# Finding the inflection point for a function

While learning about limits and derivatives, I came accross the following problem on one of Stewart's exercises book. I've been trying to wrap my head around it but I haven't got anywhere useful:

1. $\forall x f''(x)$ exists
2. $\exists c \in \mathbb{R}$ / $\forall x \neq c, f'(x) > 0 \wedge f'(c) = 0$
3. Then, $(c, f(c))$ is an inflection point.

From this I can gather that:

• $f$ is continuous
• $f$ is increasing $\forall x \neq c$.

My intuition tells me that $(3)$ is false since I might be able to come up with a function defined by parts that contradicts the statement, but I haven't found a way to prove this. Any pointers would be greatly appreciated.

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In your "2.", should it be $\exists c \in \mathbb{R}$ (c, not x)? – Isaac Sep 16 '10 at 21:02
@Isaac: I fixed it as such. – Arturo Magidin Sep 16 '10 at 21:43

Because the second derivative exists everywhere, you also know that $f'$ is continuous everywhere. Now think about the function $f'(x)$; it is positive to the left and to the right of $c$, and is $0$ at $c$, and it is continuous. Does that tell you something about $f''$, and hence about concavity?
If the signs of $f''$ to the left and to the right were different (or equal) I could use the test of the second derivative to assert something about the concavity, but as far as I understand this is not the case (I can't imply anything about the signs of $f''$ from $f'$). What am I missing here? – Federico Builes Sep 16 '10 at 23:11
@Federico: Notice that Arturo's answer doesn't really refer to $f$; it is entirely about $g = f'$ and $g' = f''$. If you re-read it in terms of $g$ (and also reread your conditions (1) and (2) in terms of $g$ and translate the definition of inflection point for $f$ in terms of a property of $g$), you are likely to find the assertion is familiar. – whuber Sep 16 '10 at 23:32
@Federico: what you are missing is that $f''$ is just the derivative of $f'$; can you conclude something about the sign of $g'$ if you know stuff about $g$? That's what you have here. Or better: instead of thinking about concavity in terms of the second derivative, remember that there is a way to describe the concavity of $f$ in terms of $f'$; the fact that you can then explain it in terms of $f''$ is a consequence of the description in terms of $f'$. – Arturo Magidin Sep 17 '10 at 3:47