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Let $f:I\rightarrow \Bbb{R}$ $2007$ times differentiable at $x_0 \in I$. Also: $f'(x_0) = f''(x_0) = ... = f^{(2006)} = 0$ but $f^{(2007)} > 0$. Prove there's $\delta> 0$ such that $f$ is strictly increasing at $(x_0-\delta, x_0 +\delta)$.

We've mentioned at class the Higher-order derivative test but I think I shouldn't use it directly.

How to prove it?

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Let $P$ be the Tailor polynomial of $f$ around $x_0$ of order $2007$. Then by assumption, $$P(x)=a(x-x_0)^{2007},$$ and so $$f(x)=a(x-x_0)^{2007}+r(x-x_0)$$where $a>0$ and$$\lim_{x\to x_0}\frac{r(x-x_0)}{(x-x_0)^{2007}}=0.$$ It follows that sufficiently close to $x_0$, $f(x)$ has the same sign as $(x-x_0)^{2007}$, and since $2007$ is odd, we are done.

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Directly from the definition of the derivative and the fact that $f^{(2006)}(x_0)=0$, there is a $\delta>0$ so that $f^{(2006)}$ is negative on $(-\delta,0)$ and positive on $(0,\delta)$.

Consider what that implies about monotonicity of $f^{(2005)}$, and conclude – using $f^{(2005)}(x_0)=0$ – that $f^{(2005)}$ is positive on $(-\delta,0)\cup(0,\delta)$.

Continue in the same fashion by induction, showing that $f^{(n)}$ is positive on $(-\delta,0)\cup(0,\delta)$ when $n$ is odd, while it is negative on $(-\delta,0)$ and positive on $(0,\delta)$ when $n$ is even.

Finally, note that $0$ is even.

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