If $f$ is a non-piecewise function defined continuous on an interval $I$, and within that interval $I$, there exists a value $x$, such that $f`(x)$ (derivative of $f$) does not exist , then at that value $x$, is a local $\max/\min$ value.
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1False. See cantor step function. – 5xum Apr 08 '14 at 11:42
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Or the examples in this question. – Git Gud Apr 08 '14 at 11:46
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so if i changed the original question to non-piecewise functions – Greg Dillon Apr 08 '14 at 11:51
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If $|x|$ is allowed, then all of these piecewise functions can be constructed from that. – Dustan Levenstein Apr 08 '14 at 11:57
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thank you all very much for your insights i really appreciate it – Greg Dillon Apr 08 '14 at 12:00
4 Answers
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This is not true.
Consider $$\begin{cases} f(x) = 1-x & \text{if } 0\le x< 1\\ f(1) = 0& \\ f(x) = \frac 12(1-x) & \text{if } 1< x\le 2\end{cases} $$
mookid
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False. Consider, for example,
$$f(x) = \begin{cases} x^3 & \text{ if } x<0 \\ x & \text{ if } x \ge 0\end{cases}.$$
Then $f$ is not differentiable at $0$, but $f$ is strictly increasing around $0$.
Dustan Levenstein
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Very not true. For example:
$$f(x) = \begin{cases}x&\text{ if } x\leq 0\\ 2x&\text{ if } x\geq0\end{cases}$$ on $I=[-1,1]$.
5xum
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I'm surprised no one has said $$f(x)=\sqrt[3] x$$ The derivative at $x=0$ is undefined (or $\infty$ if you like) but that is not an extremum of the function.