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 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. $\displaystyle x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(11)}}{2(3)}=\frac{12 \pm \sqrt{12}}{6}=\frac{12 \pm 2\sqrt{3}}{6}=\frac{6\pm \sqrt{3}}{3}$ Thanks from Azilips 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
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 Originally Posted by v8archie That just looks like an example to me, but it's hard to tell without knowing what the book says.
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 $\displaystyle f'(x) = 3x^2-12x+11$. 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, $\displaystyle x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$. Thanks from Azilips Tags rolle, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded 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 mathkid Calculus 2 October 6th, 2012 07:17 AM

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