Prove that the critical point of f satisfies the equation

[lep11]

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)^-½ (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 :/

 

[Samy_A]

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)^-½ (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?

 

[lep11]

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

 

[Samy_A]

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?

 

[lep11]

f(x₀)? i don't get it. f'(x₀)=0?

 

[Samy_A]

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 $f(x_0)=\frac{\sin x_0}{x_0}$?

 

[lep11]

so why is x₀≠0?

 

[Samy_A]

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 $a$?

 

[lep11]

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 $a$?

$x_0cosx_0-sinx_0=0 ⇔(sin x_0)/x_0=cosx_0=f(x_0)$

 

[Samy_A]

$x_0cosx_0-sinx_0=0 ⇔f(x_0[/SUB])=(sin x_0)/x_0=cosx_0$

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".

 

[lep11]

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".

$x_0 \cos x_0 = \sin x_0$ ⇔ $x_0^2 \cos^2 x_0 = \sin x_0^2$ ⇔ $x_0^2 \cos^2 x_0 =1- \cos x_0^2$ ⇔$\cos^2 x_0=\frac{1}{(1+x_0^2)^2}$ ⇔ $|f(x_0)|=\frac{1}{\sqrt{1+x_0^2}}$

 

[Samy_A]

$x_0 \cos x_0 = \sin x_0$ ⇔ $x_0^2 \cos^2 x_0 = \sin x_0^2$ ⇔ $x_0^2 \cos^2 x_0 =1- \cos x_0^2$ ⇔$\cos^2 x_0=\frac{1}{(1+x_0^2)^2}$ ⇔ $|f(x_0)|=\frac{1}{\sqrt{1+x_0^2}}$

Correct.

Now, we are left with $x_0=0$.
Assume $x_0=0$ is a critical point: does it satisfy $|f(x_0)|=\frac{1}{\sqrt{1+x_0^2}}$?

 

[lep11]

yes it does

 

[Samy_A]

yes it does

Indeed. That concludes the exercise.