11Is f"(x) continuous?
If it's so, then I have a proof.
Let f defined on [0,1] be twice differentiable such that
|f''(x)|\leq 1
for all x\in[0,1] .
If f(0) =f(1) , then show that |f'(x)|<1 for all x\in[0,1]
i think such question need mathematically rigorous analysis like the ε-δ method
so can any one explain using that
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11 Answers
39quite a good one.. solved it long back
it was a TMH question , where the answer was typed wrongly
and if my memory isnt too bad , this one has been solve din the site before.
anwaysy guys , give it a try
1soumik
i wrote na
f(x) is twice differntiable
@b555
please give the proof
11From Rolle's theorem, there exists k in [0,1] satisfying f'(k)=0
Now -1\le f"(x)\le 1. Which gives -dx\le f"(x)dx\le dx.
For the R.H.S inequality, \int_{k}^{x}{f''(x)dx}\le x-k \Rightarrow f'(x)\le x-k
Proved 1 part. Similalry for the l.h.s inequality, we integrate from x to k to arrive at our result.
Edited after kaymant sir's correction.
66@soumik,
Your statement (i) is not justified by the fact that f is symmetric about the x axis. At least not for all x in (0,1).
11H'mmm....that seems correct, I seemed to have taken it for granted.
Anyways, the proof does not change by much.
We can at least say there exists k in [0.1] such that f'(k)=0 (From Rolles theorem).
Now \int_{k}^{x}{f''(x)dx}\le x-k \Rightarrow f'(x)\le x-k.
Now it's okay, I hope.
341You can use LMVT:
\because \ f(0) = f(1) = 1, f'( \theta_1) = 0, \theta_1 \in (0,1)
Hence for any x \in (0,\theta_1)
\frac{f'(x) - f'(\theta_1)}{x-\theta_1} = f"(\theta_2) \Rightarrow |f'(x)| = |x-\theta_1| |f"(\theta_2)| \le 1
A similar analysis for x \in (\theta_1,1) completes the solution
1thanks thrprophet sir
@soumik yaar kuch samaj nahi aa raha
can u please write in a language which mass can understand
ur proof i guess only can be understood by ppl of mensa level
11Why?
See from rolle's theorem there exists k in [0,1] such that f'(k)=0.
Now f''(x)\le 1 \Rightarrow f''(x)dx\le dx
Now integrate from k to x both sides to have \int_{k}^{x}{f''(x)dx}\le x-k \Rightarrow f'(x)\le x-k.
So f'(x)<x for all x, which gives f'(x)<1.
For the part to proved f'(x)>-1, integrate from x to k, and u are on.
Understood?