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I have a function such that $f'''$ is continuous and let's suppose there's some point $a \in \mathbb{R}$ such that $f'(a) = f''(a) = 0$. Does $f'''(a)>0$ tell me anything about whether $a$ is a relative maximum/minimum or neither? I've had a look at higher-order derivative tests, which tells me that $a$ should be a strictly increasing point of inflection, but I would like to formally prove this.

Edit: As Manzano has pointed out, the general version follows from the Taylor expansion at point $a$. I'm trying to show the specific version for $f'''$ without resorting to Taylor.

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Use the Taylor expansion of $f$ at $a$. –  Manzano Oct 30 '12 at 12:55

1 Answer 1

Well, let me provide you with two theorems that might help you with this:

Theorem 1. Let's assume that $f:(a,b)\to\mathbb{R}$ has derivatives of $(2k-1)$-th order, and there exists $f^{(2k)}(x_{0})$. If

$$f^{(j)}(x_{0})=0\quad\mathrm{for}\,j=1,...,2k-1\quad\mathrm{and}\,f^{(2k)}(x_{0})\ne{0}$$

then $f$ has a local maximum ($f^{(2k)}(x_{0})<0$) or minimum ($f^{(2k)}(x_{0})>0$) in $x_{0}$.

Theorem 2. Let's assume that $f:(a,b)\to\mathbb{R}$ has derivatives of $(2k)$-th order, and there exists $f^{(2k+1)}(x_{0})$. If

$$f^{(j)}(x_{0})=0\quad\mathrm{for}\,j=1,...,2k\quad\mathrm{but}\,f^{(2k+1)}(x_{0})\ne{0}$$

then $f$ does not have neither a minimum nor a maximum in $x_{0}$.

Do you see now which one is applicable?

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The second one. But how do I prove it for the special case $k=1$ without using the Taylor expansion? –  somebody Oct 30 '12 at 17:10

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