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Finding points on a trigonometric function

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Use the function f(x) = sin x + cos x, where 0<=x<=pi to answer the following.

    a) Determine all stationary points. Classify the points as minimum or maximum
    b) locate the points of inflection
    c) determine the endpoints of the interval


    2. Relevant equations



    3. The attempt at a solution

    So I know that the first and second derivative are

    f1(x) = cos x - sin x
    f2(x) = -sin x - cos x

    Im lost at how to use this to determine my values.

    Thank you for your help!
     
  2. jcsd
  3. Mar 2, 2009 #2

    Dick

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    You are trying to solve e.g. cos(x)-sin(x)=0. Dividing both sides by cos(x) gives 1-tan(x)=0. Is that any help?
     
  4. Mar 2, 2009 #3
    I just found where cos x = sin x, put that into the function and found my value. Ive got the 1) and the 3) of the question. Now im just stuck on the point of inflection.

    Im curious though, can i divide both sides by cos x for the second derivative to find my points of inflection?

    If so that would leave me with tan x. How do I determine the point of inflection without actually pluggin in every value from 0 to pi?
     
    Last edited: Mar 2, 2009
  5. Mar 2, 2009 #4

    Dick

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    Sure. Then you get tan(x)=-1 instead of +1, right?
     
  6. Mar 2, 2009 #5
    Yup your right, so to find the point of inflection should I just do, tan x = -1 and find out what x is?
     
  7. Mar 2, 2009 #6

    Dick

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    I think so, don't you?
     
  8. Mar 2, 2009 #7
    Kind of, I get x = -0.8. But isnt that out of range of my domain since it has to be between 0 and pi?
     
  9. Mar 2, 2009 #8

    Dick

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    I think I can find a value of x between 0 and pi where tan(x)=(-1). Can't you?
     
  10. Mar 2, 2009 #9
    Yeah I got 3pi/4. Plug that into my function and I got a y = 1?

    So my point of inflection is (3pi/4, 1)?
     
  11. Mar 3, 2009 #10

    Dick

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    I don't think sin(3pi/4)+cos(3pi/4)=1.
     
  12. Mar 4, 2009 #11
    im sorry, i was in degree mode for this. In radian mode I get -0.1. Correct?
     
  13. Mar 4, 2009 #12

    Dick

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    No again. You don't NEED a calculator. To get f'' you solved sin(x)=(-cos(x)). That should make it pretty easy to figure out f(x)=sin(x)+cos(x).
     
    Last edited: Mar 4, 2009
  14. Mar 4, 2009 #13
    Well no apprantly its not that pretty easy to figure out. Your not making much sense there.
     
  15. Mar 4, 2009 #14

    Dick

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    I'm trying to point out that in this problem f''(x)=(-f(x)).
     
  16. Mar 4, 2009 #15

    lurflurf

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    helps to simplify first

    f(x) = sin(x)+cos(x)=sqrt(2)sin(x+pi/4)
     
  17. Mar 5, 2009 #16
    I dont get what you guys are saying here. Whats the problem with plugging in the values I find and using the calculator? "f''(x)=(-f(x))." is no different than me pluggin in
    3pi/4 into the original function.

    Either way I dont get how I can avoid using a calculator. Even using f''(x)=(-f(x)), f"(x)=0.1, so -f(x) = -0.1 So I get my point of inflection at (3pi/4,-0.1)
     
  18. Mar 5, 2009 #17

    HallsofIvy

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    The difficulty appears to be that you have no idea what "sine" and "cosine" are and Dick and lurflurf have been assuming that you did.

    If you did, then you would know that [itex]sin(3\pi/4)= -\sqrt{2}/2[/itex] and [itex]cos(3\pi/4)= -\sqrt{2}{2}[/itex] so that [itex]f(3\pi/4)= sin(3\pi/4)+ cos(\3pi/4)= -\sqrt{2}[/itex], not "-0.1".
     
  19. Mar 5, 2009 #18

    Dick

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    sin(3pi/4)=sqrt(2)/2, cos(3pi/4)=(-sqrt(2))/2. Even if you insist on the calculator bit, your's might need a tuneup. I get sin(3pi/4)=0.70710678118655.
    cos(3pi/4)=(-0.70710678118655). NOW do you see why the sum isn't '-.1'?
     
  20. Mar 5, 2009 #19
    Yup, your right I do see that now. The answer is 0. I dont get why my calculator is spitting out -0.1 but I get it now. Thank you!
     
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