Proof clarification for Question 68 of Spivak's 'Calculus' - exploring the function $f(x)=\alpha x + x^2 \sin(1/x)$ for $\alpha =1$

Question 68 from Chapter 11 of Spivak's Calculus is about interrogating the behavior of the function $f$ defined by: $f(x)=\alpha x + x^2\sin(1/x)$ for $x \neq 0$ and $f(0)=0$, for $\alpha$ values $\geq 1$.

I am having some difficulty following Spivak's logic in the solution manual.

To provide some relevant background:

1. $f'(x)=\alpha +2x\sin(1/x)-\cos(1/x)$ for $x \neq 0$

2. Spivak creates the function $g$ defined as $g(y)=2(\sin y)/y-\cos(y)$...where $h(x)=y=\frac{1}{x}$

3. For any $y_0$ where $g'(y_0)=0$, we have that $$|g(y_0)|=\frac{2+y_0^2}{\sqrt{4+y_0^4}}\gt 1$$

Here is the question of interest:

Show that if $\alpha=1$, then $f$ is not [strictly] increasing in any interval around $0$.

Here is Spivak's solution manual's entry:

We have $f'(x)=1+g(1/x)$. Now we clearly have $g(y) \lt 0$ for arbitrarily large $y$ (since $g(y)$ is practically $-\cos y$ for large $y$), so for arbitrarily large $y$ we have \begin{align}g(y) \lt - \frac{2+y^2}{\sqrt{4+y^4}}\lt -1 \end{align}Thus $f'(x) \lt 0$ for arbitrarily small $x$, while we also have $f'(x) \gt 0$ for arbitrarily small $x$.

Starting with the problem statement, it seems to me that our goal should be to show that for any arbitrary $\delta \gt 0$, there is an $x \in (0,\delta)$ such that $f'(x) \lt 0$. If we do that, then we can be sure that $f$ is not strictly increasing in any neighborhood around $0$.

Given this, I really do not understand what Spivak's argument is saying. Framing this from the perspective of $y$, it seems to me the Spivak's argument is saying that for any $N \gt N_0$ (where $N_0$ denotes the 'sufficiently far away' condition), we can always find a $y \gt N$ satisfying the statement:

$$g(y) \lt - \frac{2+y^2}{\sqrt{4+y^4}}\lt -1 $$

It is not clear to me why we are in a position to assert this. Spivak is presumably making use of the statement: For any $y_0$ where $g'(y_0)=0$, we have that $$|g(y_0)|=\frac{2+y_0^2}{\sqrt{4+y_0^4}}\gt 1$$

but I cannot see how. Why should it be the case that for any sufficiently far away $N$, I can find an instance of a $y_0 \gt N$ satisfying $g'(y_0)=0$? Further, even if this is true, we only know that $|g(y_0)| \gt 1$. Why could it not be the case that for all such instances of $y_0$, we have that $g(y_0) \gt 1$?

I suspect it becomes relevant to show that these $y_0$'s are local minimum/maximum, rather than merely inflection points...but I am not certain at which point this was done by Spivak.

Thanks ~

NOT AN ANSWER

I read your approach and it seems to me that it is correct, though I myself don't know if that is what the author had in mind.

I would rather follow this path. When $\alpha =1$, we have, for $x \neq 0$, $$f'(x) = 1+2x \sin\left(\frac1{x}\right)-\cos\left(\frac1{x}\right),$$ and $$f''(x) = 2\sin\left(\frac1{x}\right) - \frac2{x}\cos\left(\frac1{x}\right) -\frac1{x^2}\sin\left(\frac1{x}\right).$$ To show, for example, that $f(x)$ is not increasing in any right neighborhood of $0$, notice that for any $k=1,2,\dots$ $$f'\left(\frac1{2k\pi}\right)=0,$$ and $$f''\left(\frac1{2k\pi}\right)=-4k\pi.$$ so that $\frac1{2\pi k}$ is a local maximum. This answers part (c) of the question.

When $\alpha >1$ note that, for $0< x < \frac{\alpha-1}2$, $$f'(x) = \alpha + 2x \sin\left(\frac1{x}\right) -\cos \left(\frac1{x}\right) \geq \alpha -2x -1>0$$ which answers to question (d), i.e. for $\alpha >1$ there is a neighborhood of $0$ in which the function is strictly increasing.