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Continuity (intermediate value theorem)

  1. Sep 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Let n be a natural number.
    Prove that the equation: [tex]x^{2}(cos(x))^{n}+xsin(x)+1=0[/tex] has an infinite amount of solutions.




    3. The attempt at a solution

    I named that equation f(x)=0 and I said that f(a)<f(0)<f(b) and that f(a) x f(b) < 0.

    Should I choose n=1 and chose a=0 and b=pi ? I know that the equation is a period equation so it will have an infinite amount of solutions, but i'm not sure how to show it because I've only been dealing with equations that want you to show there is at least one solution in [a,b]. Any help would be really appreciated. Thank you.
     
  2. jcsd
  3. Sep 24, 2012 #2

    jbunniii

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    Hint: consider the values of [itex]x[/itex] for which [itex]\cos(x) = 0[/itex]. What does [itex]\sin(x)[/itex] equal for these values of [itex]x[/itex]?
     
  4. Sep 24, 2012 #3
    When cos(x)=0 then sin(x)=1 for values in the form of x= pi/2+2kpi

    So then we could have : f(3pi/2) < f(0) < f(pi/2) ??
     
  5. Sep 24, 2012 #4

    jbunniii

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    OK, and what about [itex]x = 3\pi/2 + 2k\pi[/itex]?

    Are you asking me, or telling me? What does f(x) reduce to when [itex]x = \pi/2 + 2k\pi[/itex]? How about when [itex]x = 3\pi/2 + 2k\pi[/itex]?
     
  6. Sep 24, 2012 #5
    Don't mind me I've had a really long day i wanted to say x=pi/2+kpi.
     
  7. Sep 24, 2012 #6
    Ok this is what i got: f(pi/2)=pi/2+1 and f(3pi/2)=1-3pi/2 so we can say that:

    f(3pi/2) < 0 < f(pi/2) am i correct?
     
  8. Sep 24, 2012 #7

    jbunniii

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    Yes, that's right. Now you can apply the intermediate value theorem.

    You can argue similarly for [itex]f(3\pi/2 + 2k\pi)[/itex] and [itex]f(\pi/2 + 2k\pi)[/itex], for any positive integer [itex]k[/itex].
     
  9. Sep 24, 2012 #8
    Alright I've got it thanks for your help.
     
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