Thoughts on the derivative of a function

[haruspex]

Hi, PF, sorry, I am a little bit boring:

¿Why $f(x)=\begin{cases}{x^2}&\text{if}& x\in \mathbb Q\\6(x-3)+9 & \text{if}& x\in \mathbb R \setminus \mathbb Q\end{cases}$ is differentiable only at $x=3$?

1- $\displaystyle\lim_{x \to{3}}{\dfrac{f(x)-f(3)}{x-3}}$ (*)

$\displaystyle\lim_{x\in \Bbb Q,\,x \to 3}{}\dfrac{x^2-9}{x-3}$(**)

$\displaystyle\lim_{x\in \Bbb I,\,x \to 3}{}\dfrac{6(x-3)+9-9}{x-3}$(***)

For this limit (*), it exists iff (**) and (***) are equal.

(**)$\displaystyle\lim_{x\in \Bbb Q,\,x \to 3}{}\dfrac{x^2-9}{x-3}=6$

Proof:

For $x \in\Bbb Q$, $\forall{\varepsilon}$ $\exists\;{\delta_1}$ such that if $0<|x-3|<\delta_1\Rightarrow{\left |{\dfrac{x^2-9}{x-3}-6}\right |=|x+3-6|=|x-3|<\varepsilon}$. Hence, $\delta_1=\varepsilon$(***)$\displaystyle\lim_{x\in \Bbb I,\,x \to 3}{}\dfrac{6(x-3)+9-9}{x-3}=6$

Proof:

For $x \in\Bbb R\setminus{\Bbb Q}$, $\forall{\varepsilon}$ $\exists\;{\delta_2}$ such that if $0<|x-3|<\delta_2\Rightarrow{\left |{\dfrac{(6(x-3)+9)-9}{x-3}-6}\right |=|6-6|=0<\varepsilon}$. Hence, $\delta_2$ can take any value, and the implication will be true.

Choose $\delta_1=\varepsilon$, and $f(x)$ is differentiable at $x=3$

2- ¿Why is it not continuous at $x \neq 3$?

$\displaystyle\lim_{x \to x_0,x\in \Bbb Q}{}x^2=x_0^2$

Proof :

$|x^2-x_0^2|=|x-x_0||x+x_0|$

First restriction: $|x-x_0|<1\Leftrightarrow{x_0-1<x<x_0+1}\rightarrow{|x+x_0|<2|x_0|+1}$

Hence, for $\delta=\varepsilon/(2|x_0|+1)$ this limit exists.

$\displaystyle\lim_{x \to x_0,x\in \Bbb I}{}(6(x-3)+9)=6x_0-9$

Proof:

If $0<|x-x_0|<\delta\Rightarrow{|(6x-9)-(6x_0-9)|<\varepsilon}$

$|6x-9-6x_0+9|<\varepsilon\Rightarrow{6|x-x_0|<\varepsilon}$

For $\delta=\varepsilon/6$

and $x_0^2=6x_0-9$ iff $x_0=3$

Well, this was not really homework; I left aside my textbook when I met this exercise, which captivated me. I didn't know I was facing my absolute ignorance about concepts I had already passed on september. I actually passed with a 5,6 over 10.

Yes, that’s much better.

As I wrote in post #17, I'm not entirely happy about assuming you only need to consider sequences which are either entirely of rationals or entirely of irrationals when proving differentiability at 3. Maybe you could take that on as an exercise in itself:
show that if we partition the reals into two sets, A and B, and find that
- for all sequences in A converging to x, f(x) converges to y, and
- for all sequences in B converging to x, f(x) converges to y, then
for all sequences in $\Bbb R$ converging to x, f(x) converges to y.

Wrt the epsilon-delta reasoning, the way you have written it out reflects the way you found the right value to choose for delta. As a proof to be read by someone else, it reads better if you turn it around into the form:
"Given $\epsilon>0$ choose $\delta=$ (whatever function of epsilon you have figured out)" etc.

 

[mcastillo356]

Hi, PF, that's hard work. But I know I will have to face it. Now I've turned to the textbook. I haven't become familiar enough with that area (still).
Greetings!