I don't understand how to find the absolute extrema on an open interval.

- anonymous

I don't understand how to find the absolute extrema on an open interval.

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- anonymous

on an open interval?

- anonymous

For example, if the domain is on the interval negative infinity to positive infinity. Not a closed interval.

- anonymous

well, that interval is not closed...

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## More answers

- anonymous

i understand that intuition may tell you that it is

- anonymous

f(x) = 4x^3- 3x^4 on the interval negative infintiy to positive infinity

- anonymous

okay

- anonymous

that is easier to approach

- anonymous

first note that \[\lim_{x\rightarrow \infty}f(x)=\lim_{x\rightarrow- \infty}f(x)=-\infty\]

- anonymous

do you see that? or should we discuss it?

- anonymous

My prof. taught us to find the derivative and at the critical points, weshould look how thederivative behaves, but that doesn't always work..

- anonymous

yes, but we are doing some preliminary investigation about the behavior of this function

- anonymous

my point is, this function will have an absolute max because of the result of those limits

- anonymous

oh, can you please explain that

- anonymous

the give f(x) is a polynomial, yes?

- anonymous

correct

- anonymous

of degree 4

- anonymous

so its end behavior will be the same at \[\pm\infty\]

- anonymous

yes, I understand that

- anonymous

in this case since the leading coefficient is (-1), both ends go to \[-\infty\]

- anonymous

okay
so whatever this function does, it can not have an "absolute" minimum right?

- anonymous

no, it's suppose to only have an absolute max

- anonymous

right

- anonymous

so, find the critical points

- anonymous

i.e. where \[\frac{df}{dx}=0\]

- anonymous

its at x = 1 max = 1

- anonymous

So, I understood we always look at the leading coefficient of the polynomial?

- anonymous

yes, it as well as the degree will tell you a lot

- anonymous

yes I mean that too

- anonymous

Thank you so much! Just one last question..

- anonymous

what level of calculus are you studying and what are your future plans in math?

- anonymous

this is calc 1 and the onlysemester of calc i have to take.. thankfully! I'm a pharmacy major..

- anonymous

oh ok

- anonymous

good luck then, i have had many students that were pre-pharm

- anonymous

Thank you. One last question, if the equation was f(x) = 2x^3 - 6x +2.. the polynomial is of degree 3 and leading coef is positive

- anonymous

so the behavior of function: from negative infinty to positive infinity..

- anonymous

and the critical points are -1 and 1

- anonymous

but the solution says there is no extrema and I don'tunderstand why

- anonymous

right, so it will have no "absolute" extrema

- anonymous

do you see why?

- anonymous

No I can't seem tosee it

- anonymous

its all in that one word "absolute",
absolute extreme , implies that the function achieves no values greater/less than, respectively, that value

- anonymous

if your function runs off to \[+\infty\] one way and \[-\infty\] the other, then there is no absolute max/min

- anonymous

There are however possible "local" max/min which is where many students get confused

- anonymous

Oh okay, I will look into it more... Thank you so much foryour help! I really appreciate it!

- anonymous

awesome, good luck!

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