anonymous
  • anonymous
Absolute and local extrema: Determine where the following functions have absolute and local extrema, if anywhere. f(x)=1/(xsinx), defined on the interval (o,pi)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Extrema are known by a different name, minima/maxima. Do you know how to find the minimum/maximum of an equation?
anonymous
  • anonymous
We take the derivative of the original equation and equate it to zero right?
anonymous
  • anonymous
Correct.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
i got x=0, x=-tanx
anonymous
  • anonymous
and then i don't know how to continue:(
anonymous
  • anonymous
One second. x = -tan x doesn't look right to me, but I'm doing the derivative now.
anonymous
  • anonymous
What was the derivative you arrived at?
anonymous
  • anonymous
f'(x)= [-(xsinx)^-2] (sinx+xcosx)
anonymous
  • anonymous
Alright, so to evaluate this poses an interesting problem. If x=0, both the numerator and the denominator go to 0. L'hopital's rule may help us with this issue. Alternatively, is there any way (other than x=0) to get xcosx - sinx to equal zero?
anonymous
  • anonymous
ummm, i don't know, i'm confused too.
anonymous
  • anonymous
From examination, this function does not appear to have any extrema. L'hopitals rule doesn't help eliminate the 0/0 problem, and xcosx - sinx doesnt equal 0 at any other point in (0, pi).
anonymous
  • anonymous
so does the function have no absoulte nor local extrema?
anonymous
  • anonymous
None that are so easily found, at least. If this is for calc 1 level, I would answer that it does not.
anonymous
  • anonymous
i was think about that too, thank you so much! :)
anonymous
  • anonymous
thinking*
anonymous
  • anonymous
My pleasure. Anything else I can help you with today?
anonymous
  • anonymous
yes please! if you are willing to help :D Absolute Maximum: The word "threat" was originally pronounced "threet". In linguistics, this shift in pronunciation is called a language change. It is theorized that any given language change spreads through a population according to a logistic differential equation such as the one described below. Consider a population of 1000. Let y(t) denote the number of people who have, as of time t, adopted the lanyage change. Then y'(t) = ky(t) [1- (y(t))/1000] where k is a positive constant. What is y(t) when y'(t) is maximal?
anonymous
  • anonymous
i equated the function to zero, and i got y(t)=0, y(t)=1000
anonymous
  • anonymous
So your equation is \[y'(t) = k y(t) [1 - \frac{y(t)}{1000}]\] correct?
anonymous
  • anonymous
yes :)
anonymous
  • anonymous
So, I'm solving it now, but what I'm doing is taking the derivative of both sides, and then subbing in the given value of y'(t) when you see it. Then I will set that equal to zero.
anonymous
  • anonymous
By following your instructions, i still got y(t)=0 and y(t)=1000. Is that correct?
anonymous
  • anonymous
That isn't what I'm getting. So, after taking the derivative and substituting in, I end up with: \[y''(t) = k ^{2}y(t) - \frac{k ^{2}[y(t)]^{2}}{1000} - \frac{2k ^{2}[y(t)]^{2} - \frac{2k ^{2}[y(t)]^{3}}{1000}}{1000} = 0\] Simplifying further, I end up at: \[[y(t)]^{2} - 1500 y(t) + 500000 = 0\] Could you solve from here?
anonymous
  • anonymous
can i do y(t) [y(t) -1500] = -500000 ? or do i need to do the quadratic formula?
anonymous
  • anonymous
You can factor it pretty easily, actually. And since we don't care what the expression for y(t) is, just what it equates to, you can treat it as a variable when doing so.
anonymous
  • anonymous
[y(t) - 500][y(t) - 1000] = 0
anonymous
  • anonymous
oh yeah i just got help by using the calculator:P Thanks for your help! your really a lifesaver!
anonymous
  • anonymous
My pleasure :) Just keep in mind that calculus is easy. Algebra is the "hard" part of math. Take your time doing the algebraic portions, and you'll be just fine.
anonymous
  • anonymous
our prof just simply don't teach anything and we need to read the textbook and do our assignment by ourselves. Thats really hard!
anonymous
  • anonymous
Indeed. As a graduate student, I feel your pain. You'll have a lot of bad professors, those that either don't care to teach, or just don't know how. Finding places like this can help. More importantly, check around for alternative texts (the university library almost always has a few for any subject) to help make things a bit more clear for you. Of course, for math, there is Paul's Online Math Notes :)
anonymous
  • anonymous
Anyway, good luck!
anonymous
  • anonymous
Thank you so much! :D bye!

Looking for something else?

Not the answer you are looking for? Search for more explanations.