Real function/Extrema/Higher derivatives/Fact/Proof

Proof

Under the given conditions, the Taylor formula becomes

${}f(x)-f(a)={\frac {f^{(n+1)}(c)}{(n+1)!}}(x-a)^{n+1}\,,$

with some ${}c$ (depending on ${}x$) between ${}a$ and ${}x$. Depending on whether ${}f^{(n+1)}(a)>0$ or ${}f^{(n+1)}(a)<0$ holds, we have (due to the continuity of the ${}(n+1)$-th derivative) ${}f^{(n+1)}(x)>0$ or ${}f^{(n+1)}(x)<0$ for ${}x\in [a-\epsilon ,a+\epsilon ]$, for a suitable ${}\epsilon >0$. For these ${}x$, we have ${}c\in [a-\epsilon ,a+\epsilon ]$, so that the sign of ${}f^{(n+1)}(c)$ depends on the sign of ${}f^{(n+1)}(a)$.
For ${}n$ even, ${}n+1$ is odd and therefore the sign of ${}(x-a)^{n+1}$ changes at ${}x=a$ (for ${}x<a$, the sign is negative, and for ${}x>a$, the sign is positive). Since the sign of ${}f^{(n+1)}(c)$ does not change, the sign of ${}f(x)-f(a)$ is changing. This means that there can not be an extremum.
Suppose now that ${}n$ is odd. Then ${}n+1$ is even, hence ${}(x-a)^{n+1}>0$ for all ${}x\neq a$ in the neighborhood. This means, in the neighborhood, in case ${}f^{(n+1)}(a)>0$, that ${}f(x)>f(a)$ holds, and we have an isolated minimum in ${}a$. If ${}f^{(n+1)}(a)<0$, then ${}f(x)<f(a)$ holds, and we have an isolated maximum in ${}a$.

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