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anonymous

  • one year ago

Critical numbers help? I will reward with a medal!

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  1. anonymous
    • one year ago
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    \[\sin^2x+cosx\] \[0<x<2\pi\]

  2. zepdrix
    • one year ago
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    Get that derivative? :U

  3. anonymous
    • one year ago
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    sinx (2cosx -1)

  4. zepdrix
    • one year ago
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    Mmm ok looks good! :)

  5. zepdrix
    • one year ago
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    \[\large\rm \sin x (2\cos x -1)=0\]Apply your Zero-Factor Property\[\large\rm \sin x=0\qquad\qquad\qquad 2\cos x-1=0\]

  6. zepdrix
    • one year ago
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    sine is zero at what angles?

  7. anonymous
    • one year ago
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    I'm not sure how to get that answer.

  8. zepdrix
    • one year ago
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    Don't remember your trig..? :(

  9. anonymous
    • one year ago
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    I was horrible at memorizing trig stuff. I enjoyed solving certain types of problems with them, but quite frankly I don't remember much of anything except for a few identities.

  10. anonymous
    • one year ago
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    Is it a particular formula, or is it just something you'd have to memorize?

  11. zepdrix
    • one year ago
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    Ugh OpenStudy freezing on me again.....

  12. anonymous
    • one year ago
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    It does that to me a lot.

  13. anonymous
    • one year ago
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    (this may seem like a stupid question, but which part is sine?

  14. zepdrix
    • one year ago
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    |dw:1444434263248:dw|

  15. zepdrix
    • one year ago
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    |dw:1444434276482:dw|sine is the y-coordinates

  16. zepdrix
    • one year ago
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    So we want to know which angles give us a y-coordinate of 0

  17. anonymous
    • one year ago
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    So the 180 angle

  18. zepdrix
    • one year ago
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    Hmm there's another one :d

  19. anonymous
    • one year ago
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    Oops! Didn't see that. The 0 angle

  20. zepdrix
    • one year ago
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    Oh wait... was your interval really given like this? \(\large\rm 0<x<2\pi\) with strict inequality on BOTH sides?

  21. anonymous
    • one year ago
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    That's what it looked like.

  22. zepdrix
    • one year ago
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    Oh ok :) Then I guess we can throw out the 0, since it's not in our interval. So then, \(\large\rm \sin x=0\) implies that our angle \(\large\rm x=\pi\). Ok great, we've found one critical point.

  23. zepdrix
    • one year ago
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    \[\large\rm 2\cos x-1=0\]Solving for cos x,\[\large\rm \cos x=\frac{1}{2}\]

  24. zepdrix
    • one year ago
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    |dw:1444434611166:dw|Cosine is your x-coordinates

  25. anonymous
    • one year ago
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    So 60 degrees for cosine?

  26. zepdrix
    • one year ago
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    |dw:1444434707066:dw|

  27. anonymous
    • one year ago
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    I'm not sure why I keep looking only in the first section. Thank you again for pointing that out.

  28. anonymous
    • one year ago
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    60 and 300

  29. zepdrix
    • one year ago
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    Degrees are icky :) lol You gotta try to get comfortable with radians XD

  30. zepdrix
    • one year ago
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    Ok great so we've found all of our critical points!! \[\large\rm x=\frac{\pi}{3},~\pi,~\frac{5\pi}{3}\]

  31. zepdrix
    • one year ago
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    Here is the graph in case you wanted to see it, it's pretty cool https://www.desmos.com/calculator/6sbfzaocym You can clearly see a hill top at pi/3 and 5pi/3 and the bottom of a valley at the pi.

  32. anonymous
    • one year ago
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    Wow! I never knew you could graph something like this on desmos. That's very interesting.

  33. anonymous
    • one year ago
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    My connection on this site is horrible. sorry

  34. anonymous
    • one year ago
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    Thank you for your help! (I'll probably be back on here soon with more questions)

  35. zepdrix
    • one year ago
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    Yah it's a little complicated to graph just a small interval on desmos, you have to put it into set brackets { } and the interval comes first, then colon, then your function.

  36. anonymous
    • one year ago
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    Could you help me with one last thing really quickly?

  37. zepdrix
    • one year ago
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    sure

  38. anonymous
    • one year ago
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    I know the derivative is sinx (2cosx -1) but I can't exactly figure out how to get that.

  39. zepdrix
    • one year ago
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    oh you cheated on the first part? come on broski -_-

  40. anonymous
    • one year ago
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    No, I didn't cheat.

  41. anonymous
    • one year ago
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    The assignment gives me the derivative

  42. zepdrix
    • one year ago
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    Oh :O

  43. zepdrix
    • one year ago
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    \[\large\rm y=(\sin x)^2+\cos x\]\[\large\rm y'=2(\sin x)(\sin x)'+(\cos x)'\]Start by applying power rule to the sine function. Then we have chain rule as well.

  44. zepdrix
    • one year ago
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    You really have to remember your sine and cosine derivatives to get through this one :)

  45. zepdrix
    • one year ago
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    \[\large\rm y'=2(\sin x)(\cos x)+(-\sin x)\]

  46. zepdrix
    • one year ago
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    And then factor a sin x out of each term.

  47. anonymous
    • one year ago
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    Oh okay! I see now. Thank you. :)

  48. zepdrix
    • one year ago
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    :D

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