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Define
\begin{aligned} &f(x)=x^{3}-\sin ^{2} x \tan x \\ &g(x)=2 x^{2}-\sin ^{2} x-x \tan x \end{aligned}
Find out, for each of these two functions, whether it is positive or negative for all $x \in(0, \pi / 2)$, or whether it changes sign. Prove your answer.
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Both functions tend to $-\infty$ as $x \rightarrow \frac{\pi}{2}$. Hence the only question is whether they ever become positive. We note that the derivative of the first function is $3 x^{2}-\sin ^{2} x-\tan ^{2} x$. By writing $\sin ^{2} x$ as $\frac{1}{2}-\frac{1}{2} \cos 2 x$ and making repeated use of the formula $\frac{d}{d x} \tan ^{k} x=k \tan ^{k-1} x+k \tan ^{k+1} x$, we find that the first six derivatives of this function vanish at 0 , and that the sixth derivative is
$-32 \sin 2 x-272 \tan x-1232 \tan ^{3} x-1104 \tan ^{5} x-144 \tan ^{7} x$
which is negative on $\left(0, \frac{\pi}{2}\right)$. Hence all of the first six derivatives are negative on this interval, and therefore the function itself is negative.

The same technique applies to the second function. All of its first five derivatives vanish at $x=0$ and the fifth is
\begin{aligned} -[16 \sin 2 x+16 x+& 80 \tan x+136 x \tan ^{2} x+\\ &\left.+200 \tan ^{3} x+240 x \tan ^{4} x+120 \tan ^{5} x+120 x \tan ^{6} x\right] \end{aligned}
which is negative on $\left(0, \frac{\pi}{2}\right)$. Hence this function is always negative on that interval.
by Diamond (66,887 points)

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