March 17th, 2016, 09:17 AM  #1 
Newbie Joined: Feb 2016 From: Fort Worth Posts: 8 Thanks: 0  Rolle's Theorem Help
Ok, so after I find the derivative to set f'(c)=0 on the explanations it just skips a bunch of steps saying that: 3(x^2)12x+11=0 when x=(6+sqrt3)/3 Can someone explain step by step so i can apply it to other problems? Thanks in advance. 
March 17th, 2016, 09:46 AM  #2 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics 
Pretty sure that's just the quadratic formula. $\displaystyle x = \frac{(12) \pm \sqrt{(12)^2  4(3)(11)}}{2(3)}=\frac{12 \pm \sqrt{12}}{6}=\frac{12 \pm 2\sqrt{3}}{6}=\frac{6\pm \sqrt{3}}{3}$ 
March 17th, 2016, 09:47 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
That just looks like an example to me, but it's hard to tell without knowing what the book says.

March 17th, 2016, 09:56 AM  #4 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics  Presumably it's a question where you have to verify Rolle's theorem. It's already shown that f(a) = f(b) where a, b are the two endpoints, the original function is a polynomial (and hence is differentiable and continuous everywhere), and $\displaystyle f'(x) = 3x^212x+11$.

March 17th, 2016, 11:46 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Are you asking how to solve that quadratic equation? If there is no simple factoring, I prefer to "complete the square". From $\displaystyle 3x^2 12x+ 11= 0$, subtract 11 from both sides and factor a "3" out of the left: $\displaystyle 3(x^2 4x)= 11$. Divide both sides by 3: $\displaystyle x^2 4x= \frac{11}{3}$. Adding 4 to both sides "completes the square" on the left: $\displaystyle x^2 4x+ 4= (x 2)^2= 4 \frac{11}{3}= \frac{12}{3} \frac{11}{3}= \frac{1}{3}$. Taking the square roots of both sides, $\displaystyle x 2= \pm\sqrt{\frac{1}{3}}= \frac{\sqrt{3}}{3}$. Adding 2 to both sides $\displaystyle x= 2\pm \frac{\sqrt{3}}{3}= \frac{6\pm\sqrt{3}}{3}$. If you apply those steps to the general quadratic, $\displaystyle ax^2+ bx+ c= 0$, you get the "quadratic formula" 123qwerty refers to, $\displaystyle x= \frac{b\pm\sqrt{b^2 4ac}}{2a}$. 

Tags 
rolle, theorem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Help with Rolle's Theorem [2]  Shamieh  Calculus  4  October 21st, 2013 04:18 PM 
Applying Rolle's Theorem  bobcantor1983  Calculus  1  October 1st, 2013 01:49 PM 
Rolle's Theorem Question  joeljacks  Calculus  6  October 28th, 2012 05:30 PM 
Rolle's Theorem 2  mathkid  Calculus  4  October 6th, 2012 06:35 PM 
Rolle's Theorem  mathkid  Calculus  2  October 6th, 2012 07:17 AM 