1. Feb 7, 2017

### issacnewton

1. The problem statement, all variables and given/known data
Consider the function $f(x) = x^4 \sin(\frac 1 x)$ for $x \ne 0$ and $f(x) = 0$ for $x =0$. I have to prove that $x=0$ is the critical number of this function and its derivative changes the sign indefinitely.

2. Relevant equations
Definition of the critical number

3. The attempt at a solution
$x=0$ would be the critical number if $f'(0) = 0$ or if $f'(0)$ does not exist. I have been able to show that $f'(0) = 0$ using the squeeze theorem. Now I want to show that the derivative of $f(x)$ changes sign indefinitely. I am totally stuck at this point. How would I progress from here ?

Thanks $\smallsmile$

2. Feb 7, 2017

### Staff: Mentor

As x approaches 0 from the right, look at what happens for $x = \frac 2 {\pi}, \frac 2 {3\pi}, \frac 2 {5\pi}, \dots$. What are the values of 1/x at these numbers? What are the values of $\sin(1/x)$ at these numbers?.

3. Feb 8, 2017

### issacnewton

Thanks Mark for replying. I see that the sign will be altering between positive and negative as $x$ approaches $0$ from the the right at these values. So do I just write this, or is there more formal way of writing it as a proof ?

4. Feb 8, 2017

### Staff: Mentor

No, don't write just what I said, as I was trying to steer you in the right direction by asking some questions, and showing a sequence $\{x_n \}$ whose limit is zero, and for which the sequence $\{\sin(\frac 1 {x_n}) \}$ oscillates. Have you written up something that captures this idea?

Note that a similar idea can be used to show that the sequence $\{\sin(\frac 1 {x_n}) \}$ has the same behavior as x approaches zero from the left.

5. Feb 12, 2017

### issacnewton

Ok, So I have to prove an existence of a sequence in $\mathbb{R}$, such that this sequence goes to zero as $n \rightarrow \infty$ but the sequence $\{\sin(\frac{1}{x_n})\}$ oscillates. You have already come up with the sequence. So I need to prove that $\forall~n \in \mathbb{N}~\exists k, p \in \mathbb{N}$ such that $\sin(\frac{1}{x_k}) > 0$ and $\sin(\frac{1}{x_p}) < 0$. Would this be correct ?

6. Feb 12, 2017

### Staff: Mentor

As you note, you don't have to "prove" the existence of a sequence -- I gave you one that does the trick.This sequence contains one subsequence for which $\sin(x_{n_1}) > 0$ and another subsequence for which $\sin(x_{n_2}) < 0$. IMO, you should focus first on figuring this out, and less on the fancy "mathy" formatting, such as this stuff -- $\forall~n \in \mathbb{N}~\exists k, p \in \mathbb{N}$

7. Feb 12, 2017

### issacnewton

So if $x_n = \frac{2}{\pi(2n-1)}$ is the sequence given by you, then for the subsequence $x_{n_1} = \frac{2}{\pi(4n_1 - 3)}$, we have $\sin(x_{n_1}) > 0$ and for the subsequence $x_{n_2} = \frac{2}{\pi(4n_2 - 1)}$, we have $\sin(x_{n_2}) < 0$

8. Feb 12, 2017

### Staff: Mentor

Yes, those are the subsequences I was thinking of. The whole sequence should demonstrate that f ' changes sign infinitely often as x approaches zero. "Infinitely often" as a better way to say it than "changes signs indefinitely."

9. Feb 13, 2017

### issacnewton

Ok.. so I will complete the entire proof when I get the time as I think whole process is with me. But this problem is from James Stewart's Calculus (8 ed.) and this is just a computational book. So I am surprised that they have this problem. When we start introducing subsequences, we are using machinery of real analysis.

10. Feb 13, 2017

### Staff: Mentor

Maybe you are misinterpreting what the problem is asking for. If it says "show ..." rather than "prove ..." a less rigorous explanation is called for. Can you post a picture of the problem statement?

11. Feb 13, 2017

### issacnewton

Ok, I am putting snapshot of the problem.

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12. Feb 21, 2017

### issacnewton

So I will present the proof here. Consider a sequence $x_n = \frac{2}{\pi(2n-1)}$ where $n\in\mathbb{N}$. Then we have a subsequence $x_{n_1} = \frac{2}{\pi(4n_1 - 3)}$, such that $f'(x_{n_1}) = \frac{32}{\pi^3(4n_1-3)^3} >0$ for all $n_1 \in \mathbb{N}$ and we have another subsequence $x_{n_2} = \frac{2}{\pi(4n_2 - 1)}$ such that $f'(x_{n_2}) = \frac{-32}{\pi^3(4n_2-1)^3} <0$ for all $n_2 \in \mathbb{N}$. So this proves that, the derivative changes sign infinitely often. I hope this is correct.

13. Feb 21, 2017

### issacnewton

I also have to prove that $f$ has neither a local maximum nor a local minimum at $0$. Here I present my proof. Suppose $f$ has a local maxima at $x=0$. Then we have some open interval $(a,b)$ containing $x=0$ such that $f(x)\leqslant f(0)$, for all $x \in (a,b)$. This implies $f(x) \leqslant 0$, for all $x \in (a,b)$. Now $a<0<b$. By Archimedean Property, there exists $n_1 \in \mathbb{N}$ such that $\frac 1 n_1 < \frac{\pi b}{2}$. But we have $4n_1-3 \geqslant n_1 >0$, which implies that $\frac{1}{4n_1-3} < \frac{\pi b}{2}$. Hence we have $0< \frac{2}{\pi(4n_1-3)} < b$. So $\frac{2}{\pi(4n_1-3)} \in (a,b)$. This leads us to $f\left(\frac{2}{\pi(4n_1-3)}\right) \leqslant 0$. But $\forall n_1 \in \mathbb{N}$, we have $f\left(\frac{2}{\pi(4n_1-3)}\right) = \frac{16}{\pi^4(4n_1-3)^4} > 0$. So we reach a contradiction. Hence $f$ doesn't have a local maxima at $x=0$. Now suppose that $f$ has a local minima at $x=0$. Again, there exists an open interval $(a,b)$ containing $x=0$ such that $f(0)\leqslant f(x)$, for all $x \in (a,b)$. Since we have $a<0<b$, we can come up with some $n_2 \in \mathbb{N}$, due to Archimedean property, such that $\frac 1 n_2 < \frac{\pi b}{2}$. But $(4n_2-1) > n_2 > 0$, so we get $\frac{1}{4n_2-1} < \frac{\pi b}{2}$. So $0< \frac{2}{\pi(4n_2-1)} < b$. This means that $\frac{2}{\pi(4n_2-1)} \in (a,b)$. Because of our assumption, this means that $f\left(\frac{2}{\pi(4n_2-1)}\right) \geqslant 0$. But $\forall n_2 \in \mathbb{N}$, we have that $f\left(\frac{2}{\pi(4n_2-1)}\right) = \frac{-16}{\pi^4(4n_2-1)^4} < 0$. So again we reach a contradiction, which means that $f$ can not have a local minima at $x=0$. I hope my proof is correct.