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
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.
Not to put too fine of a point on it but
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 and the results follows as stated