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# The Rolle's constant for the function $y=x^{2}$ on $[-2,2]$ is

$\begin{array}{1 1}(1)\frac{2\sqrt{3}}{3}&(2)0\\(3)2&(4)-2\end{array}$

Let $f(x) =x^2$
$f'(x) =2x$
$f'(c) =0 => 2c =0$
$c=0$
Hence 2 is the correct answer.