What value(s) c, if any, are predicted by the Mean Value Theorem for the function f(x)=(x-2)2 on the interval [0,2]?
\(f'(c)=\dfrac{f(2)-f(0)}{2-0}=\dfrac{(2-2)^2-(0-2)^2}{2}=-2\\ f'(c) = 2c -4\\ \therefore 2c - 4 = -2\\ \begin{array}{rll}2c-4&=&-2\\2c &=&2\\c&=&1\end{array}\)
+77 
