$f$ not increasing in neighborhood of $0$

Question

Let $f: \mathbb R \to \mathbb R$, \begin{eqnarray} f(x)= \begin{cases} x-x^2&\quad\text {if } x \in \mathbb Q \cr x+x^2&\quad\text{if } x \notin \mathbb Q \cr \end{cases} \end{eqnarray}

Show that $f'(0)=1$ but $f$ isnt increasing in the neighborhood of $0$.

I derivated both of $f(x)$ and both result in $f'(0)=1$ (not sure if both are needed, or just the one where $x \in \mathbb Q$).

Can anyone help if I've misunderstood anything, is it about being continuous/differentiable? Since $x+x^2$ is over $x-x^2$, after $0$ it isnt increasing, but how do I prove it

Answer 1

Using the definition of derivative,

$$f'(0)=\lim_{\delta \rightarrow 0}\frac{f(\delta)-f(0)}{\delta}=\lim_{\delta \rightarrow 0}\frac{f(\delta)}{\delta}$$

$$\frac{f(\delta)}{\delta}=\begin{cases} 1-\delta &, \delta \in \mathbb{Q} \\ 1+ \delta &, \delta \notin \mathbb{Q}\end{cases}$$

$$f'(0)=1$$

We consider any neighborhood of $0$ of radius $\epsilon>0$, since rational numbers are dense, there is a rational number $q \in (0, \min(1,\epsilon))$. Notice that $f(q)>0$.

Since $q>0 \implies 8q^2>0 \implies -4q^2 < 4q^2$,

we have $$1+4q-4q^2 < 1+ 4q +4q^2=(2q+1)^2$$

which implies that

$$\sqrt{1+4(q-q^2)}<2q+1$$

and hence, $$\frac{-1+\sqrt{1+4f(q)}}{2} < q$$

Since irrational numbers are dense, we can choose irrational $y \in \left( \frac{-1+\sqrt{1+4f(q)}}{2}, q\right)$.

Hence $y< q$, but

$$y > \frac{-1+\sqrt{1+4f(q)}}{2}$$

$$2y+1 > \sqrt{1+4f(q)}$$

$$4y^2+4y+1 > 1+4f(q)$$

$$f(y) > f(q)$$

Remark: $f$ is not a $C^1$ function.