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on an open interval?

well, that interval is not closed...

i understand that intuition may tell you that it is

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

okay

that is easier to approach

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

do you see that? or should we discuss it?

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

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

oh, can you please explain that

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

correct

of degree 4

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

yes, I understand that

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

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

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

right

so, find the critical points

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

its at x = 1 max = 1

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

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

yes I mean that too

Thank you so much! Just one last question..

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

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

oh ok

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

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

and the critical points are -1 and 1

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

right, so it will have no "absolute" extrema

do you see why?

No I can't seem tosee it

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

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

awesome, good luck!