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Ok so I have a problem that I know satisfies all the rules of the mean value theorem, and I have to find "c" that satisfies the conclusion of the mean value theorem, and everything I do I can only get one answer ( 1/3 ) but the online submission form says I'm doing something wrong and won't accept the answer! Any suggestions? Here is the problem below:

$$ f(x) = 3x^2 − 2x + 1 $$ On the interval $$[0, 2] $$

and the answer I keep getting by taking the derivative and solving for X is

$$ 1/3 $$

Please help! Thank you :)

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  • $\begingroup$ How did you get $1/3$? $\endgroup$
    – wckronholm
    Apr 11 '14 at 3:58
  • $\begingroup$ derived then solved for x $\endgroup$
    – maribov
    Apr 11 '14 at 3:58
  • $\begingroup$ Solved for $x$ in what equation? $\endgroup$
    – wckronholm
    Apr 11 '14 at 3:58
  • $\begingroup$ did i do it wrong? $\endgroup$
    – maribov
    Apr 11 '14 at 3:59
  • $\begingroup$ derivitive of 3x^2-2x+1 or 6x-2 $\endgroup$
    – maribov
    Apr 11 '14 at 3:59
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$$\frac{f(2)-f(0)}{2-0}=\frac{9-1}{2}=4$$ So solve $$f'(c)=6c-2=4$$ to get $$c=1$$

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  • $\begingroup$ Ohhhhhhhhhhhhhhhhh i see, thank you! i was setting 6c-2 to 0, i see what i did wrong $\endgroup$
    – maribov
    Apr 11 '14 at 4:00
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MVT says:$f(2) - f(0) = f'(c)(2 - 0)$ translates $9 - 1 = 2(6c - 2)$. So$4 = 6c - 2$, and $c = 1$

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