# Thread: differentiable function /bounded sequence

1. ## differentiable function /bounded sequence

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

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

2. Originally Posted by serious331
Suppose that the function$\displaystyle f:R \rightarrow R$ is differentiable and that there is a bounded sequence $\displaystyle \{x_n\}$ with $\displaystyle {x_n} \neq {x_m}$ if $\displaystyle {n \neq m}$, such that $\displaystyle f(x_n) =0$ for every index n. Show that there is a point $\displaystyle x_0$ at which $\displaystyle f(x_0) = 0$ and $\displaystyle f'(x_0) = 0$.

---> 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 {$\displaystyle y_n$} 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:

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

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

3. Originally Posted by southprkfan1
You are on the right track:

Let {$\displaystyle y_n$} 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:

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

But we know a) The limit exists (by assumption) and b) there's a sequence {$\displaystyle y_n$} that converges to y and that f($\displaystyle y_n$) = 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, $\displaystyle f(y)=0$ and $\displaystyle f'(y) = 0$ right?

4. 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, $\displaystyle f(y)=0$ and $\displaystyle f'(y) = 0$ right?
Well that was the question, right?