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March 17th, 2016, 09:17 AM   #1
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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.
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March 17th, 2016, 09:46 AM   #2
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Pretty sure that's just the quadratic formula.

$\displaystyle x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(11)}}{2(3)}=\frac{12 \pm \sqrt{12}}{6}=\frac{12 \pm 2\sqrt{3}}{6}=\frac{6\pm \sqrt{3}}{3}$
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March 17th, 2016, 09:47 AM   #3
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That just looks like an example to me, but it's hard to tell without knowing what the book says.
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March 17th, 2016, 09:56 AM   #4
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Quote:
Originally Posted by v8archie View Post
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 $\displaystyle f'(x) = 3x^2-12x+11$.
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March 17th, 2016, 11:46 AM   #5
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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, $\displaystyle x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$.
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