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Question

Cosider the quadratic function f(x) = Ax² +Bx +C, where A,B,and C are real numbers with A≠ 0. Show that when the Mean Value Theorem is applied is applied to f on the interval [a,b],the number c guaranteed by the theorem is the midpoint of the interval.

 

f(b)-f(a)

_______  =  ?????????

b-a

Answer 1

The mean value theorem says f'(c) = [f(b) - f(a)] / [b - a] for some c between a and b.

 

f(x) = Ax² + Bx + C

 

f(a) = Aa² + Ba + C

 

f(b) = Ab² + Bb + C

 

 

[f(b) - f(a)] / [b - a] = [A(b² - a²) + B(b - a)] / [b - a] = A(b + a) + B

 

 

 

f'(x) = 2Ax + B, so f'(c) = 2Ac + B

 

 

2Ac + B = A(b + a) + B

 

 

c = ½(a + b)