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Prove that the critical point of f satisfies the equation

  • Thread starter lep11
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  • #1
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Homework Statement


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.)


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 :/
 

Answers and Replies

  • #2
Samy_A
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Homework Statement


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.)


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 :/
How do you find the critical points of a differentiable function?
 
  • #3
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How do you find the critical points of a differentiable function?
By taking the derivative; f'(x)=0?
in this case f'(x)=(xcosx-sinx)/x2=0
 
  • #4
Samy_A
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By taking the derivative; f'(x)=0?
f'(x)=(xcosx-sinx)/x2=0
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?
 
  • #5
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f(x0)? i don't get it. f'(x0)=0?
 
  • #6
Samy_A
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f(x0)? i don't get it. f'(x0)=0?
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}##?
 
  • #7
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so why is x0≠0?
 
  • #8
Samy_A
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so is x0=0?
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##?
 
  • #9
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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##?
##x_0cosx_0-sinx_0=0 ⇔(sin x_0)/x_0=cosx_0=f(x_0)##
 
  • #10
Samy_A
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##x_0cosx_0-sinx_0=0 ⇔f(x_0[/SUB])=(sin x_0)/x_0=cosx_0##
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".
 
  • #11
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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".
##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}}##
 
  • #12
Samy_A
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##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}}##
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}}##?
 
  • #13
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yes it does
 
  • #14
Samy_A
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yes it does
Indeed. That concludes the exercise.
 

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