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anonymous

  • 4 years ago

determine the concavity of the graph of f(x) = 3sin(x)+2(cos(x))^2 at x=pie

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  1. TuringTest
    • 4 years ago
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    you're gonna need to find f''(x) first see how you do finding f'(x) hint: remember to use the chain rule on 2(cos(x))^2

  2. anonymous
    • 4 years ago
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    okay im going to work through it as far as i can when i get stuck ill tell u

  3. anonymous
    • 4 years ago
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    derivative of sin is cos right?

  4. TuringTest
    • 4 years ago
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    yes

  5. TuringTest
    • 4 years ago
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    this may be handy http://tutorial.math.lamar.edu/pdf/Calculus_Cheat_Sheet_Derivatives_Reduced.pdf chain rule is critical on the next term

  6. anonymous
    • 4 years ago
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    when doing the product rule for 2(cos(x))^2 my f(x)= 2x^2 and g(x) is cos(x) right?

  7. TuringTest
    • 4 years ago
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    chain rule, but yes to your substitution chain rule is f(g(x))'=f'(g(x))*g'(x)

  8. anonymous
    • 4 years ago
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    okay i got 4(cos(x))^2(-sin(x))

  9. TuringTest
    • 4 years ago
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    close, but since you took the derivative of the cos^2 it's just cos to the first power f'(x)=3cosx-4cos(x)sin(x)

  10. anonymous
    • 4 years ago
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    i took the derivative of cosx and the derivative of 2x^2 and then it was 4x(cosx)

  11. TuringTest
    • 4 years ago
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    yeah like I tried to amend, your substitution was a little off... it should actually be f(x)=2[g(x)]^2 g(x)=cosx chain rule is f(g(x))'=f'(g(x))*g'(x)

  12. anonymous
    • 4 years ago
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    oh i see what i did wrong

  13. TuringTest
    • 4 years ago
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    I shouldn't have glossed over that :/

  14. anonymous
    • 4 years ago
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    instead of replacing the x in 2x^2 with cosx i just left it alone. so it turn 4cosx

  15. TuringTest
    • 4 years ago
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    exactly

  16. anonymous
    • 4 years ago
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    great thanks for being patient in helping me understand that.

  17. TuringTest
    • 4 years ago
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    then times the derivative of cosx...

  18. anonymous
    • 4 years ago
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    -4cosxsinx

  19. TuringTest
    • 4 years ago
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    sure, thanks for listening so you get f(x)=2[g(x)]^2 g(x)=cosx chain rule is [f(g(x))]'=f'(g(x))*g'(x)=4[g(x)]g'(x)=-4cosxsinx yup :)

  20. anonymous
    • 4 years ago
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    so i have f'(x) = 3cosx-4cosxsinx

  21. TuringTest
    • 4 years ago
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    and now we gotta do it again this time we will need the product rule for the second term

  22. anonymous
    • 4 years ago
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    thats right cause f'(x) gets increase/decrease and f"(x) gets concavity

  23. TuringTest
    • 4 years ago
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    exactly

  24. anonymous
    • 4 years ago
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    okay i got -3sin(x)+4sin(x)^2-4cos(x)^2

  25. TuringTest
    • 4 years ago
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    me too :) now plug in x=pi what do you get and what does it say about the concavity there?

  26. anonymous
    • 4 years ago
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    should i use a calculator?

  27. TuringTest
    • 4 years ago
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    no it's a special value what's sin(pi) ? cos(pi) ?

  28. TuringTest
    • 4 years ago
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    remember your unit circle...

  29. anonymous
    • 4 years ago
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    oh okay yes

  30. anonymous
    • 4 years ago
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    pi = 180degrees (1,0) i believe so 0

  31. TuringTest
    • 4 years ago
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    what is zero?

  32. anonymous
    • 4 years ago
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    sin(pi)

  33. TuringTest
    • 4 years ago
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    right and cos?

  34. anonymous
    • 4 years ago
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    but i meant (-1,0)

  35. TuringTest
    • 4 years ago
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    right

  36. anonymous
    • 4 years ago
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    -1

  37. TuringTest
    • 4 years ago
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    so what is the concavity?

  38. anonymous
    • 4 years ago
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    concave down

  39. TuringTest
    • 4 years ago
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    right, and the number?

  40. anonymous
    • 4 years ago
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    lol let me work through it

  41. anonymous
    • 4 years ago
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    -4

  42. TuringTest
    • 4 years ago
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    exactly :D

  43. anonymous
    • 4 years ago
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    damn bro god bless you.

  44. TuringTest
    • 4 years ago
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    happy to help :)

  45. anonymous
    • 4 years ago
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    lol i have more problems :) id be great if you could help me understand

  46. TuringTest
    • 4 years ago
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    I've gotta get something to eat at some point, but until/after that I can help. Just post them and if I'm here I'll help, but others are good at this stuff too.

  47. anonymous
    • 4 years ago
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    oh great okay ima post one up right now

  48. TuringTest
    • 4 years ago
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    I mean you should post them separately just to be clear

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