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Homework Help: Math question

  1. Dec 24, 2004 #1
    I'm not quite sure what the question is asking:
    "If [tex]f(x)=e^xsin(x)[/tex], then the number of zeros of [tex]f[/tex] on the interval [o,2pi] is?"

    I took the derivative of this and found where it was equal to zero:
    [tex]f(x)=e^xsin(x)[/tex]
    [tex]f'(x)=e^xsin(x)+e^xcos(x)[/tex]
    [tex]0=sin(x)+cos(x)[/tex]

    I got zero. However, tha answer is 3, any suggestions?
     
  2. jcsd
  3. Dec 24, 2004 #2
    No, it wants where the function is zero, not where the gradient of the curve is zero. Where is sin x zero in that interval?
     
  4. Dec 24, 2004 #3
    you mean when [tex]0=e^xsin(x)[/tex]?
     
  5. Dec 24, 2004 #4

    dextercioby

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    [tex] e^{x}\sin x=0\Rightarrow \sin x=0 [/tex]
    Solve the last equation on the interval [itex] [0,2\pi] [/itex]

    Daniel.
     
  6. Dec 24, 2004 #5

    Yes, yes I do (sorry had to lengthen my post).
     
  7. Dec 24, 2004 #6
    it's is zero when x=0, x=pi, x=2pi

    so it hits three times!

    thanks! Also, why doesnt the e^x make a difference?
     
  8. Dec 24, 2004 #7

    dextercioby

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    It never annulates.Not even for complex arguments.

    Daniel.

    EDIT:Cause it never annulates,it does not affect the zero-s of the function.Plot the graph of 'f'.U'll see quite an interesting behavior.It has no limit for x->+infty.At minus infty it goes to zero.
     
    Last edited: Dec 24, 2004
  9. Dec 24, 2004 #8
    so e^x only increases the sinX amplitude, never now far it streatchs, so it doesnt effect how many times sinX crosses the x-axis
     
  10. Dec 25, 2004 #9

    HallsofIvy

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    "Annulates"??? I assume you mean "is never equal to 0" but the only definition I can find of "annulate" is "ring shaped".

    UrbanXrises: It's not so much that it is 'always increasing'. In order to solve AB= 0, you solve A= 0 and B= 0. Since ex is never 0, The only solutions of exsin(x)= 0 are where sin(x)= 0.
     
    Last edited by a moderator: Dec 25, 2004
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