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I know that if $f(x)$ is a cusp or a point of inflection, then it's not a local max/min when $f'(x)=0$.

But what about in a piece wise function?

Given $f(x) = 0$ when $x = 0$, and $f(x) = \sin(1/x)$ for all other values of $x$, it's clear that $f(0)$ is not a local max or min. However, $f'(0) = 0$. Is it because $f''(0)$ also equals $0$ making $x=0$ a point of inflection?

Edit: Sorry maybe I should have asked instead: is $f'(x)$ differentiable at $0$?

BlackAdder
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ben
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  • How are you proving that $f'(0)=0$? – Sigur Nov 15 '13 at 01:04
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    Do you mean $f(x) = x^2 \sin (\frac{1}{x})$ for $x \neq 0$? Otherwise, you function won't even be differentiable at $x = 0$ to begin with. – user43378 Nov 15 '13 at 01:11
  • While we're on this point... Let $f(0) = 0$. Otherwise, if $f(x) = sin(\frac{1}{x})$ is not even continuous. $f(x) = x\cdot sin(\frac{1}{x})$ is continuous but not differentiable and $f(x) = x^{2}\cdot sin(\frac{1}{x})$ is differentiable. As @user43378 said, this is probably the function you meant, since $lim_{x\rightarrow 0} |\frac{f(x)-f(0)}{x}| \leq lim_{x\rightarrow 0} |x| \cdot |sin(\frac{1}{x})| \leq lim_{x\rightarrow 0} |x| = 0$. So this works... – Christopher K Nov 15 '13 at 01:22

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For the first point, if $f''(a) = 0$, then that is a necessary but not sufficient condition for $(a,f(a))$ to be a point of inflexion. However, the function you described is not differentiable at $0$. In fact, it is not even continuous there for I can always choose $N$ large enough such that $\forall n\geq N$, $\frac{2}{\pi\cdot n} < \varepsilon$. So for some $n>N$, $f(x)=1$ and for some $m>N$, $f(x) = -1$. So just let $\varepsilon \leq 1$ and we have a contradiction.

EDIT: I corrected an error in my earlier post regarding the conditions for a point of inflexion.

Christopher K
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    Sorry to nitpick, but I don't want the OP to be confused. A point of inflection is a point on the curve, not just an $x$-value, and more importantly, $f''$ must change sign at $a$. It is not enough to just have $f''(a)=0$, even if $f'(a)=0$ also (consider $f(x)=x^4$,$a=0$). – Stefan Smith Nov 15 '13 at 02:02
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    @StefanSmith, thank you for your due diligence. I still remember how when I took calculus, I got that definition mixed up. Apparently, it seems old habits die hard. – Christopher K Nov 15 '13 at 02:07