Can relative maximum and minimum exist when the derivative....

[MidgetDwarf]

Can relative maximum and minimum points exist when a function is defined at say x=c, however the derivative does not exist or tends to infinity? Ie the graph of. F (x)= |x|, for x=c=o. If I am correct the relative minimum is at o, can it also be the abs minimum?

I recalled the theorem by memory.

Let the function f be defined on the closed interval from a to b and have a relative maximum or minimum at x=c, where c is in the open interval from a to b.

If the derivative exist as a finite number at x=c, then the derivative is o at that point.

I went through the derivation of the proof, and the theorem (by analyzing my derivation), does not say what happens when.

1) when the derivative fails to exist at a point c.
2) at the endpoints, or that there is always a max or min when the derivative equals 0.

I sometimes tutor my little neighbor next door for free, because it helps me practice things I havnt used in a while. Today I told him, that a relative max or min can occur if the function is defined at that point, does not have to be necessary differential at that point. I use y=|x| as an example.

I felt bad, because I believed I gave him the wrong answer. Sorry for the long post.

 

[pasmith]

The definitions of absolute and relative maxima and minima have nothing to do with derivatives:

$f: [a,b] \to \mathbb{R}$ has an absolute maximum (minimum) at $c \in [a,b]$ if and only if $f(c) \geq f(x)$ ($f(c) \leq f(x)$) for all $x \in [a,b]$.

$f: [a,b] \to \mathbb{R}$ has a relative maximum (minimum) at $c \in [a,b]$ if and only if there exists a $\delta > 0$ such that for all $x \in [a,b]$, if $|x - c| < \delta$ then $f(c) \geq f(x)$ ($f(c) \leq f(x)$).

Thus $f(x) = |x|$ has a absolute minimum at x = 0; this is obvious from the fact that $|x| \geq 0$ for all $x \in \mathbb{R}$.

There is, of course, a theorem which says that if a function is differentiable at a relative extremum then its derivative must vanish there. But the converse does not hold: $f: [-1,1] \to \mathbb{R} : x \mapsto x^3$ has $f'(0) = 0$ but zero is neither a maximum nor a minimum of any type.

This theorem can help you find extrema, but you must always check that a point where the derivative vanishes really is a relative extremum, and you must also check whether the end points and any points where the derivative does not exist are extrema.

 

[MidgetDwarf]

The definitions of absolute and relative maxima and minima have nothing to do with derivatives:

$f: [a,b] \to \mathbb{R}$ has an absolute maximum (minimum) at $c \in [a,b]$ if and only if $f(c) \geq f(x)$ ($f(c) \leq f(x)$) for all $x \in [a,b]$.

$f: [a,b] \to \mathbb{R}$ has a relative maximum (minimum) at $c \in [a,b]$ if and only if there exists a $\delta > 0$ such that for all $x \in [a,b]$, if $|x - c| < \delta$ then $f(c) \geq f(x)$ ($f(c) \leq f(x)$).

Thus $f(x) = |x|$ has a absolute minimum at x = 0; this is obvious from the fact that $|x| \geq 0$ for all $x \in \mathbb{R}$.

There is, of course, a theorem which says that if a function is differentiable at a relative extremum then its derivative must vanish there. But the converse does not hold: $f: [-1,1] \to \mathbb{R} : x \mapsto x^3$ has $f'(0) = 0$ but zero is neither a maximum nor a minimum of any type.

This theorem can help you find extrema, but you must always check that a point where the derivative vanishes really is a relative extremum, and you must also check whether the end points and any points where the derivative does not exist are extrema.

Thanks. This is exactly what I was thinking. The requirement is that the function needs to exist at that point? That is why I used y=|x|, thanks for the insight and fast response, I now feel better I did not give him miss information.

 

[mfb]

The function has to exist, sure.
There are useful statements about the maxima and minima of functions that are differentiable at one point or in some interval, but the concept of maxima and minima is much more general and can be used for non-differentiable functions as well.

 

[WWGD]

For a contrived example of MFB's post, take $f=\chi \mathbb Q$ (Take it, please!) nowhere-differentiable (since, e.g., nowhere continuous) , with Max reached at $\mathbb Q$ and min reached at $\mathbb R-\mathbb Q$.