## 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)

1. anonymous

Extrema are known by a different name, minima/maxima. Do you know how to find the minimum/maximum of an equation?

2. anonymous

We take the derivative of the original equation and equate it to zero right?

3. anonymous

Correct.

4. anonymous

i got x=0, x=-tanx

5. anonymous

and then i don't know how to continue:(

6. anonymous

One second. x = -tan x doesn't look right to me, but I'm doing the derivative now.

7. anonymous

What was the derivative you arrived at?

8. anonymous

f'(x)= [-(xsinx)^-2] (sinx+xcosx)

9. 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?

10. anonymous

ummm, i don't know, i'm confused too.

11. 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).

12. anonymous

so does the function have no absoulte nor local extrema?

13. anonymous

None that are so easily found, at least. If this is for calc 1 level, I would answer that it does not.

14. anonymous

i was think about that too, thank you so much! :)

15. anonymous

thinking*

16. anonymous

17. 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?

18. anonymous

i equated the function to zero, and i got y(t)=0, y(t)=1000

19. anonymous

So your equation is $y'(t) = k y(t) [1 - \frac{y(t)}{1000}]$ correct?

20. anonymous

yes :)

21. 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.

22. anonymous

By following your instructions, i still got y(t)=0 and y(t)=1000. Is that correct?

23. 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?

24. anonymous

can i do y(t) [y(t) -1500] = -500000 ? or do i need to do the quadratic formula?

25. 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.

26. anonymous

[y(t) - 500][y(t) - 1000] = 0

27. anonymous

oh yeah i just got help by using the calculator:P Thanks for your help! your really a lifesaver!

28. 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.

29. 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!

30. 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 :)

31. anonymous

Anyway, good luck!

32. anonymous

Thank you so much! :D bye!

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