Thoughts on the derivative of a function

[PeroK]

Well, here is my output. Sorry for the excessive time taken to post, and, forgive language mistakes.

I think you need to talk to your tutor and sort out what you need to be able to do before you tackle a problem such as this.

 

[haruspex]

Well, here is my output. Sorry for the excessive time taken to post, and, forgive language mistakes.

Any algebra which may have been in that image seems to have evaporated. That makes your thoughts hard to interpret.

There is not really any such thing as a "piecewise function". You can have a piecewise definition of a function, but that is a property of the way you choose to define it, not of the function.
Perhaps you mean piecewise continuous. Such a function is one which can be described in a piecewise manner such that each sub function is continuous in its own interval.
Either way, the pieces need to be intervals, not isolated points.

Please try to do what I asked you to do in post #33. The answers must be in the form of algebraic expressions, not just verbiage.

 

[PeroK]

Here's an example of some real analysis. Suppose we have:
$$f(x) =\bigg\{ \begin{matrix} g(x) & x < a \\ h(x) & x \ge a \end{matrix}$$ Where $g$ and $h$ are continuous functions on $\mathbb R$ with $g(a) = h(a)$. We want to show that $f$ is continuous at $a$.

Proof: Let $\epsilon > 0$. As $g$ is continuous at $a$, there exists $\delta_1$ such that: $$|x - a| < \delta_1 \ \Rightarrow \ |g(x) - g(a)| < \epsilon$$ And, as $h$ is continuous at $a$, there exists $\delta_2 > 0$ such that $$|x - a| < \delta_2 \ \Rightarrow \ |h(x) - h(a)| < \epsilon$$ Letting $\delta = min\{\delta_1, \delta_2 \}$ we have for $x \ge a$:
$$|x - a| < \delta \ \Rightarrow \ |f(x) - f(a)| = |h(x) - h(a)| < \epsilon$$ And, for $x < a$:$$|x - a| < \delta \ \Rightarrow \ |f(x) - f(a)| = |g(x) - g(a)| < \epsilon$$ And, it follows that $f$ is continuous at $a$.

Now, suppose that $g(a) \ne h(a)$. We show that $f$ is not continuous at $a$.

Proof: let $\epsilon = |g(a) - h(a)|/2$. As $g$ is continuous at $a$ we can find $\delta_0$ such that:
$$|x - a| < \delta_0 \ \Rightarrow \ |g(x) - g(a)| < |g(a) - h(a)|/2$$ Now, by the triangle inequality we have for $a -\delta_0 < x < a$:
$$|f(x) - f(a)| = |g(x) - h(a)| \ge |g(a) - h(a)| - |g(x) - g(a)| > |g(a) - h(a)| - |g(a) - h(a)|/2$$ Hence:
$$|f(x) - f(a)| > |g(a) - h(a)|/2 = \epsilon$$
In summary, we have found $\epsilon > 0$ and $\delta_0 > 0$ such that:$$a -\delta_0 < x < a \ \Rightarrow \ |f(x) - f(a)| > \epsilon$$ And that is enough to show that $f$ is not continuous at $a$, because:

For any $\delta > 0$ we can find $x$ with $|x- a| < \delta$ and $|f(x) - f(a)| > \epsilon$.

And you can see that real analysis is hard work. It's taken a lot of effort to prove something that is rather obvious. The idea is that you use the same techniques to prove things that are not so obvious.

In any case, this is the sort of work I would expect you to produce for your question.

 

[MidgetDwarf]

Teines que aprender la definición de Continuity y Differentiation. Cuando digo que aprendas la definición, arriéndela usando epsilon/delta y también la forma usando sequences.

Estas asciendo las cosas muy complicadas. Cuando aprendas la definición, usa las (las dos formas) para probar problemas fáciles de matemáticas. Cuando hagas esto por un tiempo, attempta el problema que escribiste aquí.

Tables mira el libro Abbot: Understanding Analysis. También puedes mirar Bartle: Elements Of Real Analysis. No mi recuerdo si Abbot tiene la definición de Continuity usando sequences. Si Abbot no lo teine, mira la definición de Bartle usando sequences. También ley cuando dice como enseñar que una función no es continuous.


Sorry for writing in Spanish. It is clear that English is not the OP native language. I suggested that he first learn the definition of continuity (both epsilon/delta and sequence), practice on more simple problems, then attempt the problem he posted. Since it is a waste of his time attempting this problem when he does not know how to do a basic epsilon/delta proof. I also recommend that he read Abbot (a gentle and well motivated book), and maybe look at Bartle for sequence characterization of continuity (I do not remember if Abbot contains it).

At this point. It seems he is being hardheaded...

 

[mcastillo356]

Another approach. First of all, my tutor has given me a non worthposting answer. So I'll keep on trying: a theorem: $$\displaystyle{ \lim_{x\to c}g(x)=L\iff \lim_{n\to\infty}g(x_n)=L\text{ for all sequence }\{x_n\}_{n\in \mathbb N}\text{ so that }\lim_{n\to\infty }x_n=c }$$

Now,

Hi, PF
(...) every irrational number in $\mathbb R$ is the limit of a sequence of rational numbers, and every rational number in $\mathbb R$ is the limit of a sequence of irrational numbers (...)

Be $\{a_n\}_{n=1}^{+\infty}$ a sequence s.t.
1.) $a_n$ is irrational for all natural.
2.) $a_n \neq 3$ for all natural
3.) $\displaystyle \lim_{n \to +\infty} a_n = 3$.

Be $\{b_n\}_{n=1}^{+\infty}$ a sequence s.t.
1.) $b_n$ is rational for all natural.
2.) $b_n \neq 3$ for all natural
3.) $\displaystyle \lim_{n \to +\infty} b_n = 3$.


$\displaystyle \lim_{n \to +\infty} \dfrac{f(a_n) - f(3)}{a_n-3} =\lim_{n \to +\infty} \dfrac{6\cdot(a_n-3) + 9 - 9}{a_n-3} = 6$

$\displaystyle \lim_{n \to +\infty} \dfrac{f(b_n)-f(3)}{b_n-3} = \lim_{n \to +\infty} \dfrac{b_n^2 -9}{b_n-3} = \lim_{n \to +\infty} \dfrac{(b_n -3) \cdot (b_n +3)}{b_n-3} = \lim_{n \to +\infty} b_n + 3 = 6$

Comments:
This theorem has got a proof in terms of topological issues
If I'm right, I've proven that $f(x)$ is differentiable if and only if $x=3$

Greetings!

 

[PeroK]

Another approach. First of all, my tutor has given me a non worthposting answer. So I'll keep on trying: a theorem: $$\displaystyle{ \lim_{x\to c}g(x)=L\iff \lim_{n\to\infty}g(x_n)=L\text{ for all sequence }\{x_n\}_{n\in \mathbb N}\text{ so that }\lim_{n\to\infty }x_n=c }$$

Now,


Be $\{a_n\}_{n=1}^{+\infty}$ a sequence s.t.
1.) $a_n$ is irrational for all natural.
2.) $a_n \neq 3$ for all natural
3.) $\displaystyle \lim_{n \to +\infty} a_n = 3$.

Be $\{b_n\}_{n=1}^{+\infty}$ a sequence s.t.
1.) $b_n$ is rational for all natural.
2.) $b_n \neq 3$ for all natural
3.) $\displaystyle \lim_{n \to +\infty} b_n = 3$.


$\displaystyle \lim_{n \to +\infty} \dfrac{f(a_n) - f(3)}{a_n-3} =\lim_{n \to +\infty} \dfrac{6\cdot(a_n-3) + 9 - 9}{a_n-3} = 6$

$\displaystyle \lim_{n \to +\infty} \dfrac{f(b_n)-f(3)}{b_n-3} = \lim_{n \to +\infty} \dfrac{b_n^2 -9}{b_n-3} = \lim_{n \to +\infty} \dfrac{(b_n -3) \cdot (b_n +3)}{b_n-3} = \lim_{n \to +\infty} b_n + 3 = 6$

Comments:
This theorem has got a proof in terms of topological issues
If I'm right, I've proven that $f(x)$ is differentiable if and only if $x=3$

Greetings!

This is not a proof either in content or format. It's a useful start.

The definition of continuity says for all sequences. What about all the sequences that are a mixture of rational and irrational numbers?

 

[Office_Shredder]

I don't think the goal is to prove that theorem, the goal is to use the theorem to solve the problem. If that's the case,I think you correctly have shown f is continuous at 3 in your post. The other parts can be done similarly.

 

[PeroK]

I don't think the goal is to prove that theorem, the goal is to use the theorem to solve the problem. If that's the case,I think you correctly have shown f is continuous at 3 in your post. The other parts can be done similarly.

Are you sure that constitutes a valid proof that $f$ is differentiable at $x = 3$? Note that it's not showing that $f$ is continuous.

 

[mcastillo356]

I think there is something missing... I must think it over.
I will take some time.

 

[bhobba]

I think you need to talk to your tutor and sort out what you need to be able to do before you tackle a problem such as this.

Yes. My initial reaction is that you use the theorem that iff a function is continuous at a point, every subsequence must converge to the point.

Thanks
Bill

 

[mcastillo356]

I think you need to talk to your tutor and sort out what you need to be able to do before you tackle a problem such as this.

Well, my decission is to write a last post, in a week. If I still show no progress, let me know. The alternative would be to return to the textbook; leave the exercise.

 

[mcastillo356]

Hi, PF

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

(1)- Why the function has derivative at $x=3$?

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

This derivate turns into two:

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

(b)-$\displaystyle\lim_{x \in \mathbb I, x \to{3}}{\dfrac{(6(x-3)+9)-9}{x-3}}=6$

Proof for (a)

For $x\in \mathbb Q\;\forall{\epsilon_1>0}\;\exists\;{\delta_1}$ such that if $0<|x-3|<\delta_1\Rightarrow{|g(x)-9|=|x^2-9|<\epsilon_1}$:

$\delta_1=\mbox{mín}\left ({1,\dfrac{\epsilon_1}{3}}\right )$

Proof of (b)

For $x\in \mathbb I\;\forall{\epsilon_2>0}\;\exists\;{\delta_2}$ such that if $0<|x-3|<\delta_2\Rightarrow{|h(x)-9|=|(6(x-3)+9)-9|<\epsilon_2}$:

$\delta_2=\dfrac{\epsilon_2}{6}$

$\delta=\mbox{mín}\left ({1,\dfrac{\epsilon_1}{3},\dfrac{\epsilon_2}{6}}\right )$


If we separate the domain of some function into two subsets, and approaching to a point $a$ by each one, they reach the same value, the function approaches to this value as it comes near $a$ by all the domain.

Proof:

$f:\mathbb R\rightarrow{\mathbb R}$

$h:\mathbb Q\rightarrow{\mathbb R}$

$g:\mathbb R\setminus{\mathbb Q}\rightarrow{\mathbb R}$

(2)- Why is not differentiable if $x\neq 3$?

Now, suppose that $g(a)\neq h(a)$ (this happens iff $x\neq 3$). We show that $f$ is not continuous at $a$.

Proof

Let $\epsilon=|g(a)-h(a)|/2$. As $g$ is continuous at $a$ we can find $\delta_{\mathbb Q}$ such that

$|x-a|<\delta_{\mathbb Q}\Rightarrow{|g(x)-g(a)|<|g(a)-h(a)|/2}$

Now, by the triangle inequality we have for $|x-3|<\delta_{\mathbb Q}$

$|f(x)-f(a)|=|g(x)-h(a)|\geq |g(a)-h(a)|-|g(x)-g(a)|>|g(a)-h(a)|-|g(a)-h(a)|/2$

Hence

$|f(x)-f(a)|>|g(a)-h(a)|/2=\epsilon$

In summary, we have found $\epsilon>0$ and $\delta_{\mathbb Q}>0$ such that

$|x-a|<\delta_{\mathbb Q}\Rightarrow{|f(x)-f(a)|>\epsilon}$

And that is enough to show that $f$ is not continuous at $a$ (if $a\neq 3$), because for any $\delta_{\mathbb Q}>0$ we can find $\epsilon>0$ such that $|x-a|<\delta_{\mathbb Q}$ and $|f(x)-f(a)|>\epsilon$

Proven it is not continous, it is not differentiable at that points

 

[haruspex]

If we separate the domain of some function into two subsets, and approaching to a point a by each one, they reach the same value, the function approaches to this value as it comes near a by all the domain.

Proof:

What proof?

Now, suppose that g(a)≠h(a)

There is no point a at which g and h are both defined.

 

[mcastillo356]

It's clear I must study further. I thoght I had clear the concepts of function, continuity, derivative, proof... I must return to the textbook studied and the subject I passed a year ago. I feel very thankful for your support and advice at this thread. Thank you, PeroK, haruspex, ...Loving support. Loving indeed.

 

[mcastillo356]

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.

 

[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!