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Prove that the critical point of f satisfies the equation

  1. Apr 16, 2016 #1
    1. The problem statement, all variables and given/known data
    a.) Let f(x)=(sin x)/x, x≠0. Define f(0) such that f is continuous at x=0.
    b.) Prove that if x0 is critical point of function f (f(0) defined as in part a), then |f(x0)|=1/(1+x02) (Hint: use the basic properties of sine and cosine with given information.)


    3. The attempt at a solution
    a.)Easy. Let f(0)=1, because (sin x)/x approaches 1 when x--->0 so f(0) has to be equal to 1.
    b.) I have no idea how to begin :/
     
  2. jcsd
  3. Apr 16, 2016 #2

    Samy_A

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    How do you find the critical points of a differentiable function?
     
  4. Apr 16, 2016 #3
    By taking the derivative; f'(x)=0?
    in this case f'(x)=(xcosx-sinx)/x2=0
     
  5. Apr 16, 2016 #4

    Samy_A

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    Correct (except for x=0, but let's leave that aside for now).
    So what can you deduce from that about f(x0), where x0 is a critical point of f?
     
  6. Apr 16, 2016 #5
    f(x0)? i don't get it. f'(x0)=0?
     
  7. Apr 16, 2016 #6

    Samy_A

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    Yes, ##f(x_0)##, that's what the question is about.
    You already know that for a critical point ##x_0 \neq 0##, ##f'(x_0)=\frac{x_0 \cos x_0 -\sin x_0}{x_0^ 2}=0##.
    Can you simplify that last equation?
    And what does it mean for ##f(x_0)=\frac{\sin x_0}{x_0}##?
     
  8. Apr 16, 2016 #7
    so why is x0≠0?
     
  9. Apr 16, 2016 #8

    Samy_A

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    No, we actually exclude x=0 for now (because you have to compute f'(0) in a different way than what you did in post #3 for x ≠ 0).

    So, again ##x_0 \neq 0## is a critical point of ##f##, and that gives us ##f'(x_0)=\frac{x_0 \cos x_0 -\sin x_0}{x_0^ 2}=0##.
    When you have an equality like ##\frac{a}{b}=0##, where ##b \neq 0##, what does this tell you about ##a##?
     
  10. Apr 16, 2016 #9
    ##x_0cosx_0-sinx_0=0 ⇔(sin x_0)/x_0=cosx_0=f(x_0)##
     
  11. Apr 16, 2016 #10

    Samy_A

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    Yes, that is correct.
    But you are not finished yet. The question was to prove that for a critical point ##x_0##, ##|f(x_0)|=\frac{1}{\sqrt{1+x_0^2}}##.

    Hint: you have that ##f(x_0)=\cos x_0##. Now use ##x_0 \cos x_0 = \sin x_0## to express ##\cos x_0## in function of ##x_0##. Squaring ##x_0 \cos x_0 = \sin x_0## may be helpful. And remember the hint in the question: "use the basic properties of sine and cosine".
     
  12. Apr 16, 2016 #11
    ##x_0 \cos x_0 = \sin x_0## ⇔ ##x_0^2 \cos^2 x_0 = \sin x_0^2## ⇔ ##x_0^2 \cos^2 x_0 =1- \cos x_0^2## ⇔##\cos^2 x_0=\frac{1}{(1+x_0^2)^2}## ⇔ ##|f(x_0)|=\frac{1}{\sqrt{1+x_0^2}}##
     
  13. Apr 16, 2016 #12

    Samy_A

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    Correct.

    Now, we are left with ##x_0=0##.
    Assume ##x_0=0## is a critical point: does it satisfy ##|f(x_0)|=\frac{1}{\sqrt{1+x_0^2}}##?
     
  14. Apr 16, 2016 #13
    yes it does
     
  15. Apr 16, 2016 #14

    Samy_A

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    Indeed. That concludes the exercise.
     
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