Mean Value Theorem
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Let f(x)= sqrt(2x-x2) on the interval [0,2]. Find all numbers c, 0<c<2, which satisfy the conclusion of the mean value theorem.
f(x)= sqrt(2x-x2)
f'(x)=(1-x)/sqrt(2x-x2)
f(x) is differential and continous on (0,2),
Thus f'(c)= (f(2)-f(0))/(2-0)
f'(c)=(1-c)/sqrt(2c-c2)=(0-0)/2=0
(1-c)=0 => c=1
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