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*HELP* Finding x in [_,_] of Possible Inflection Points

  1. Jan 15, 2009 #1
    The solutions of f''(x) = 0 are the x-coordinates of possible inflection points on the curve y = f(x), where f'' denotes the second derivative of f. The second derivative of f(x) = esinx is:

    f'' (x) = esinxcos2x - esinxsinx

    Find all x in [0,2π] of possible inflection points.

    Uhhhmmmmm...:surprised

    also: our hint for this question is:
    "keep in mind that the function g(t) = et is never zero for any real number t"
    :bugeye: what is going on here?
    Thanks in Advance:!!)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 15, 2009 #2

    Mark44

    Staff: Mentor

    Factor [tex]e^{sin(x)}[/tex] from both terms in f''(x) and then set the remaining factor to 0. As a hint in solving the resulting equation cos^2(x) = 1 - sin^2(x).
     
  4. Jan 15, 2009 #3

    HallsofIvy

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    F"(t)= [itex]e^{sin(t)}(cos^2(x)- sin(x))[/itex] and, as your hint tells you, [itex]e^{sin(t)}[/itex] is never 0. In order to have F"(t)= [itex]e^{sin(t)}(cos^2(x)- sin(x))= 0[/itex] you must have [itex]cos^2(x)- sin(x)= 0[/itex].
     
  5. Jan 15, 2009 #4
    Uhm ok so...
    f''(x) = esinx (½(1+cos2x))-esinxsinx

    f''(x) =esinx (½ + ½cos2x - sinx)
    so what should i do now?
     
  6. Jan 15, 2009 #5
    so
    cos2x = sinx or (½+½cos2x) = sinx ?? is this what you mean?
    okay my main problem is that i dont understand what the question is asking for what does the last part of the question mean the 'find all x in [0,2π] of possible inflection points'?
    what am I trying to find here?
     
  7. Jan 15, 2009 #6

    HallsofIvy

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    ??No one has suggested that you convert [itex]cos^2(x)= (1/2)(1+ cos(2x))[/itex]. What has been suggested is that you solve [itex]cos^2(x)- sin(x)= 0[/itex]. Mark44 suggested you use [itex]cos^2(x)= 1- sin^2(x)[/itex] so that [itex]cos^2(x)- sin(x)= 0[/itex] becomes [itex]1- sin^2(x)- sin(x)= 0[/itex]. That is a quadratic equation in sin(x). If you let y= sin(x) you have [itex]y^2+ y- 1= 0[/itex]. Solve that that for y and then solve sin(x)= y for x.
     
  8. Jan 15, 2009 #7
    OKAY so:
    f'' (x) = esinx(cos2x - sinx)
    let cos2x - sinx = 0
    1-sin2(x) - sinx = 0
    let y = sinx
    y2 + y - 1 = 0
    Quadratic Equation Gives 2 Answers:
    0.6180
    and
    -1.6180 <-- DO NOT use because it is not in the interval in question
    sin(x) = y
    sin(x) = 0.6180
    x = sin-1(0.6180)
    x = 38.17
    is that it for this problem?
     
  9. Jan 15, 2009 #8

    Mark44

    Staff: Mentor

    Not quite.
    You're correct to discard the value -1.6180, but for the other, you have found one value of x for which sin(x) = .6180. There are two values in the interval of interest, [0, 2pi]. Both values should be in radians, not degrees.
     
  10. Jan 15, 2009 #9
    so ok since:
    sin(x) = 0.6180
    x = 0.6662 rad
    so I drew my sin curve but...
    ok so am I supposed to just for example do:
    x = 0.6662 + π
    ??
     
  11. Jan 15, 2009 #10

    Mark44

    Staff: Mentor

    No, that's not it. Think about what the graph of y = sin(x) looks like between 0 and 2pi. As a hint, sin(pi/6) = 1/2 (or sin(30 deg.) = 1/2). What's another value x, in radians, for which sin(x) = 1/2? It's not pi + pi/6.
     
  12. Jan 15, 2009 #11
    oh ok I have to find the where the line
    sinx = y = 0.6180 crosses with the curve y = sinx
    that is two times in this interval once at the inverse of sin(0.6180) and ???
    okay i can see it on my graph but how do i actually calculate this value of x?
     
  13. Jan 15, 2009 #12
    With a calculator?
     
  14. Jan 15, 2009 #13
    lol omg *embarrassed* YES
     
  15. Jan 15, 2009 #14
    Didn't know anyone got so excited about calculus
     
  16. Jan 15, 2009 #15
    can you help me:redface: please
     
  17. Jan 15, 2009 #16
    Help you with what? I thought your question has been answered
     
  18. Jan 15, 2009 #17
    No...this question hasnt:
    oh ok I have to find the where the line
    sinx = y = 0.6180 crosses with the curve y = sinx
    that is two times in this interval once at the inverse of sin(0.6180) and ???
    okay i can see it on my graph but how do i actually calculate this value of x?
     
  19. Jan 15, 2009 #18
    You would use your calculator and type in arcsin(.618)?
     
  20. Jan 15, 2009 #19
    haha, yes I did taht but it only gives me the one value of x, but there are two intersections so there is another value of x that gives sin(x) = 0.6180
    if you just do sin-1 (0.6180) = 0.6662 ---> only one value, i need 2 values
     
  21. Jan 15, 2009 #20
    OK let's pick a simpler example. If you had to find sin(x) = 1/2 how would you figure out x?
     
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