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anonymous
 5 years ago
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)
anonymous
 5 years ago
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)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Extrema are known by a different name, minima/maxima. Do you know how to find the minimum/maximum of an equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We take the derivative of the original equation and equate it to zero right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then i don't know how to continue:(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0One second. x = tan x doesn't look right to me, but I'm doing the derivative now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What was the derivative you arrived at?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f'(x)= [(xsinx)^2] (sinx+xcosx)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright, 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
 5 years ago
Best ResponseYou've already chosen the best response.0ummm, i don't know, i'm confused too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From 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
 5 years ago
Best ResponseYou've already chosen the best response.0so does the function have no absoulte nor local extrema?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0None that are so easily found, at least. If this is for calc 1 level, I would answer that it does not.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i was think about that too, thank you so much! :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0My pleasure. Anything else I can help you with today?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes 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
 5 years ago
Best ResponseYou've already chosen the best response.0i equated the function to zero, and i got y(t)=0, y(t)=1000

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So your equation is \[y'(t) = k y(t) [1  \frac{y(t)}{1000}]\] correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, 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
 5 years ago
Best ResponseYou've already chosen the best response.0By following your instructions, i still got y(t)=0 and y(t)=1000. Is that correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That 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
 5 years ago
Best ResponseYou've already chosen the best response.0can i do y(t) [y(t) 1500] = 500000 ? or do i need to do the quadratic formula?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0[y(t)  500][y(t)  1000] = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yeah i just got help by using the calculator:P Thanks for your help! your really a lifesaver!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0My 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
 5 years ago
Best ResponseYou've already chosen the best response.0our prof just simply don't teach anything and we need to read the textbook and do our assignment by ourselves. Thats really hard!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Indeed. 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
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you so much! :D bye!
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