Hello I have some question about the Rolle's theorem

Here is the question:

Use Rolle's theorem to show that f(x)=x^3+3x-5 has exactly one real root.

what is my limit!?I need some limit right!?

Thanks

NaNa

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- May 26th 2009, 01:46 PMNaNaRolle's Theorem
Hello I have some question about the Rolle's theorem

Here is the question:

Use Rolle's theorem to show that f(x)=x^3+3x-5 has exactly one real root.

what is my limit!?I need some limit right!?

Thanks

NaNa - May 26th 2009, 02:15 PMTheEmptySet
No note quite.

Suppose by way of contradiction that

$\displaystyle f(a)=f(b)=0,a \ne b$ note that $\displaystyle f'(x)=3x^2+3=3(x^2+1)$

Now if you apply Rolle's theorem

$\displaystyle f(a)-f(b)=f'(c)(b-a) \iff 0=f'(c)(b-a)$

Now we have a problem since $\displaystyle b \ne a \implies b-a \ne 0$

That mean that $\displaystyle f'(c)=0 \implies 0=3(c^2+1)$ this is a contradiction because there is no real number c that fits the bill.

So this is Half of the proof. We have show that there cannot be more than two roots, it is still up to you to show that there is at least one.

Good luck. - May 26th 2009, 02:46 PMCalculus26
Not to put too fine of a point on it but

http://www.mathhelpforum.com/math-he...158023cd-1.gif

is a statement of the mean value theorem which granted is prooved using Rolle's Thm

But we don't need the MVT for this problem

Rolles theorem states if f(a) = f(b) = 0 under the proper continuity and differentiability conditions then there is at least one point c where

f ' (c) = 0

As stated assume f(a) = f(b) = 0

Therefore http://www.mathhelpforum.com/math-he...c598f2ef-1.gif and the results follows as stated - May 26th 2009, 02:48 PMTheEmptySet
- May 26th 2009, 02:56 PMCalculus26
When I have time I read your posts because I've always been impressed

with you replies --- I probably wouldn't even have read it if you weren't so damn good.