prove that there exists \xi

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Peter

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Peter

#1 h Nov 1, 2005, 12:43 AM • 3 Y

Y by Mathuzb, Adventure10, Mango247

The function $f: \mathbb{R}\rightarrow\mathbb{R}$ is twice differentiable and satisfies $f(0)=2,f'(0)=-2,f(1)=1$ .
Prove that there is a $\xi \in ]0,1[$ for which we have $f(\xi)\cdot f'(\xi)+f''(\xi)=0$ .

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wolfman

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wolfman

#2 h Apr 6, 2006, 2:38 AM • 2 Y

Y by Adventure10, Mango247

I know this is really old, but is this the exact statement of the problem?
We don't assume the continuity of f''?
Also, is it possible that there's a typo and f' got switched with f''?

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10000th User

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10000th User

#3 h Apr 6, 2006, 2:44 AM • 2 Y

Y by Adventure10, Mango247

Uh... did you read the question carefully? It says f is twice differentiable...

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wolfman

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wolfman

#4 h Apr 6, 2006, 2:50 AM • 2 Y

Y by Adventure10, Mango247

as I understand it, that means that the first and second derivatives exist.
So f' must be continuous, but f'' may not be.

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10000th User

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10000th User

#5 h Apr 6, 2006, 2:54 AM • 2 Y

Y by Adventure10, Mango247

That's why I said if you read and thought carefully because differentiability DOES imply continuity. Do think you know the difference between derivative and differentiability?

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wolfman

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wolfman

#6 h Apr 6, 2006, 3:05 AM • 2 Y

Y by Adventure10, Mango247

Every definition I can find says that differentiable means the derivative EXISTS, not that it must be continuous. So,
f' exists, and f'' exists.
f' is continuous, since it's derivative exists.
Why must f'' be continuous?
Unless you're working under a different definition of "differentiable".

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wolfman

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wolfman

#7 h Apr 6, 2006, 3:07 AM • 2 Y

Y by Adventure10, Mango247

And yes, I know that if a function is differentiable, it's continuous.
And of course I know the difference between the derivative of a function and differentiability.

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kokosai

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kokosai

#8 h Apr 6, 2006, 3:12 AM • 3 Y

Y by Adventure10, Moussore, Mango247

10000th User, there's no need to be unnecessarily pejorative; let's just focus on the mathematics. I think the question is that we don't know that the second derivative is differentiable, so how do we know that it is continuous? The function and its derivative are differentiable and therefore continuous, but does that imply that the function's second derivative is continuous?

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10000th User

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10000th User

#9 h Apr 6, 2006, 3:33 AM • 2 Y

Y by Adventure10, Mango247

My questions are legitimate and certainly in no way being derogatory, and I do not understand your use of the word 'pejorative' but I'll let it slide.

The way I see this (this is hard problem BTW), you do not need the continuity of $f''$ . Yes I agree what you are saying, that the continuity of $f''$ is ambiguous.

[EDIT] I just talked to my cousin and he said that $f''$ is continuous up to discontinuities of second kind.... I'm sorry I don't know what discontinuities of second kind means. Perhaps I should learn more from him...

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wolfman

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wolfman

#10 h Apr 6, 2006, 5:41 AM • 2 Y

Y by Adventure10, Mango247

actually, now I think I have a counterexample.
Forgive my current ignorance of Latex.

try

f(x) = 1/2*x^2 - 2x + 2 for 0 < x < 1/2

5/2*x^2 - 4x + 5/2 for 1/2 < x < 1

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wolfman

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wolfman

#11 h Apr 6, 2006, 5:47 AM • 2 Y

Y by Adventure10, Mango247

darn, I'm wrong.

f''(1/2) would not exist.

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pou

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pou

#12 h Apr 6, 2006, 9:24 AM • 2 Y

Y by Adventure10, Mango247

Peter VDD wrote:

The function $f: \mathbb{R}\rightarrow\mathbb{R}$ is twice differentiable and satisfies $f(0)=2,f'(0)=-2,f(1)=1$ .
Prove that there is a $\xi \in ]0,1[$ for which we have $f(\xi)\cdot f'(\xi)+f''(\xi)=0$ .

Did you gave enough information? $f'(1) = -1/2$ or not ?

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Peter

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Peter

#13 h Apr 6, 2006, 12:12 PM • 2 Y

Y by Adventure10, Mango247

The problem is correct as it stands.

The midvalue theorem works for any derivative, even if not continuous, if that's what you're asking. Not knowing this costed me an alike problem last year too, so it does come back from time to time.

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spix

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spix

#14 h Apr 6, 2006, 1:39 PM • 7 Y

Y by Mathuzb, Davrbek, XbenX, S117, Adventure10, Mango247, Cognoscenti

The problem is correct.
Take $g(x)=\frac{1}{2}f^2(x)+f'(x)$ ...clearly $g(0)=0$ .
Now take the function $h(x)=\frac{x}{2}-\frac{1}{f(x)}$ notice that $h(1)=h(0)=-\frac{1}{2}$ so by Rolle`s theorem there exists $c\in (0,1)$ such that $h'(c)=0\longrightarrow f^2(c)+2f'(c)=0\longrightarrow g(c)=0$ so now apply Rolle on $(0,c)$ for $g$ .

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hpe

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hpe

#15 h Apr 6, 2006, 1:49 PM • 2 Y

Y by Adventure10, Mango247

Peter VDD wrote:

The function $f: \mathbb{R}\rightarrow\mathbb{R}$ is twice differentiable and satisfies $f(0)=2,f'(0)=-2,f(1)=1$ .
Prove that there is a $\xi \in ]0,1[$ for which we have $f(\xi)\cdot f'(\xi)+f''(\xi)=0$ .

Let $g(x) = f'(x) + \frac{1}{2}f^2(x)$ . Then $g(0) = 0$ by assumption, and the challenge is to show that $g'(\xi) = f''(\xi) + f(\xi)f'(\xi) = 0$ for some $\xi \in ]0,1[$ . Clearly this will follow if we can show that $g(c) = 0$ for some $c \in ]0,1[$ , by Rolle's theorem. Note that $g$ is defined on $[0,1]$ and is differentiable there.

Let's also set $h(x) = \frac{2}{x+1}$ . Then $h(0) = 2, \, h(1) = 1, \, h'(x) + \frac{1}{2}h^2(x) = 0$ .

Suppose now $g(x) \ne 0$ for all $x \in ]0,1[$ . This means that either $g(x)> 0$ or $g(x) < 0$ for all $x\in ]0,1[$ .

Suppose $g(x) > 0$ for all $x \in ]0,1[$ , i.e. $f'(x) + \frac{1}{2}f^2(x) > 0$ on $]0,1[$ . Then $f(x)>h(x)$ on some maximal interval $]0,d[$ , since $f(1) = h(1)$ , and $f(d) = h(d), \, f'(d)\le h'(d)$ . But then $-\frac{1}{2}f^2(d) < f'(d) \le h'(d) = -\frac{1}{2}h^2(d) = -\frac{1}{2}f^2(d)$ , which is impossible.

Hence $g(x) > 0$ on $]0,1[$ is impossible. For the same reason, $g(x)<0$ on $]0,1[$ is also impossible.

Hence there exists $c \in ]0,1[$ such that $g(c) = 0$ and $\xi \in ]0,c[$ such that $g'(\xi) = f(\xi) f'(\xi) + f''(\xi) = 0$ .

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spix

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spix

#16 h Apr 6, 2006, 2:54 PM • 2 Y

Y by Adventure10, Mango247

10000th User wrote:

[EDIT] I just talked to my cousin and he said that $f''$ is continuous up to discontinuities of second kind.... I'm sorry I don't know what discontinuities of second kind means. Perhaps I should learn more from him...

Since $f''$ is the derivative of a function it has Darboux property so it`s discontinuities are of the second kind. Discontinuities of the second kind are those that are not of the first kind aka jump discontinuities.

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hpe

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hpe

#17 h Apr 6, 2006, 2:57 PM • 2 Y

Y by Adventure10, Mango247

spix wrote:

The problem is correct ...

Nice and elegant

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sanyalarnab

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sanyalarnab

#18 h Mar 20, 2024, 1:57 PM

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Also MTRP Subjective P2.4(Seniors)

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chakrabortyahan

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#19 h Mar 20, 2024, 2:08 PM • 2 Y

Y by sanyalarnab, mqoi_KOLA

LOL...but I shall post a solution. I LOVED THE PROBLEM....
Take the function $g(x) = \frac{1}{f(x)}-x/2$ If $f$ has no zero in $[0,1]$ then $g$ is cont and diff in the domain and so $\exists c\in(0,1)$ so that $g(c) = 0$ as $g(0) = g(1)$ Now take $h(x) = f'(x) +f(x)^2/2$ Note that $h(c) = 0$ and also $h(0) = 0$ so apply rolles's theorem to get the desired result. Now say $S$ be the set of zeroes of $f$ in the domain $(0,1)$ as $S$ is bounded so it has an infimum $a$ and a supremum $b$ and as $f$ is cont so $f(a) = f(b) = 0$ Now note that as $f(0) = 2$ so $f'(a)$ needs to be $\le 0$ and similarly $f'(b)\ge 0$ so as $h$ is continuous by IVP $h(e) = 0$ for some $e\in[a,b]$ now again apply Rolle's theorem on $(0,e)$

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mqoi_KOLA

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mqoi_KOLA

#20 h Mar 24, 2025, 12:27 PM

Y by

chakrabortyahan wrote:

LOL...but I shall post a solution. I LOVED THE PROBLEM....
Take the function $g(x) = \frac{1}{f(x)}-x/2$ If $f$ has no zero in $[0,1]$ then $g$ is cont and diff in the domain and so $\exists c\in(0,1)$ so that $g(c) = 0$ as $g(0) = g(1)$ Now take $h(x) = f'(x) +f(x)^2/2$ Note that $h(c) = 0$ and also $h(0) = 0$ so apply rolles's theorem to get the desired result. Now say $S$ be the set of zeroes of $f$ in the domain $(0,1)$ as $S$ is bounded so it has an infimum $a$ and a supremum $b$ and as $f$ is cont so $f(a) = f(b) = 0$ Now note that as $f(0) = 2$ so $f'(a)$ needs to be $\le 0$ and similarly $f'(b)\ge 0$ so as $h$ is continuous by IVP $h(e) = 0$ for some $e\in[a,b]$ now again apply Rolle's theorem on $(0,e)$

what was the motivation behind taking $g(x) = \frac{1}{f(x)}-x/2$ ? did you take gx as that to make g(1)=g(0)?

This post has been edited 1 time. Last edited by mqoi_KOLA, Mar 24, 2025, 12:55 PM

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mqoi_KOLA

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mqoi_KOLA

#21 h Mar 24, 2025, 1:19 PM

Y by

chakrabortyahan wrote:

LOL...but I shall post a solution. I LOVED THE PROBLEM....
Take the function $g(x) = \frac{1}{f(x)}-x/2$ If $f$ has no zero in $[0,1]$ then $g$ is cont and diff in the domain and so $\exists c\in(0,1)$ so that $g(c) = 0$ as $g(0) = g(1)$ Now take $h(x) = f'(x) +f(x)^2/2$ Note that $h(c) = 0$ and also $h(0) = 0$ so apply rolles's theorem to get the desired result. Now say $S$ be the set of zeroes of $f$ in the domain $(0,1)$ as $S$ is bounded so it has an infimum $a$ and a supremum $b$ and as $f$ is cont so $f(a) = f(b) = 0$ Now note that as $f(0) = 2$ so $f'(a)$ needs to be $\le 0$ and similarly $f'(b)\ge 0$ so as $h$ is continuous by IVP $h(e) = 0$ for some $e\in[a,b]$ now again apply Rolle's theorem on $(0,e)$

why should f(e)=0? how do u guarantee that?

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Alphaamss

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Alphaamss

#22 h Mar 25, 2025, 4:31 AM • 1 Y

Y by mqoi_KOLA

mqoi_KOLA wrote:

chakrabortyahan wrote:

LOL...but I shall post a solution. I LOVED THE PROBLEM....
Take the function $g(x) = \frac{1}{f(x)}-x/2$ If $f$ has no zero in $[0,1]$ then $g$ is cont and diff in the domain and so $\exists c\in(0,1)$ so that $g(c) = 0$ as $g(0) = g(1)$ Now take $h(x) = f'(x) +f(x)^2/2$ Note that $h(c) = 0$ and also $h(0) = 0$ so apply rolles's theorem to get the desired result. Now say $S$ be the set of zeroes of $f$ in the domain $(0,1)$ as $S$ is bounded so it has an infimum $a$ and a supremum $b$ and as $f$ is cont so $f(a) = f(b) = 0$ Now note that as $f(0) = 2$ so $f'(a)$ needs to be $\le 0$ and similarly $f'(b)\ge 0$ so as $h$ is continuous by IVP $h(e) = 0$ for some $e\in[a,b]$ now again apply Rolle's theorem on $(0,e)$

why should f(e)=0? how do u guarantee that?

Note that $h(a)=f'(a)+(f(a))^2/2=f'(a)\leq0$ and $h(b)=f'(b)+(f(b))^2/2=f'(b)\geq0$ , then by IVP we will get $h(e)=0$ for some $e\in(a,b)$ since $h$ is continuous.
(When $a=b$ , take $e=a=b$ , then $0\leq h(b)=h(e)=h(a)\leq0$ , i,e,. $h(e)=0$ .)

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