Thoughts on the derivative of a function

[mcastillo356]

Homework Statement

Prove that
$f(x) = \left \{ \begin{matrix} x^2 & \mbox{if }\;x\in \mathbb Q \\ 6(x-3)+9 & \mbox{if }\;x\in \mathbb R\setminus \mathbb Q\end{matrix}\right.$
has a derivative only at $x=3$.

Relevant Equations

Analysis, algebra, number sets

(a) $f(x)$ is continuous only at $x=3$:
1- If $x\in\mathbb Q$, $f(x)=9$ at $x=3$; around, there is $\mathbb Q$
2- If $x\in \mathbb R\setminus \mathbb Q$, this is the set of irrational numbers.

Intuitively, if $x$ was in $\mathbb R$, $x^2$ and $6(x-3)+9$ would meet at $x=3$; but, around $x=3$, there are $\mathbb Q$ and $\mathbb R\setminus Q$.
(b) If $f:\mathbb R \to \mathbb R$, it's easy to prove that $f(x)$ has a derivative at 3: 6.

Greetings!

 

[haruspex]

(a) $f(x)$ is continuous only at $x=3$:
Intuitively, if $x$ was in $\mathbb R$, $x^2$ and $6(x-3)+9$ would meet at $x=3$; but, around $x=3$, there are $\mathbb Q$ and $\mathbb R\setminus Q$.
(b) If $f:\mathbb R \to \mathbb R$, it's easy to prove that $f(x)$ has a derivative at 3: 6.

Neither of those is a proof.
Start by quoting the definitions of continuity and differentiability.

 

[vela]

Can you explain why f isn't continuous for $x \ne 3$?

 

[mcastillo356]

Can you explain why f isn't continuous for $x \ne 3$?

If $x\in \mathbb Q$, the graph of $x^2$ is a sequence of points; between those points, gaps.
If $x\in \mathbb R\setminus\mathbb Q$, the same problem.

Neither of those is a proof.
Start by quoting the definitions of continuity and differentiability.

Continuity at an inner point
We say that a function $f$ is continous at an inner point $c$ of its domain if
$$\displaystyle\lim_{x \to {c}}{f(x)}=f(c)$$
For example, the domain of $f(x)=\sqrt{4-x^2}$ is the closed interval $[-2,2]$, formed by the inner points of $(-2,2)$, the -2 at left, and the 2 at the outer right.
Differentiability
The derivative of a funtion $f$ is another function $f'$ defined this way
$$f'(x)=\displaystyle\lim_{h \to{0}}{\dfrac{f(x+h)-f(x)}{h}}$$
at each point $x$ where the limit exists (it is a finite number). If $f'(x)$ exists, $f$ is differentiable at $x$.
It's a translation from spanish . I guess the domain is a single point: 3. I'm I wrong?
Greetings!

 

[PeroK]

If $x\in \mathbb Q$, the graph of $x^2$ is a sequence of points; between those points, gaps.
If $x\in \mathbb R\setminus\mathbb Q$, the same problem.

Continuity at an inner point
We say that a function $f$ is continous at an inner point $c$ of its domain if
$$\displaystyle\lim_{x \to {c}}{f(x)}=f(c)$$
For example, the domain of $f(x)=\sqrt{4-x^2}$ is the closed interval $[-2,2]$, formed by the inner points of $(-2,2)$, the -2 at left, and the 2 at the outer right.
Differentiability
The derivative of a funtion $f$ is another function $f'$ defined this way
$$f'(x)=\displaystyle\lim_{h \to{0}}{\dfrac{f(x+h)-f(x)}{h}}$$
at each point $x$ where the limit exists (it is a finite number). If $f'(x)$ exists, $f$ is differentiable at $x$.
It's a translation from spanish . I guess the domain is a single point: 3. I'm I wrong?
Greetings!

It's not a question of being wrong. This is pure mathematics where you are expected to prove things formally.

It may be "obvious" that $f$ is continuous at $x = 3$ and discontinuous elsewhere, but you are being asked to prove that formally.

Does this make sense, given your recent course material?

 

[Office_Shredder]

What is $\lim_{c\to x} f(c)$ when we restrict to $c\in \mathbb{Q}$ (note, this says nothing about where x lives!) What is $\lim_{c\to x} f(c)$ when $c\notin \mathbb{Q}$? What is required for $\lim_{c\to x}f(c)$ to exist when c can be any real number?

 

[FactChecker]

For a formal proof, start with the $\epsilon$, $\delta$ definition of continuity and show that the function is not continuous except at $x=3$. Then, do you have a theorem that a function with a derivative at a point must be continuous at that point?

 

[haruspex]

I guess the domain is a single point: 3

No, the domain of a function is the set of points for which its value is defined. You are told the domain is $\mathbb R$:

$f(x) = \left \{ \begin{matrix} x^2 & \mbox{if }\;x\in \mathbb Q \\ 6(x-3)+9 & \mbox{if }\;x\in \mathbb R\setminus \mathbb Q\end{matrix}\right.$

To show it is continuous at 3 you need to show that:
- for any ε>0 there exists δ > 0 s.t. if |x-3|<δ then |f(x)-f(3)|<ε.

You are also asked to show it is not continuous anywhere else.

 

[mcastillo356]

To show it is continuous at 3 you need to show that:
- for any ε>0 there exists δ > 0 s.t. if |x-3|<δ then |f(x)-f(3)|<ε.

$|x^2-9|=|(x+3)(x-3)|=|x+3|\cdot|x-3|$
$x\in[2,4]\Rightarrow{|x+3|\leq 7}$
$|x^2-9|\leq 7\cdot |x-3|\leq7\cdot \dfrac{\epsilon}{7}=\epsilon$
$\delta=\dfrac{\epsilon}{7}$
$\lim_{x \to{3}}{x^2}=9=f(3)$$|6(x-3)+9-9|=|6x-18|=6\cdot |x-3|$
$\delta=\dfrac{\epsilon}{6}$
$\lim_{x \to{3}}{(6(x-3)+9)}=9=f(3)$

Therefore, $f$ is continuous at $x=3$

For a formal proof, start with the $\epsilon$, $\delta$ definition of continuity and show that the function is not continuous except at $x=3$.

Sorry, any hint?

Does this make sense, given your recent course material?

 

[haruspex]

$|x^2-9|=|(x+3)(x-3)|=|x+3|\cdot|x-3|$
$x\in[2,4]\Rightarrow{|x+3|\leq 7}$
$|x^2-9|\leq 7\cdot |x-3|\leq7\cdot \dfrac{\epsilon}{7}=\epsilon$
$\delta=\dfrac{\epsilon}{7}$
$\lim_{x \to{3}}{x^2}=9=f(3)$$|6(x-3)+9-9|=|6x-18|=6\cdot |x-3|$
$\delta=\dfrac{\epsilon}{6}$
$\lim_{x \to{3}}{(6(x-3)+9)}=9=f(3)$

Therefore, $f$ is continuous at $x=3$

I'm afraid you don't show any grasp of how continuity works.
Of course $\lim_{x \to{3}}{x^2}=9=f(3)$, but you need to show $\lim_{x \to{3}}f(x)=f(3)$.
I spelt out exactly the form of proof that is needed in post #8. It is called an ε-δ proof.
The task for you is to figure out an algorithm by which to select a value of δ for a given value of ε. Then show that if x is within δ of 3 then regardless of whether x is rational or irrational, f(x) is within ε of 9.
There are no shortcuts.

 

[PeroK]

$|x^2-9|=|(x+3)(x-3)|=|x+3|\cdot|x-3|$
$x\in[2,4]\Rightarrow{|x+3|\leq 7}$
$|x^2-9|\leq 7\cdot |x-3|\leq7\cdot \dfrac{\epsilon}{7}=\epsilon$
$\delta=\dfrac{\epsilon}{7}$
$\lim_{x \to{3}}{x^2}=9=f(3)$$|6(x-3)+9-9|=|6x-18|=6\cdot |x-3|$
$\delta=\dfrac{\epsilon}{6}$
$\lim_{x \to{3}}{(6(x-3)+9)}=9=f(3)$

Therefore, $f$ is continuous at $x=3$

This looks reasonable to me, but it's not what you were asked to prove. In fact, you may even be allowed to assume what you've proved here. I.e. for this question you may be allowed to assume that all polynomials of a real variable are continuous everywhere.

In any case, whether you assume this or prove it first, the specific aspect of this question is a function that is defined as two different polynomials, one on the rationals and one on the irrationals.

To show the discontinuity, I would think of sequences of rationals and irrationals converging to an arbitrary point $a$.

 

[haruspex]

$|x^2-9|=|(x+3)(x-3)|=|x+3|\cdot|x-3|$
$x\in[2,4]\Rightarrow{|x+3|\leq 7}$
$|x^2-9|\leq 7\cdot |x-3|\leq7\cdot \dfrac{\epsilon}{7}=\epsilon$
$\delta=\dfrac{\epsilon}{7}$
$\lim_{x \to{3}}{x^2}=9=f(3)$$|6(x-3)+9-9|=|6x-18|=6\cdot |x-3|$
$\delta=\dfrac{\epsilon}{6}$
$\lim_{x \to{3}}{(6(x-3)+9)}=9=f(3)$

Therefore, $f$ is continuous at $x=3$

Ah.. reading through the above again I see what you were trying to say. What you meant was,
Given ε > 0, choose x ∈ (3-ε/7, 3+ε/7)∩(2,4).
if x rational then |f(x)-9|=|x²-9| ... etc. ... < ε.
If x irrational then |f(x)-9|=|6(x-3)| ... etc. ... < ε.

 

[mcastillo356]

Well, the truth is that trying to make sense of my erratic steps, now I guess I've taken a first restriction for $\delta$: $x \in{[2,4]}$, and then a second one: $|x-3|\leq \dfrac{\epsilon}{7}$. Now it's enough to choose $\delta\leq \dfrac{\epsilon}{7}$. This implies that for any $\epsilon>0$, we take $\delta=\mbox{min}\left(3,\dfrac{\epsilon}{7}\right)$.
And I think there is another mistake: there are two $\delta$: to ensure this piecewise (is piecewise a right word?) function is continuous at $x=3$, we must choose the shortest one.

This looks reasonable to me, but it's not what you were asked to prove. In fact, you may even be allowed to assume what you've proved here. I.e. for this question you may be allowed to assume that all polynomials of a real variable are continuous everywhere.

Let's assume it for a while.

In any case, whether you assume this or prove it first, the specific aspect of this question is a function that is defined as two different polynomials, one on the rationals and one on the irrationals.

To show the discontinuity, I would think of sequences of rationals and irrationals converging to an arbitrary point $a$.

Is it possible to prove the discontinuity with my background?(Basic knowledge of limits, continuity, and differentiation)
Thanks!

 

[Office_Shredder]

I think you should think about post #6 that I put up earlier. It let's you prove continuity/discontinuity much easier for every point (and is more educational for how to think about things like this in the future)

 

[PeroK]

Is it possible to prove the discontinuity with my background?(Basic knowledge of limits, continuity, and differentiation)
Thanks!

Yes, it's just a little awkward, that's all. Let's call the two polynomial functions p(x) and q(x).

To prove continuity at x=3, as you say, you should have two deltas covering the rational and irrational cases.

To prove the discontinuity at other points, a: if a is rational, you can consider a sequence of irrational points converging to a. And vice versa. Then use the fact that p(a)≠q(a), which means that |p(a)−q(a)|=b>0. Then use the convergence of the sequence to p(a) or q(a). You're effectively using the fact that a sequence cannot converge to two separate limits.

For differentiability at x=3, you can use the same idea of using the differentiability of each polynomial and the two deltas approach again.

That would be my strategy.

 

[WWGD]

Yet another perspective, hoping to help.
If your point $x_0$ is in $\mathbb Q$, your function takes the value $x_0^2$. Now, there will be points$x_i$ in $\mathbb R-\mathbb Q$ that are indefinitely close to $x_0$ ( Why?). In the latter, the function will take values $6x_i-9$. For any $\epsilon>0$ , there will be points $x_i$ satisfying $|x_0-x_i|< \delta$ which must also satisfy $|f(x_0)-f(x_i)| < \epsilon$, because of continuity.

 

[haruspex]

I think you should think about post #6 that I put up earlier. It let's you prove continuity/discontinuity much easier for every point (and is more educational for how to think about things like this in the future)

While that works, it might not be quite obvious that only considering sequences within $\mathbb{Q}$ and sequences entirely outside $\mathbb{Q}$ is enough. Seems like a lemma is needed to show that covers sequences that are arbitrary mixes.

 

[Office_Shredder]

While that works, it might not be quite obvious that only considering sequences within
$\mathbb{Q}$ and sequences entirely outside $\mathbb{Q}$ is enough. Seems like a lemma is needed to show that covers sequences that are arbitrary mixes.

I agree, but it's pretty straightforward and it saves having to do an epsilon delta proof for every time you see this type of problem.

 

[mcastillo356]

Yes, it's just a little awkward, that's all. Let's call the two polynomial functions p(x) and q(x).

To prove continuity at x=3, as you say, you should have two deltas covering the rational and irrational cases.

To prove the discontinuity at other points, a: if a is rational, you can consider a sequence of irrational points converging to a. And vice versa. Then use the fact that p(a)≠q(a), which means that |p(a)−q(a)|=b>0. Then use the convergence of the sequence to p(a) or q(a). You're effectively using the fact that a sequence cannot converge to two separate limits.

For differentiability at x=3, you can use the same idea of using the differentiability of each polynomial and the two deltas approach again.

That would be my strategy.

Well, let's start proving discontinuity at other points:

$\left\{x_n=\left(1+\dfrac{1}{n}\right)^n\right\}_{n=0}^{\infty}=e$

$\left\{x_n=\dfrac{e}{n}\right\}_{n=0}^{\infty}=0$

$p(e)=6(e-3)+9=6e-9$
$q(0)=0$

Well, is this a good start?

 

[mcastillo356]

No, it is not. Let's forget it

 

[PeroK]

Well, let's start proving discontinuity at other points:

$\left\{x_n=\left(1+\dfrac{1}{n}\right)^n\right\}_{n=0}^{\infty}=e$

$\left\{x_n=\dfrac{e}{n}\right\}_{n=0}^{\infty}=0$

$p(e)=6(e-3)+9=6e-9$
$q(0)=0$

Well, is this a good start?

What is that?! It's an original approach at least.

The obvious way to start is to let $a \ne 3$ and, first, let $a$ be rational, so that $f(a) = a^2$

Now, if we take a sequence of irrational points, $x_n$, that converges to $a$, then we have $f(x_n) = 6x_n - 9$. Can you make progress from there?

 

[haruspex]

Well, let's start proving discontinuity at other points:

This part is where it is easier to see how to use the approach in post #6.
For a≠3, consider a sequence of rationals converging to a and a sequence of irrationals converging to a.

 

[mcastillo356]

Hi, PF, what about if I link this thread at the spanish maths forum "Rincón Matemático"?. I've realized my lack of background, and my gaps in English. To accomplish them I would also appreciate very much your help. I know it's unusual, it's just an idea to keep on moving, the next step to give a try, from my point of view. RM is ready. What is your opinion?.
Greetings!

 

[MidgetDwarf]

As someone pointed out. Maybe an argument using sequences instead of epsilon/delta argument would be easier, and a bit more intuitive.

A good start, it acquaint yourself with the definition of continuity using epsilon/delta and the one using sequences. There is another formulation using neighborhoods, but I believe, its less intuitive than the others for beginning students.

After practicing them a bit, you could prove that they are all equivalent.My suggestion is to practice on two easier problems. Ie., showing that a function is continuous and showing that a function is discontinuous. Then return to this one. It is apparent that you lack an understanding of continuity and differentiation. So trying to work this problem out at this stage may be wasted time...

 

[mcastillo356]

Hi,PF

What is $\lim_{c\to x} f(c)$ when we restrict to $c\in \mathbb{Q}$ (note, this says nothing about where x lives!) What is $\lim_{c\to x} f(c)$ when $c\notin \mathbb{Q}$? What is required for $\lim_{c\to x}f(c)$ to exist when c can be any real number?

I've been at the spanish forum, RM: we've came to the conclusion that some background on sequences and topology is needed to face the question, and I have none of them.
My decision is to give a last chance: the quote above. Is there an easier answer to the statement of the thread, based on an undergraduate's skills and lacks?.
I've also translated and wrote down the quote, but RM leads me again to topology.
My question: can you give me some more hints on Office_Shredder's quote?.
If I should provide more information, please tell me.

Greetings!

 

[PeroK]

Hi,PF
I've been at the spanish forum, RM: we've came to the conclusion that some background on sequences and topology is needed to face the question, and I have none of them.
My decision is to give a last chance: the quote above. Is there an easier answer to the statement of the thread, based on an undergraduate's skills and lacks?.
I've also translated and wrote down the quote, but RM leads me again to topology.
My question: can you give me some more hints on Office_Shredder's quote?.
If I should provide more information, please tell me.

Greetings!

Sequences are usually the first thing to study in a course on real analysis. If you know the definition of a limit in terms of sequences, then this is fairly elementary. It's introductory pure mathematics and requires no topology.

Are you taking a course in real analysis?

 

[mcastillo356]

Sequences are usually the first thing to study in a course on real analysis. If you know the definition of a limit in terms of sequences, then this is fairly elementary. It's introductory pure mathematics and requires no topology.

Are you taking a course in real analysis?

I'm enrolled in real analysis, in the first college year of the Physics career. It's the only subject I've chosen (I had that chance, and I took this subject). At this moment, I've only taken a quick look at sequences. So I guess I need some more time.

Greetings. Thanks, PF!

 

[haruspex]

Hi,PF
I've been at the spanish forum, RM: we've came to the conclusion that some background on sequences and topology is needed to face the question, and I have none of them.
My decision is to give a last chance: the quote above. Is there an easier answer to the statement of the thread, based on an undergraduate's skills and lacks?.
I've also translated and wrote down the quote, but RM leads me again to topology.
My question: can you give me some more hints on Office_Shredder's quote?.
If I should provide more information, please tell me.

Greetings!

If you consider a sequence of rational numbers, x_i converging to x, what does the sequence f(x_i) look like? What does it converge to?
Then do the same for a sequence of irrationals converging to x.

 

[mcastillo356]

Hi, PF
It suffices to prove that 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? I've got some stuff on internet. It might be useful, but I need time to check it.

 

[PeroK]

Hi, PF
It suffices to prove that 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? I've got some stuff on internet. It might be useful, but I need time to check it.

I think you can assume that. This question doesn't expect you to prove that.

 

[mcastillo356]

If you consider a sequence of rational numbers, x_i converging to x, what does the sequence f(x_i) look like? What does it converge to?
Then do the same for a sequence of irrationals converging to x.

$f(x_i)$ is another sequence; it converges to $f(x)$
The same for irrationals.
Is it that easy?. It's already proven: the polynomial functions converge to $9$ at $x=3$

 

[PeroK]

$f(x_i)$ is another sequence; it converges to $f(x)$
The same for irrationals.
Is it that easy?. It's already proven: the polynomial functions converge to $9$ at $x=3$

If you have enrolled in a course on real analysis, you perhaps ought to take it seriously.

 

[haruspex]

$f(x_i)$ is another sequence; it converges to $f(x)$
The same for irrationals.
Is it that easy?. It's already proven: the polynomial functions converge to $9$ at $x=3$

You do not seem to have understood what I asked you to do.

If you consider a sequence of rational numbers, x_i converging to x, what does the sequence f(x_i) look like?

I.e. write out the expression for that sequence, using what f looks like for rationals. Do the same for a sequence of irrationals. Post what you get.

 

[Office_Shredder]

As a more concrete example, what is
$$\lim_{n\to \infty} f(\pi/n))$$
And what is
$$\lim_{n\to \infty} f(1/n)$$

 

[mcastillo356]

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

 

[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.