My Math Forum Rolle's Theorem

 Calculus Calculus Math Forum

 October 5th, 2012, 07:20 PM #1 Senior Member   Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Rolle's Theorem y = f(x) = (x-3) (x+1)^(2) on [-1,3] closed interval Can Rolle's Theorem be applied to this function on [1,3] I said Yes because I set each to zero and got zero for both therefore if f '(a) = f '(b) then f '(c) must exist please someone check to see if I did this properly Next it says if Rolle's Theorem can be applied, find all values, C, in the open interval (-1,3) such that f ' (c) = 0 I need some help with this second question
October 5th, 2012, 08:55 PM   #2
Senior Member

Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
Thanks: 521

Math Focus: Calculus/ODEs
Re: Rolle's Theorem

Rolle's theorem states:

Quote:
 Let $f$ be a function that is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f(a)=f(b)=0$, then there exists a number $c$ in $(a,b)$ such that $f'(c)=0$.
Since:

$f(-1)=f(3)=0$, and the function is continuous and differentiable for all reals (and so it is thus on the given interval) we know by Rolle's theorem there is at least one critical number on the given interval.

To find the critical number(s), we equate the derivative to zero, and solve for $x$.

$f(x)=(x-3)(x+1)^2$

Using the product rule, we find:

$f'(x)=(x-3)$$2(x+1)$$+(1)(x+1)^2=(x+1)$$2(x-3)+(x+1)$$=(x+1)(3x-5)=0$

We discard the root $x=-1$ as it is an end-point of the interval, and we find:

$c=\frac{5}{3}$

 October 6th, 2012, 07:17 AM #3 Senior Member   Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: Rolle's Theorem Follow

 Tags rolle, theorem

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Shamieh Calculus 4 October 21st, 2013 04:18 PM bobcantor1983 Calculus 1 October 1st, 2013 01:49 PM joeljacks Calculus 6 October 28th, 2012 05:30 PM mathkid Calculus 4 October 6th, 2012 06:35 PM maximus101 Real Analysis 4 February 11th, 2011 08:42 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top