# Thread: More calculus ideas and proofs.

1. ## More calculus ideas and proofs.

1. For a continuous, differentiable function f(x), and an interval [a,b], supose that f(a) and f(b) have opposite signs and that f' does not equal 0. Show that f(x) is exactly one root between a and b.

2. Show that for any numbers a and b, the inequality |sin(b) - sin(a)| is less than or equal to |a - b| is true.

2. 1. Let's suppose it has more than one (it clearly has at least one by Bolzano), which would be, at least 2.

$x_1$ and $x_2$ denote 2 of our roots

We are supposing that
$x_1\in(a,b)$ and $x_2\in(a,b)$ with $x_1\neq{x_2}$ and $f(x_1)=f(x_2)=0$

Then, by Rolle's Theorem, there exists $c\in{(x_1,x_2)}$ such that: $f'(c)=0$

But $f'(x)$ can't be 0 in (a,b)

Thus it is absurd to suppose it has more than 1 root

3. Originally Posted by PaulRS
1. Let's suppose it has more than one (it clearly has at least one by Bolzano), which would be, at least 2.

$x_1$ and $x_2$ denote 2 of our roots

We are supposing that
$x_1\in(a,b)$ and $x_2\in(a,b)$ with $x_1\neq{x_2}$ and $f(x_1)=f(x_2)=0$

Then, by Rolle's Theorem, there exists $c\in{(x_1,x_2)}$ such that: $f'(c)=0$

But $f'(x)$ can't be 0 in (a,b)

Thus it is absurd to suppose it has more than 1 root
What do you mean by "Bolzano" is that how you say it in Ugurarian (that word probably does not exist ).

4. Originally Posted by Prophet
2. Show that for any numbers a and b, the inequality |sin(b) - sin(a)| is less than or equal to |a - b| is true.
Consider $\sin x$ on $[a,b]$ then by mean value there is $x_0\in (a,b)$ so that $\sin a - \sin b = \cos x_0 (a-b)$ so $|\sin a - \sin b| = |\cos x_0||a-b|$ but $|cos x_0|\leq 1$ that means $|\sin a - \sin b|\leq |a-b|$.

5. Originally Posted by ThePerfectHacker
What do you mean by "Bolzano"
I was talking about Bolzano's Theorem

Sorry, I am used to call it Bolzano directly

6. Originally Posted by PaulRS
I was talking about Bolzano's Theorem

Sorry, I am used to call it Bolzano directly
There is something called the Bolzano-Weierstrass Theorem which is one of the most important theorems in Analysis.

It says if $x_n$ is a bounded sequence we can find a convergent sub-sequence.

What you used: "If $f(x)$ is continous on $[a,b]$ and $f(a)<0,f(b)>0$ there exists a $c\in(a,b)$ so that $f(c)=0$" is called Intermediate-Value theorem.

(The intermediate-value theorem can be easily proven using Bolzano-Weierstrass theorem so maybe that is why you referred to it as Bolzano theorem. But if so, then how can you possibly leave out 'Weierstrass', he might get insulted).

7. Originally Posted by ThePerfectHacker
What you used: "If $f(x)$ is continous on $[a,b]$ and $f(a)<0,f(b)>0$ there exists a $c\in(a,b)$ so that $f(c)=0$"
Well, that is what in Uruguay we commonly call Bolzano's Theorem.
I thought that it was broadly called so.
PlanetMath: Bolzano's theorem