Let F(x)= px^2+qx+r be a quadratic function defined on a closed interval [a,b]. (Mean Value Theorem)?

Question

"Let F(x)= px^2+qx+r be a quadratic function defined on a closed interval [a,b]. Show that there is exactly one point c, in (a,b) at which f satisfies the conclusion of the Mean Value Theorem."

It is one of the practice questions that I have trouble figuring out. Anyone have any ideas?

Answer 1

The MVT says that there exists c in (a,b) such that F'(c) = (F(b)-F(a)) / (b-a)



F'(c) = 2pc + q



(F(b) - F(a)) / (b-a) = (p(b^2 - a^2) + q(b - a)) / (b-a) = p(b+a) + q



So 2pc + q = p(b+a) + q => c = (b+a)/2

Answer 2

The mean value theorem states that there is some x in the range [a,b] such that



f'(x) = ( f(b) - f(a) ) / (b - a) = C (some constant)



But since f is quadratic, f' is linear, hence there can only be one solution to the above relation:



f'(x) = 2px + q = C



Thus x = (C - q)/(2p) is the unique location where the mean value theorem is satisfied.

Answer 3

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