - x^3$, then find the number of positive integral values of 'x' which satisfies $f(f(f(x))) > f(f(-x))$.

The first thing that I saw in the above question was that the function f(x) is decreasing for all values of x, and it's derivative is $-2 -3x^2$, which is less than 0 for all real values of x.

Now, in my opinion, this implies, that $f(f(f(x)))$ is a decreasing function, while, $f(f(-x))$ is an increasing function. Is that correct?

Also, how do I proceed from here on?

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If $f$ is a decreasing function, then $x < y$ if and only if $f(x) > f(y)$. So your inequality can be simplified as $$f(f(f(x))) > f(f(-x)) \iff$$ $$f(f(x)) < f(-x) \iff$$ $$30 - 2x - x^3 = f(x) > -x \iff$$ $$x^3 + x - 30 < 0 \iff$$ $$(x-3)(x^2+3x+10)<0$$ The second factor is always positive, and so the inequality holds only when $x-3 < 0$ i.e. when $x < 3$.

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  • $\begingroup$ Thanks a lot! I don't know what I missed that. :) $\endgroup$ – Brahmnoor Singh Jul 22 '16 at 10:35

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