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question with derivative
- Thread starter orir
- Start date
I don't understand the question - it would help a lot if you would show what you have done (or tried to do)!f is a differntiable function at [0,1] which maintains \(\displaystyle 0<=f'(x)<=1 \) to every \(\displaystyle x\in[0,1] \).
i need to prove that there's a point \(\displaystyle x\in[0,1] \) so
\(\displaystyle f'(x)=3x/(sqrt(3x^2+6)) \)
Is the problem stating that f'(0) = 0 and f'(1) = 1 ??
And then if the derivative is continuous it must somewhere have every value in [0,1] ??
And then in particular there is someplace it takes the value 3x/sqrt(3x^2 + 6) ??
OR is the actual question different from that - I would wonder if you are supposed to find the antiderivative of f'(x) ??
I don't understand the question - it would help a lot if you would show what you have done (or tried to do)!
Is the problem stating that f'(0) = 0 and f'(1) = 1 ??
And then if the derivative is continuous it must somewhere have every value in [0,1] ??
And then in particular there is someplace it takes the value 3x/sqrt(3x^2 + 6) ??
OR is the actual question different from that - I would wonder if you are supposed to find the antiderivative of f'(x) ??
no, i'm not supposed to find the antiderivative because we haven't learnt Integrals yet.
and, the problem states that f'(x) is also in [0,1] for every \(\displaystyle x\in[0,1]\).
Let \(\displaystyle g(x) = \dfrac {3x}{\sqrt{3x^2 + 6}}\)no, i'm not supposed to find the antiderivative because we haven't learnt Integrals yet.
and, the problem states that f'(x) is also in [0,1] for every \(\displaystyle x\in[0,1]\).
Evaluating at 0 and 1 gives \(\displaystyle g(0) = 0, \ \ \ g(1) = 1 \)
Use g'(x) to show that g(x) in monotonic increasing on [0,1].
Thus for any specified value of G in [0,1], you can solve for for a unique x, such that g(x) = G
The fact that g(x) is the derivative of some function f(x) does not enter in this "proof."
Let \(\displaystyle g(x) = \dfrac {3x}{\sqrt{3x^2 + 6}}\)
Evaluating at 0 and 1 gives \(\displaystyle g(0) = 0, \ \ \ g(1) = 1 \)
Use g'(x) to show that g(x) in monotonic increasing on [0,1].
Thus for any specified value of G in [0,1], you can solve for for a unique x, such that g(x) = G
The fact that g(x) is the derivative of some function f(x) does not enter in this "proof."
i didn't get your way...
D
Deleted member 4993
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i didn't get your way...
Did you calculate g'(x) ?
Did you show that g'(x) > 0 for 0≤ x ≤ 1?
Please share your work.....
Actually, I think that Dr. Phil's answer is a little spare in that it does not mention Darboux's Theorem. When I read his answer, I grasped his logic, but one step in that logic seemed to me plausible but unproved. (Such doubt was foolish of me, I know.) I had to do a few minutes research on wikipedia to learn about Darboux's Theorem. Given that theorem, however, it became obvious that his answer is perfectly rigorous. Of course, it may very well be that the actual problem said that f(x) was continuously differentiable, in which case Darboux's Theorem is irrelevant.
My approach to math tends to be "practical" rather than rigorous! Thanks to JeffM for finding justification for my proposed method.:wink:Actually, I think that Dr. Phil's answer is a little spare in that it does not mention Darboux's Theorem. When I read his answer, I grasped his logic, but one step in that logic seemed to me plausible but unproved. (Such doubt was foolish of me, I know.) I had to do a few minutes research on wikipedia to learn about Darboux's Theorem. Given that theorem, however, it became obvious that his answer is perfectly rigorous. Of course, it may very well be that the actual problem said that f(x) was continuously differentiable, in which case Darboux's Theorem is irrelevant.
The question still remains, "What was the actual question?" I have made an interpretation of what Orir asked, but have I gotten it right?
Orir, can you expand on the question please - especially by telling us what you have tried, or what you think the approach should be? If you have translated to English, perhaps we could see the original language as well?
The question makes sense and differentiability is not needed. I will try to generalize the problem: given continuous functions \(\displaystyle g(x),h(x)\) on \(\displaystyle [a,b]\) with \(\displaystyle g([a,b])\subseteq h([a,b])=[c,d]\), show there is a point \(\displaystyle t\in[a,b]\) such that \(\displaystyle g(t)=h(t)\).
The proof is the intermediate value theorem. Since \(\displaystyle h(x)\) is continuous and \(\displaystyle g(x)\) has image, say \(\displaystyle [p,q]\) lying inside \(\displaystyle [c,d]\), the functions \(\displaystyle h(x),g(x)\) must cross at some point \(\displaystyle t\). Formally, one would let \(\displaystyle D(x) = h(x)-g(x)\). It is not hard to show \(\displaystyle D(x)\neq 0\) for all \(\displaystyle x\in[a,b]\) results in a contradiction.
The proof is the intermediate value theorem. Since \(\displaystyle h(x)\) is continuous and \(\displaystyle g(x)\) has image, say \(\displaystyle [p,q]\) lying inside \(\displaystyle [c,d]\), the functions \(\displaystyle h(x),g(x)\) must cross at some point \(\displaystyle t\). Formally, one would let \(\displaystyle D(x) = h(x)-g(x)\). It is not hard to show \(\displaystyle D(x)\neq 0\) for all \(\displaystyle x\in[a,b]\) results in a contradiction.
Daon and Dr. PhilThe question makes sense and differentiability is not needed. I will try to generalize the problem: given continuous functions \(\displaystyle g(x),h(x)\) on \(\displaystyle [a,b]\) with \(\displaystyle g([a,b])\subseteq h([a,b])=[c,d]\), show there is a point \(\displaystyle t\in[a,b]\) such that \(\displaystyle g(t)=h(t)\).
The proof is the intermediate value theorem. Since \(\displaystyle h(x)\) is continuous and \(\displaystyle g(x)\) has image, say \(\displaystyle [p,q]\) lying inside \(\displaystyle [c,d]\), the functions \(\displaystyle h(x),g(x)\) must cross at some point \(\displaystyle t\). Formally, one would let \(\displaystyle D(x) = h(x)-g(x)\). It is not hard to show \(\displaystyle D(x)\neq 0\) for all \(\displaystyle x\in[a,b]\) results in a contradiction.
As the problem is given, f'(x) need not be continuous. I now see that my first post was also too spare. I did not explain that the part of Dr Phil's answer that seemed plausible but unproved to me was that the intermediate value theorem applied to f'(x).
My immediate reaction was that there must be a theorem that says f(x) is differentiable entails that f'(x) is continuous, which in turn entails that the intermediate value theorem applies. (This is analysis, which I never studied because my college did not require it for history majors.) According to wikipedia, however, it is false that f(x) is differentiable implies that f'(x) is continuous. Hooray, I just learned something not taught to me in my rather lop-sided education. Immediately, Dr. Phil's answer looked flawed, but almost the next sentence in wiki said that, according to Darboux's Theorem, f(x) is differentiable entails that the intermediate value theorem applies to f'(x). Hooray, I just learned something else new. Good day.
To summarize, Dr. Phil's answer is rigorous. daon is correct that the intermediate value theorem is required in the proof. And I am correct that Darboux's theorem is also required to show that the intermediate value theorem applies to f'(x) because it is not necessarily true that f'(x) is continuous. This is a win-win-win. Of course we still cannot be certain what the actual problem was. It may be that Darboux's Theorem is not necessary for that problem.