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Say we are given the derivative of a function say, $$f'(x)=\begin{cases} 5 & x<3 \\ -5 & x>3 \end{cases}$$ Notice that the derivative has opposite signs on either side of $x=3$, so you would expect an extrema to occur in $f$ at $x=3$ (specifically a maximum in this case), however the derivative is undefined at $x=3$, so is there still an extrema?

This is just an example of the general case: if the derivative of a function is opposite signs on either side of $x=\rho$, but the derivative is undefined at $x=\rho$, does the function still have an extrema?

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  • $\begingroup$ My guess would be yes because the function $f=|x|$ experiences this, but I would like conformation. $\endgroup$ – wfish454 Tutorials May 22 '16 at 1:43
  • $\begingroup$ When you said "extrema", do you mean local extrema or global extrema? $\endgroup$ – BigbearZzz May 22 '16 at 2:01
  • $\begingroup$ Local extrema, my bad $\endgroup$ – wfish454 Tutorials May 22 '16 at 2:23
  • $\begingroup$ It's possible that $\lim_{x \nearrow 3} f(x) < f(3) < \lim_{x \searrow 3} f(x)$. In this case $f$ does not have a local extremum at $x = 3$. $\endgroup$ – littleO May 22 '16 at 2:49
  • $\begingroup$ We say that because $ \ f \ ' ( x ) \ $ is not defined at $ \ x \ = \ 3 \ $ , this value of $ \ x \ $ is a critical point for the function, which does not automatically imply that it is an extremum. The whole question of extrema has to be handled with some care in any event: we also know that a critical point with $ \ f \ ' (x) \ = \ 0 \ $ is not necessarily an extremum, as with $ \ x \ = \ 0 \ $ for $ \ f ( x ) \ = \ x^3 \ $ . The anti-derivative for your function might not be continuous (as Graham Kemp's reply discusses), so some caution in analysis is always advisable. $\endgroup$ – colormegone May 22 '16 at 3:55
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At any local maximum $x$, $\lim_{t \to 0^+} \frac{f(x+t)-f(x)}{t}\leq 0$ and $\lim_{t \to 0^-} \frac{f(x+t)-f(x)}{t}\geq 0$ (if these exist, you can further generalize this using the $\lim\sup$ and $\lim\inf$ in place of the respective limits), and the reverse holds for a minimum.

This is easy to verify, as we approach a maximum from the right $f(x+t)-f(x)\leq 0$, $t\geq 0$ so the inequality must hold. The other inequality holds by similar logic.

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  • $\begingroup$ This, while correct, doesn't seem to answer the question. The question is, I believe, "If the derivative of a function is opposite signs on either side of a point $x=\rho$ but the derivative is undefined there, does the function necessary have an extrema?". $\endgroup$ – BigbearZzz May 22 '16 at 1:59
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If $f'(x)= \begin{cases} -5 & : x< 3\\ +5&: x>3 \\ \textsf{undef} & : x=3\end{cases}$ , then all we know about $f$ is that: $$f(x)=5\lvert x-3\rvert +\begin{cases} c_1 & : x < 3 \\ c_2 & : x>3 \\ c_3 & : x=3\end{cases}$$

Where $c_1,c_2,c_3$ are arbitrary constants, but while $c_1,c_2$ will be definite (though not necessarily equal), $c_3$ need not be.

So there could be a local extrema at the point $x=3$, but there need not be one.

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