Given: $0 \leq f(x) \leq x^2$ for all $x$. Prove that $f$ is differentiable at $ x=0$, and find $f '(0)$. Give a counterexample of a function which satisfies the hypothesis, but which is not continuous for $x \neq 0$.
How can I prove the differentiability? I am not aware of any "squeeze" rules that could apply in this case. Also, I cannot come up with the counterexample.
We have to show that
$$ \lim_{h \to 0} \frac{ f(h) - f(0)}{h} $$
exists. Notice
$$ 0 \leq \frac{f(h)-f(0)}{h} \leq \frac{ h^2 - f(0)}{h} = h $$
Since $f(0) = 0$ Why? $0 \leq f(0) \leq 0^2 $
Now, as $h \to 0$, we see that the quotient tends to $0$ by squeeze rule. In particular, it must be differentiable at the origin.
Heres an example: $f(x) = x^2$ if $x$ is rational. $f(x) = -x^2$ if $x$ is irrational.
Show that this works by looking at the difference quotient, and proving that $f(x) = x$ if $x$ is rational, and $-x$ if $x$ is irrational is continuous at $0$ by the squeeze theorem.
The derivative at $0$ of $f(x)$ is $Lim_{h \rightarrow 0}\frac{f(h)-f(0)}{h}$. Now, you can bound $f(0)$ below by $0$, and $f(h)$ can be bounded above by $h^2$. This is un upper bound for $f'(x)$.
