The Mean Value Theorem, question 11

A student finds the $c$ that the Mean Value Theorem guarantees for the function below on $[0, 2]$ , as shown. Find the first line that is wrong.

f(x) = x^{3}

$\frac{f(2) - f(0)}{2 - 0} = \frac{8 - 0}{2} = 4$
The slope of the secant line over [0, 2].
$f'(c) = 3c^{2} = 4$
The theorem gives a c where the tangent has that slope.
$c = \tfrac{4}{3}$
Divide both sides by 3.
$\tfrac{4}{3} \in (0, 2)$
Check that c lies inside the interval.

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