differentiable function /bounded sequence

**serious331** · May 1st 2010, 03:46 PM

Suppose that the function $f:R \rightarrow R$ is differentiable and that there is a bounded sequence $\{x_n\}$ with ${x_n} \neq {x_m}$ if ${n \neq m}$ , such that $f(x_n) =0$ for every index n. Show that there is a point $x_0$ at which $f(x_0) = 0$ and $f'(x_0) = 0$ .

---> According to Bolzano–Weierstrass theorem, a bounded sequence has a convergent subsequence, but How can this be used to prove the above question.

**southprkfan1** · May 1st 2010, 04:50 PM

Originally Posted by serious331

Suppose that the function $f:R \rightarrow R$ is differentiable and that there is a bounded sequence $\{x_n\}$ with ${x_n} \neq {x_m}$ if ${n \neq m}$ , such that $f(x_n) =0$ for every index n. Show that there is a point $x_0$ at which $f(x_0) = 0$ and $f'(x_0) = 0$ .

---> According to Bolzano–Weierstrass theorem, a bounded sequence has a convergent subsequence, but How can this be used to prove the above question.

You are on the right track:

Let { $y_n$ } be the convergent subsequence and suppose it converges to y. Then I claim 2 things:

1. f(y) = 0
2. f'(y) = 0

1. is easy and follows from continuity of f.

For 2, we use first principles:

$f'(y) = \lim_{x\to y} \frac {f(x) - f(y)}{x-y} = \lim_{x\to y} \frac {f(x)}{x-y}$

But we know a) The limit exists (by assumption) and b) there's a sequence { $y_n$ } that converges to y and that f( $y_n$ ) = 0 for every point...

**serious331** · May 1st 2010, 05:02 PM

Originally Posted by southprkfan1

You are on the right track:

Let { $y_n$ } be the convergent subsequence and suppose it converges to y. Then I claim 2 things:

1. f(y) = 0
2. f'(y) = 0

1. is easy and follows from continuity of f.

For 2, we use first principles:

$f'(y) = \lim_{x\to y} \frac {f(x) - f(y)}{x-y} = \lim_{x\to y} \frac {f(x)}{x-y}$

But we know a) The limit exists (by assumption) and b) there's a sequence { $y_n$ } that converges to y and that f( $y_n$ ) = 0 for every point...

Thank you so much southparkfan..

So all I had to do was to show that for a point at which the subsequence converges, $f(y)=0$ and $f'(y) = 0$ right?

**southprkfan1** · May 1st 2010, 05:06 PM

Originally Posted by serious331

Thank you so much southparkfan..

So all I had to do was to show that for a point at which the subsequence converges, $f(y)=0$ and $f'(y) = 0$ right?

Well that was the question, right?

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