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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)

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  1. freckles
    • one year ago
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    So first step is to differentiate to find critical numbers.

  2. anonymous
    • one year ago
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    wait, whats the derivative of cot?

  3. freckles
    • one year ago
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    the derivative of cot(x) w.r.t. x is -csc^2(x).

  4. anonymous
    • one year ago
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    oh i know

  5. freckles
    • one year ago
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    you have to use chain rule there since you have x/2 inside though

  6. freckles
    • one year ago
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    \[\frac{d}{dx}\cot(u(x))=u'(x) \cdot [-\csc^2(u(x))] \\ \frac{d}{dx}\cot(u(x))=-u'(x) \cdot \csc^2(u(x))\]

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

  8. freckles
    • one year ago
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    yeah and we set that =0 to find critical numbers

  9. freckles
    • one year ago
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    \[1-\csc^2(\frac{x}{2}) \cdot \frac{1}{2} =0 \\ 2 -\csc^2(\frac{x}{2})=0 \\ \csc^2(\frac{x}{2})=2 \\ \sin^2(\frac{x}{2})=\frac{1}{2}\] that should make it easier for you to solve that last step was just taking reciprocal of both sides

  10. anonymous
    • one year ago
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    wait...how did you get it to become sin^2 (x/2) = 1/2 ?

  11. freckles
    • one year ago
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    well first step I multiplied 2 on both sides

  12. freckles
    • one year ago
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    second step I added csc^2(x/2) on both sides

  13. anonymous
    • one year ago
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    I got everything until the last part

  14. freckles
    • one year ago
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    last step I took the reciprocal of both sides

  15. anonymous
    • one year ago
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    so you did: (1/2)*(2) = 1/csc*csc^2(x/2)

  16. freckles
    • one year ago
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    I don't think that is what I did.

  17. freckles
    • one year ago
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    I'm not sure what that is.

  18. anonymous
    • one year ago
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    lol could you show me the work for how you got what you got in the last part

  19. freckles
    • one year ago
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    \[\text{ if } \frac{1}{a}=\frac{1}{b} \text{ then } a=b\]

  20. anonymous
    • one year ago
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    yea thats true

  21. freckles
    • one year ago
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    \[\csc^2(\frac{x}{2})=2 \\ \frac{1}{\sin^2(\frac{x}{2})}=2 \\ \frac{1}{\sin^2(\frac{x}{2})}= \frac{1}{\frac{1}{2}} \\ \text{ which means } \sin^2(\frac{x}{2})=\frac{1}{2}\]

  22. anonymous
    • one year ago
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    oh you just switch out csc to 1/sin

  23. anonymous
    • one year ago
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    k i get it

  24. freckles
    • one year ago
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    yes those are reciprocal functions

  25. freckles
    • one year ago
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    csc(u) is the reciprocal sin(u) and vice versa

  26. freckles
    • one year ago
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    which means csc(u)=1/sin(u)

  27. anonymous
    • one year ago
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    so now we find the critical points.. if its sin(x/2)^2 = 1/2

  28. anonymous
    • one year ago
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    then would it be pi/6?

  29. anonymous
    • one year ago
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    Oh wait, that doesnt fit the interval

  30. freckles
    • one year ago
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    \[\sin^2(\frac{x}{2})=\frac{1}{2} \implies \sin(\frac{x}{2})=\frac{\sqrt{2}}{2} \text{ or } \sin(\frac{x}{2})=-\frac{\sqrt{2}}{2}\]

  31. anonymous
    • one year ago
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    how did you get a sqrt of 2 in the second part, wasnt there a 1 there before?

  32. freckles
    • one year ago
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    \[\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}} =\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2} \\ \]

  33. anonymous
    • one year ago
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    oh okay

  34. freckles
    • one year ago
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    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
    • one year ago
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    Yea thats a good thing to keep a note of

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

  38. anonymous
    • one year ago
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    wow, didnt think of splitting the equation into 2 separate parts. But yell ill solve for u on the interval

  39. anonymous
    • one year ago
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    I dont know how to find where pi/8 and the other value is onthe unit circle

  40. freckles
    • one year ago
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    \[\text{ we are solving } \sin(u)=\frac{\sqrt{2}}{2} \text{ or } \sin(u)=\frac{-\sqrt{2}}{2} \text{ on } u \in [\frac{\pi}{8},\frac{7\pi}{8}] \\ u=\frac{\pi}{4}, \frac{3\pi}{4}\]

  41. freckles
    • one year ago
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    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
    • one year ago
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    so x= 2pi/4 and 6pi/4

  43. freckles
    • one year ago
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    reducing x=pi/2 or x=3pi/2

  44. freckles
    • one year ago
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    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
    • one year ago
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    do i plug the x values in the derivative or the original function?

  46. freckles
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    easier in general not just you lol

  49. freckles
    • one year ago
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    \[g(x)=x+\cot(\frac{x}{2}) \\ g(\frac{\pi}{4})=\frac{\pi}{4}+\cot(\frac{\pi}{8}) \\ g(\frac{\pi}{2})=\frac{\pi}{2}+\cot(\frac{\pi}{4}) \\ g(\frac{3\pi}{2})=\frac{3\pi}{2}+\cot(\frac{3\pi}{4}) \\ g(\frac{7\pi}{4})=\frac{7\pi}{4}+\cot(\frac{7 \pi}{8})\]

  50. freckles
    • one year ago
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    so you just might want to whip out your calculator and plug those into see which gives the highest and lowest output

  51. anonymous
    • one year ago
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    okay, I see

  52. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    half angle-identity *

  55. freckles
    • one year ago
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    but either way will suffice

  56. anonymous
    • one year ago
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    got it, thanks!

  57. freckles
    • one year ago
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    cool stuff

  58. freckles
    • one year ago
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    peace

  59. anonymous
    • one year ago
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    Step1- find critical values and replace them in original function Step2- replace endpoints as well Step3- check your highest and lowest values Done :)

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