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There is an exercise in Stephen Abbott's Understanding Analysis that states:

Exercise 5.3.7. (b) Show that the function $g(x)=\begin{cases} x/2+x^2\sin(1/x)&\text{ if }x\neq0\\\ 0&\text{ if }x=0 \end{cases}$ is differentiable on $\mathbb{R}$ and satisfies g'(0)\geq0. Now, prove that $g$ is not increasing over any open interval containing $0$.

First of all, I know that for $x=0$, g'(0)=\lim_{x\to0}\frac{x/2+x^2\sin(1/x)}{x}=\lim_{x\to0}\left[\frac{1}{2}+x\sin\left(\frac{1}{x}\right)\right]=\frac{1}{2}\geq0, and for $x\neq0$, g'(x)=\frac{1}{2}+2x\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right). Hence, g'(x)=\begin{cases} 1/2+2x\sin(1/x)-\cos(1/x)&\text{ if }x\neq0\\\ 1/2&\text{ if }x=0, \end{cases} and $g$ is differentiable on $\mathbb{R}$.

However, I do not know how to formally show that if $(a,b)$ is an open interval containing $0$, then $g$ is not increasing on it; the idea I have is to keep in mind that as $x$ approaches $0$, it oscillates 'faster,' and you can thus always find two different points $x,y\in(a,b)$ such that g'(x)>0 and g'(y)<0. Is this a valid assertion? If so, how can I go about showing it? Thanks in advance.

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Yes that would work. Since $g$ is differentiable on $(a,b)$ it would be increasing on $(a,b)$ if and only if g'\ge0 over $(a,b)$. So, you only need to find a point $c$ in $(a,b)$ where g'(c)<0.

Try a point $c$ where $\cos(1/c)=1$. These points would be of the form ${1\over 2n\pi}$. We have \textstyle g'({\textstyle{1\over 2n\pi}})= {1\over2}+ {1\over n\pi}\cdot\sin(2n\pi) -\cos(2n\pi)={1\over2}+0-1={-1\over2}.

As you can select $n$ so large that ${1\over 2n\pi}$ is in $(a,b)$, you are done.

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Yes, this is the right way. In your expression for g'(x), you see that the term $\cos(1/x)$ oscillates infinitely many times between $-1$ and $+1$ on any open interval containing zero, and because of this fact, such an interval contains infinitely many little subintervals where g'>0 and infinitely many where g'<0.