Vtyjkyuk

We have this function

$$f(x) =
begincases
x^2, & textif $x$ $in mathbbQ$ \[2ex]
0, & textif $x$ $inmathbbI$
endcases$$

We have to determine where the function is differentiable and calculate the derivative. Then we need to determine where is not differentiable and prove this with $epsilon - delta$

Attempt:

We define:
$$text$h(x)=0$ and $g(x)=x²$, then $h(0)=0=g(0)=f(0)$$$
Now we have:
$$text$h(x)$ ≤ $f(x)$ ≤ $g(x)$, for all x$inmathbbR$$$
$$textthat is equal to: $h(x)-h(0)$ ≤$ f(x) - f(0) $≤$ g(x) - g(0)$$$
$$textAnd without losing generality assuming that $x>0$:$$
$$text$frach(x)-h(0)x$ ≤ $fracf(x)-f(0)x$ ≤ $fracg(x)-g(0)x$$$
$$textFinally: $lim_xto 0 frach(x)-h(0)x$ ≤ $lim_xto 0 fracf(x)-f(0)x$ ≤ $lim_xto 0 fracg(x)-g(0)x$$$
$$textBy the squeeze theorem we have that:$$
$$lim_xto 0 fracf(x)-f(0)x=0$$
$$textSo $f$ is differentiable at $c=0$ and $f'(0)=0$$$

The problem comes when we have to show that is not differentiable at any other point. I managed to do it with limits but I don't know how to put in into $epsilon - delta$

$$textLet $xinmathbbQ$ and $x≠0$. Then $f(x)=x²$. Now we know that exist a sequence $(y_n)_ninmathbbN$ of irrational numbers such that:$$
$$lim_nto inftyy_n = x$$
$$textAlso we have that $f(y_n)=0$ for all $n$, because $y_n$ is irrational, but:$$
$$lim_nto inftyf(y_n) = f(x) = x²$$
$$textSo we can see that:$$
$$lim_nto inftyf(y_n) ≠ f(y_n)$$
$$textThis implies that $f$ is not continuous at $mathbbI$, therefore $f$ is not differentiable at $mathbbI$$$
$$textThe same thing works to prove that $f$ is not continuous at $mathbbQ$ so $f$ is not differentiable at any point, except $x=0$.$$

As I said the problem comes when I have to write that second thing with $text$epsilon - delta$ $, cause I don't really know how to start. Someone has any ideas? Thanks to everyone.

Allow $(x_n)$ be a sequence of rational numbers such that $x_n to sqrt2$.

Fix $epsilon > 0$. First notice that $|x_n - sqrt2| < delta implies |x_n + sqrt2| = |x_n - sqrt2 + 2sqrt2| leq |x_n - sqrt2| + 2sqrt2 < delta + 2sqrt2$.

So $|x_n - sqrt2||x_n + sqrt2| < delta(delta + 2sqrt2)$.

Allowing $delta < fracepsilon1 + 2sqrt2$ for small enough $epsilon$ (i.e. if $epsilon < 1 + 2sqrt2$) produces

$|f(x_n) - 2| = |x_n-sqrt2||x_n + sqrt2| < delta(delta + 2sqrt2) < delta(1+2sqrt2) < fracepsilon1+2sqrt2(1+2sqrt2) = epsilon$

Thus, $lim_x_n to sqrt2 f(x_n) = 2$, however $f(sqrt2) = 0$.

It is not continuous, thus not differentiable.

If a function is differentiable, it is continuous. Since $f$ is discontinuous for $x neq 0$ it cannot be differentiable for $x neq 0$.

If $x neq 0$ and $x$ is irrational you can find rationals $q_n to x$ and then $f(q_n) = q_n^2$ but $f(x) = 0$. Then $f(q_n)-f(x) over q_n - x = q_n^2 over q_n-x$.

If $x neq 0$ and $x$ is rational you can find irrationals $alpha_n to x$ and then $f(alpha_n) = 0$ but $f(x) = x^2 neq 0$. Then $f(alpha_n)-f(x) over alpha_n - x = x^2 over alpha_n-x$.

In particular, the ratio is not bounded as $n to infty$, so it is straightforward to use an $epsilon$-$delta$
argument to show that it is not differentiable at $x$ (if it was differentiable,
the ratio would be bounded).