## anonymous one year ago NEED HELP! In finding the absolute extrema of the function on the given interval of [pi/4, 7pi/4] when g(x) = x + cot(x/2)

1. freckles

So first step is to differentiate to find critical numbers.

2. anonymous

wait, whats the derivative of cot?

3. freckles

the derivative of cot(x) w.r.t. x is -csc^2(x).

4. anonymous

oh i know

5. freckles

you have to use chain rule there since you have x/2 inside though

6. freckles

$\frac{d}{dx}\cot(u(x))=u'(x) \cdot [-\csc^2(u(x))] \\ \frac{d}{dx}\cot(u(x))=-u'(x) \cdot \csc^2(u(x))$

7. anonymous

yea I was about to type that... so basically the derivative is g'(x) = 1+ -csc^2(x/2) * 1/2

8. freckles

yeah and we set that =0 to find critical numbers

9. freckles

$1-\csc^2(\frac{x}{2}) \cdot \frac{1}{2} =0 \\ 2 -\csc^2(\frac{x}{2})=0 \\ \csc^2(\frac{x}{2})=2 \\ \sin^2(\frac{x}{2})=\frac{1}{2}$ that should make it easier for you to solve that last step was just taking reciprocal of both sides

10. anonymous

wait...how did you get it to become sin^2 (x/2) = 1/2 ?

11. freckles

well first step I multiplied 2 on both sides

12. freckles

second step I added csc^2(x/2) on both sides

13. anonymous

I got everything until the last part

14. freckles

last step I took the reciprocal of both sides

15. anonymous

so you did: (1/2)*(2) = 1/csc*csc^2(x/2)

16. freckles

I don't think that is what I did.

17. freckles

I'm not sure what that is.

18. anonymous

lol could you show me the work for how you got what you got in the last part

19. freckles

$\text{ if } \frac{1}{a}=\frac{1}{b} \text{ then } a=b$

20. anonymous

yea thats true

21. freckles

$\csc^2(\frac{x}{2})=2 \\ \frac{1}{\sin^2(\frac{x}{2})}=2 \\ \frac{1}{\sin^2(\frac{x}{2})}= \frac{1}{\frac{1}{2}} \\ \text{ which means } \sin^2(\frac{x}{2})=\frac{1}{2}$

22. anonymous

oh you just switch out csc to 1/sin

23. anonymous

k i get it

24. freckles

yes those are reciprocal functions

25. freckles

csc(u) is the reciprocal sin(u) and vice versa

26. freckles

which means csc(u)=1/sin(u)

27. anonymous

so now we find the critical points.. if its sin(x/2)^2 = 1/2

28. anonymous

then would it be pi/6?

29. anonymous

Oh wait, that doesnt fit the interval

30. freckles

$\sin^2(\frac{x}{2})=\frac{1}{2} \implies \sin(\frac{x}{2})=\frac{\sqrt{2}}{2} \text{ or } \sin(\frac{x}{2})=-\frac{\sqrt{2}}{2}$

31. anonymous

how did you get a sqrt of 2 in the second part, wasnt there a 1 there before?

32. freckles

$\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}} =\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2} \\$

33. anonymous

oh okay

34. freckles

i didn't do all those steps I just remember that sqrt(1/2) =1/sqrt(2) or I also know I can write sqrt(2)/2

35. anonymous

Yea thats a good thing to keep a note of

36. anonymous

wait... how do you solve for that x...? I was assuming it could be pi/4, 3pi/4...but it just doesnt look right with how the x is placed in the equation

37. freckles

$\text{ we have two equations \to solve } \text{ Let } u=\frac{x}{2} \\ \sin(u)=\frac{\sqrt{2}}{2} \text{ or } \sin(u)=\frac{-\sqrt{2}}{2} \\ u=\frac{\pi}{4}+2n \pi , \frac{-\pi}{4}+ 2n \pi \text{ or } u=\frac{2 \pi}{3}+2 n \pi , \frac{5\pi}{4}+2 \pi n \\ \text{ or condensing this a little } \\ u=\frac{\pi}{4}+\pi n \text{ or } \frac{-\pi}{4} + n \pi \\ \text{ now remember we wanted } \frac{\pi}{4} \le x \le \frac{7 \pi}{4} \\ \text{ but we solved for } u \\ \\ \text{ recall } u=\frac{x}{2} \\ \text{ so } x=2u \\ \text{ so we need to find } u \text{ above such that } \\ \frac{\pi}{4} \le 2u \le \frac{ 7\pi}{4} \\ \text{ dividing 2 on all sides gives } \\ \frac{\pi}{8} \le u \le \frac{ 7\pi}{8}$ so before we go further and solve for x can you solve for u on the interval [pi/8,7pi/8]

38. anonymous

wow, didnt think of splitting the equation into 2 separate parts. But yell ill solve for u on the interval

39. anonymous

I dont know how to find where pi/8 and the other value is onthe unit circle

40. freckles

$\text{ we are solving } \sin(u)=\frac{\sqrt{2}}{2} \text{ or } \sin(u)=\frac{-\sqrt{2}}{2} \text{ on } u \in [\frac{\pi}{8},\frac{7\pi}{8}] \\ u=\frac{\pi}{4}, \frac{3\pi}{4}$

41. freckles

now since u=x/2 we need to solve: $\frac{x}{2}=\frac{\pi}{4} \text{ and also solve } \frac{x}{2}=\frac{3\pi}{4}$

42. anonymous

so x= 2pi/4 and 6pi/4

43. freckles

reducing x=pi/2 or x=3pi/2

44. freckles

now you have critical points on the given interval so plug in the critical numbers and the endpoints into f to see which gives you highest output and which gives you lowest output

45. anonymous

do i plug the x values in the derivative or the original function?

46. freckles

also for the above equation if it would have made things easy we could have also used the half angle identity for sin $\sin^2(\frac{x}{2})=\frac{1}{2} \\ \frac{1}{2}(1-\cos(x))=\frac{1}{2} \\ \text{ multiply both sides by } 2 \\ 1-\cos(x)=1 \\ \text{ subtract 1 on both sides } -\cos(x)=0 \\ \cos(x)=0 \\ \text{ this probably would have been easier for you \to solve }$

47. freckles

no you plug into the original function the derivative will only tell you if the slope is positive,0, negative ,or non-existing

48. freckles

easier in general not just you lol

49. freckles

$g(x)=x+\cot(\frac{x}{2}) \\ g(\frac{\pi}{4})=\frac{\pi}{4}+\cot(\frac{\pi}{8}) \\ g(\frac{\pi}{2})=\frac{\pi}{2}+\cot(\frac{\pi}{4}) \\ g(\frac{3\pi}{2})=\frac{3\pi}{2}+\cot(\frac{3\pi}{4}) \\ g(\frac{7\pi}{4})=\frac{7\pi}{4}+\cot(\frac{7 \pi}{8})$

50. freckles

so you just might want to whip out your calculator and plug those into see which gives the highest and lowest output

51. anonymous

okay, I see

52. anonymous

This problem required a lot of work...I thought it wouldve been easy, I guess I have to review the concept behind this problem

53. freckles

well I think the only hard part was the way we solved for the critical numbers in the beginning the half-identity made the equation faster to solve

54. freckles

half angle-identity *

55. freckles

but either way will suffice

56. anonymous

got it, thanks!

57. freckles

cool stuff

58. freckles

peace

59. anonymous

Step1- find critical values and replace them in original function Step2- replace endpoints as well Step3- check your highest and lowest values Done :)