relative and absolute extrema

• Apr 27th 2006, 02:43 PM
lingyai
relative and absolute extrema
I'm learning calculus from a book called "Forgotten Calculus". I'm getting confused by something. The author states that a critical point can be, but need not be, both a relative extreme and an absolute extreme at the same time. Fair enough. But in the case of two simple exercises, I find that either she's not being optimally precise in her wording of the solution, or I'm missing something fundamental.

Consider these two functions:

a) y = f(x) = 1/4x^4 + x and
b) y= f(x) = x^10 + 2

I can find the critical points easily enough. For each there is only one. For a) it's (-1, -3/4) and for b) it's (0, 2).

She says that both are relative minima. Yes, but I think the more complete answer would be to call them absolute minima. My reasoning is that in each case, if the critical point is the ONLY critical point of the function (and having established, via the first derivative test, that each one is a minimum), then by definition each must be an ABSOLUTE minimum. As a check I graphed each function with a TI 83 with the window set quite tall and wide and as far as I can see, indeed, each is an absolute minimum.

Am I right? If not, why not?

I'd be grateful for any advice.
• Apr 27th 2006, 03:01 PM
Jameson
You need to look at the end behavior, so check the $\displaystyle \lim_{x\rightarrow\infty}f(x)$ and $\displaystyle \lim_{x\rightarrow-\infty}f(x)$. If your relative extrema still has a magnitude greater than these, than you can call it an absolute max/min. But if not, it is only a max/min relative to the points around it.
• Apr 27th 2006, 03:07 PM
ThePerfectHacker
Quote:

Originally Posted by lingyai
I'm learning calculus from a book called "Forgotten Calculus". I'm getting confused by something. The author states that a critical point can be, but need not be, both a relative extreme and an absolute extreme at the same time. Fair enough. But in the case of two simple exercises, I find that either she's not being optimally precise in her wording of the solution, or I'm missing something fundamental.

Consider these two functions:

a) y = f(x) = 1/4x^4 + x and
b) y= f(x) = x^10 + 2

I can find the critical points easily enough. For each there is only one. For a) it's (-1, -3/4) and for b) it's (0, 2).

She says that both are relative minima. Yes, but I think the more complete answer would be to call them absolute minima. My reasoning is that in each case, if the critical point is the ONLY critical point of the function (and having established, via the first derivative test, that each one is a minimum), then by definition each must be an ABSOLUTE minimum. As a check I graphed each function with a TI 83 with the window set quite tall and wide and as far as I can see, indeed, each is an absolute minimum.

Am I right? If not, why not?

I'd be grateful for any advice.

The rigorous meaning of relative minima $\displaystyle c$ of a function is a point such as there exists an open interval $\displaystyle (a,b)$ such as, $\displaystyle f(x)\geq f(c)$ for all $\displaystyle a<x<b$. The rigorous meaning a an absolute minima $\displaystyle c$ of a function on an interval is when $\displaystyle f(c)$ is less than an other value on that interval. Thus, an absolute minima is a relative minima, but a relative minima is not necessarily an absolute minima.

The same for absolute/relative maxima. You are right about saying it is an absolute minima, by the author is also right it being a relative minima.
• Apr 27th 2006, 03:10 PM
ThePerfectHacker
Quote:

Originally Posted by Jameson
You need to look at the end behavior, so check the $\displaystyle \lim_{x\rightarrow\infty}f(x)$ and $\displaystyle \lim_{x\rightarrow-\infty}f(x)$. If your relative extrema still has a magnitude greater than these, than you can call it an absolute max/min. But if not, it is only a max/min relative to the points around it.

That is not necessarily true. You can have an oscilating function perhaps? Not having a limit but exceeding the relative maxima.

You can perhaps use this for "well-behaved" functions, but if you function behaves like the House on C-SPAN then you need to reconsider.
• Apr 27th 2006, 03:21 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
You are right about saying it is an absolute minima, by the author is also right it being a relative minima.

Warning: I'm going to be a stuffy prig again! :rolleyes:

The word "minima" refers to the plural. What you meant to say was "minimum" because you were referring to a single point.

Okay, pet peeve fit over! :D

-Dan

(I know, I know! Who really cares!)
• Apr 27th 2006, 03:22 PM
lingyai
Follow-up question
First, thank you both. As for your reply, PerfectHacker: it's understood. But what do you think of my reasoning that if we know there is only 1 critical point, and that it is, say, a minimum, it follows that that it must be an absolute minimum?
• Apr 27th 2006, 03:34 PM
ThePerfectHacker
Quote:

Originally Posted by lingyai
First, thank you both. As for your reply, PerfectHacker: it's understood. But what do you think of my reasoning that if we know there is only 1 critical point, and that it is, say, a minimum, it follows that that it must be an absolute minimum?

Absosultey positively not, consider,
Code:

.              .  .            .   .  ..    .   . .  . .
This is supposed to look like a graph, to lazy to use my program :). Look only one relative maximum (thank you topsquark) in the middle but not a absolute.