Question

\(f(x) =\begin{cases}h(x) ~~~if~~~ x \in \mathbb Q\\g(x) ~~~if~~~x\notin \mathbb Q\end{cases}\)
Prove that f(x) is differentiable at x = a if and only if g(a) =h(a) and g'(a) =h'(a)
Please, help

Answer 1

@xapproachesinfinity

Answer 2

My attempt:
suppose f(x) is diff. at x =a, then f'(x) is defined and \(f'(a) = lim_{x\rightarrow a}\dfrac{f(x)-f(a)}{x-a}\)

Answer 3

since f(x) is a piecewise function,
\(f'(x) =\begin{cases}h'(x) ~~~if~~x\in\mathbb Q\\g'(x) ~~~if ~~~x\notin \mathbb Q\end{cases}\)

Answer 4

Moreover, f(x) is diff. at x=a, hence f(x) is continuous at x =a, that is h(x) = h(a)

Answer 5

yeah but that what we are trying to prove here
if h'(a)=g'(a) then f must be diff at x=a

Answer 6

and g(x) = g(a) AND h(a) =g(a)

Answer 7

yes the first is the continuity case
well f diff at means that g(a)=h(a)

Answer 8

@xapproachesinfinity we go both ways, right? I am proving --> direction

Answer 9

iff and only mean both ways

Answer 10

so are you trying contradiction?

Answer 11

Nope, why? I go directly.

Answer 12

hmm you made a supposition so i thought you want to go with contradiction

Answer 13

f is diff. hence, f is conti. at x =a , that is h(x) = g(x) at a. It means h(a) =g(a)

Answer 14

i'm not sure that is a good proof hehe

Answer 15

what if we start backwards?

Answer 16

Let finish this way, please

Answer 17

we just have h(a) = g(a) , still need h'(a) =g'(a)

Answer 18

well if f is diff on a and since it is peice-wise
then h and g must be diff as well
and the only that can happen is when h'(a)=g'(a) is there are not equal then
our assumption must be wrong

Answer 19

if there are not*

Answer 20

that is way i think the proof is weak lol

Answer 21

ok, kid, I have a proposition for that. It says: f( x) is diff. at x =a iff \(h'|_Q(a) = g'|_{R\Q}(a)\)

Answer 22

hence h'(a) =g'(a)

Answer 23

Now, other way.

Answer 24

hmm is that R\Q ?

Answer 25

\(\notin \mathbb Q= \mathbb R/\mathbb Q\)

Answer 26

yep i know i'm just making sure that's what you mean
i didn't know such proposition
if that's true then your proof makes some sense

Answer 27

the other way
we prove that both h and g are differential at a an equal
and that there a continuity with both conditions we can say f is diffrentiable

Answer 28

By definition \(h'(a) = lim_{x\rightarrow a}\dfrac{h(x) -h(a) }{x-a} =lim_{x\rightarrow a}\dfrac{g(x) -g(a) }{x-a}= g'(a)\)

Answer 29

that also means there both continuous at x=a
so lim h(x)=lim (g(x))

Answer 30

they are*

Answer 31

and.... oohhooh... old man is stuck.

Answer 32

well if you have h and g diff at a
then that implies they are also continuous at a
so g(a)=h(a)

Answer 33

to be honest i feel there is messing links haha

Answer 34

hey kid, I saw @kmeis002 is typing but why it takes toooo long to pop up?? hehehe

Answer 35

i was gonna ask the same thing lol
perhaps he/she has something to say :)

Answer 36

hehehe, let me tag someone else but it is soooooo impolite when other is helping me.

Answer 37

hehehe why impolite?
tag jim

Answer 38

Let \(x = a\) where \( a \in \mathbb{Q} \) then \( f(a) = h(a) \). For \( f\) to be continous at a, for any open ball \(f(B_\delta(a)) \subset B_\epsilon (f(a)) \). Assume \(g(a) \neq h(a)\). Since the reals are dense then \(\exists b \in B_\delta(a) \quad \text{s.t.} \quad b \notin \mathbb{Q}\). This implies we can choose an open ball of some radius such that \(\exists b \: \text{s.t.} \:f(b) \notin B_\epsilon(f(a)) \quad \text{and} \quad b \in B_\delta(a) \). Therefore, its not continuous. I believe you can show the same for the case of the derivatives to show the derivatives are continuous at a only if they are equal.

Sorry, just trying to formulate, and multi-task. So it may not be correct at the moment, just throwing it out there.

Answer 39

ohoh... he is playing with balls.

Answer 40

what are the balls? no idea here lol

Answer 41

kid, you don't know this stuff, it is from advance cal

Answer 42

darn i feel bad now :)

Answer 43

Open balls are collection of numbers based on a radius r. Just an open set around a point.

Answer 44

hmm kinda heard something similar in algebra or something

Answer 45

@kmeis002
question, if \(b\in B_\delta (a)\) and \(a\in \mathbb Q\), then b can be in Q , can be not in Q also. How can you assume that b is not in Q?

Answer 46

For any open ball around a rational, there exists some element that is not rational.

Answer 47

because R is so dense

Answer 48

oh, got you \(\exists\), not all

Answer 49

but it goes nowhere, right?

Answer 50

Nowhere, how so? I'm a bit confused at this question

Answer 51

I admit, my post isn't fully complete, I'm pondering at the moment.

Answer 52

we need prove if f is diff. then f is conti. to get h(a) = g(a) . YOur balls give me f is NOT conti. hence.... I... stuck.

Answer 53

Mainly the explanation of the fact that at \(b, f(b) = g(b)\). Then we just need to show that we arrive at the contradiction (not yet complete).

My post is hinting that if we assume the opposite, the function is not continuous. The same argument (incomplete) can be constructed for \(f'(a)\), which may take care of the differentiable. In fact, that would be all you would need to show since that implies continuity at a.

Answer 54

Ok, I am waiting, please, complete the stuff.

Answer 55

I'll see what I can do, if its anything at all. I apologize if my replies are not in the most timely matter. I have to take care of something for a moment, but if rigorous explanation arises I will share it.

Answer 56

Thank you for the help. :)

Answer 57

Alright, bear with me, but I think I have a half decent explanation to follow. If its not that great, I apologize. My formal training is not in pure mathematics so my proof skills go as far as my hobby does.

Answer 58

(Im am going to post these piecemeal so you dont have to wait for one long text)
First we have to assume that \(g'(x), h'(x)\) are continuous about \(a\). And as discussed, if we can show f'(a) is continuous, then f(a) is continuous. A similar argument can be made for the \(g(a) = h(a)\) condition, if you please.

Assume the opposite is true \(g'(a) \neq h'(a)\). This means \(|g'(a) -h'(a)| = \xi >0 \). Since \(g'(a), h'(a)\) are continuous about \(a\), there must exist an interval of radius \(\delta\), \( \left ( B_\delta(a) \right ) \) such that any value in the interval \( \alpha \in B_\delta(a) \) has the property \(|g'(\alpha) -h'(\alpha)| > 0\). Basically, we can choose an interval where \(g'(x) \neq h'(x)\) for some interval by the definition of continuity.

Since the reals are dense, there must exist a \(b \in B_\delta(a) \) such that \(b \in \mathbb{R}\setminus \mathbb{Q}\)

Answer 59

If we assume \(a \in \mathbb{Q}\) then at \(a\) we get that \(f'(a) = h'(a)\) and at the \(b\) we know that \(f'(b) = g'(b) \). In fact, we know that since \(|g'(a)-h'(a)|>0\) we can also that that the arbitrary interval can be chosen to have the property \(|g'(b) - h'(a)| > 0 \) (essentially from the Intermediate Value Theorem, if g(a) and h(a) are not equal, one is greater and one is less than on some interval until they equal each other. That is if they ever do). This property will help us show that \(f'(a)\) is not continuous and therefore \(f(a)\) is not differentiable at a.

Answer 60

Finally, from the epsilon delta definition of a limit and since \(f'(b) = g'(b)\) and \( f'(a) = h'(a)\), we can show that \(|g'(b) -b'(a)| =|f'(b) -f'(a)| >0 \). Since \(b\) is some arbitrary irrational number in the interval, it can be expressed as \(b\) as any \( x \in B_\delta(a) \) that is irrational. Then \(|f'(x) -f'(a)| >\epsilon > 0\). That is, we can choose a \(b\) to make the difference around \(f'(a)\) not arbitrarily small. Infact, the smaller the interval, the closer it is to \(\xi\) (a constant value defined by \(g'(a), h'(a)\).

This shows that \(f'(x)\) is not continuous about \(a\) if \(g'(a) \neq h'(a) \). Proving your proposition you posted above. I think with some effort, it can be formalized and made more compact, but that is where my thoughts were going. I wanted to try to write them in a more intuitive way.

Answer 61

Question: how |g'(a) -h'(a)|>0?? is it not that we assume that g'(a) = h'(a)?

Answer 62

The proof is by contradiction, in a sense. We assume \(g'(a) \neq h'(a)\) and show that this produces a discontinuous \(f'(a)\), implying that \(f'(a)\) is continuous iff \(g'(a) = h'(a)\)

Answer 63

oh, yeah,

Answer 64

My explanation got hairy on the third post. I got sloppy with the limit and differed to an English explanation. I think you can show that for some \(0<|x-a|<\delta\) produces a \(|f'(x)-f(a)|=\xi>\epsilon \)

Answer 65

*deferred

Answer 66

Thank you for the exercise!

Answer 67

Much appreciate :)