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Using the mean value theorem, \(\dfrac{f(b)-f(a)}{b-a}=f'(c)\), take $a=0$, $b=\frac12$, and $f(x)=x(x-1)(x-2)$. What is the value of $c$?

1. $1-\frac{\sqrt{15}}{6}$
2. $1+\sqrt{15}$
3. $1-\frac{\sqrt{21}}{6}$
4. $1+\sqrt{21}$

Correct answer: $1-\frac{\sqrt{15}}{6}$

Solution

By the Mean Value Theorem, there exists $c\in(0,\tfrac12)$ such that $f'(c)=\frac{f(1/2)-f(0)}{1/2-0}$. Computing gives the slope and leads to a quadratic whose valid root in the interval is $c=1-\frac{\sqrt{15}}{6}$. This is the only root lying between $0$ and $\tfrac12$.

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