The mean value theorem states that there is a point x in the interval for which
f'(x) = [f(x2) - f(x1)]/(x2-x1)
With f'(x) < 2 for the entire interval,
[f(x2) - f(x1)]/(x2-x1) cannot exceed 2,
which means
f(x2) - f(x1) < 2*3 = 6
f(x2) = f(3) < 6 + f(x1) = 10
f(3) < 10
Suppose that f(0) = 4 and f'(x)< or equal to 2 for x>0. Apply the MVT to the interval [0 , 3] to prove that f(3) < or equal to 10
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