How do we show that this is true?

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By inspection, I see an application of the Mean Value Theorem. Also, I know that #a < (a+b)/2 <b# for all #a,b in RR#,
and since the polynomial f(x) is well-behaved, I do know that it is continuous and differentiable for all #x#, I can apply the MVT straightaway.

However, I don't know if this would be a 'rigorous' proof that I could write down in an exam. Please help me!

Thanks in advance :)

2 Answers
May 7, 2018

I do not think that the MVT is the good approach.

In fact, as #f'(x)# is continuous, using the fundamental theorem of calculus we can state that:

#f(b)-f(a) = int_a^b f'(t)dt#

so that using the MVT we have:

#f(b)-f(a) = (b-a) f'(xi)# with #xi in (a,b)#

but we cannot determine that necessarily:

#xi = (a+b)/2#

We can however demonstrate the equality directly:

#f'(x) = 2Ax+B#

so that:

#f'((a+b)/2) = 2A(a+b)/2 +B = (a+b)A+B#

and:

#(f(b)-f(a))/(b-a) = (Ab^2+bB+C -Aa^2-aB-C)/(b-a)#

#(f(b)-f(a))/(b-a) = (A(b^2-a^2) +B(b-a))/(b-a)#

#(f(b)-f(a))/(b-a) = (A(b-a)(b+a)+B(b-a))/(b-a)#

#(f(b)-f(a))/(b-a) = A(b+a)+B = f'((a+b)/2)#

May 7, 2018

Please see below.

Explanation:

The mean value theorem tells us that there is a solution to #f'(x) = (f(b)-f(a))/(b-a)#, but it does not tell us what the solution is. Only solving the equation can do that.

Calculate #f'(x)#.

Calculate #f'((a+b)/2)#. (You should get #Aa+Ab+B#.)

Calculate #(f(b)-f(a))/(b-a)#. (You should get #Aa+Ab+B#.)