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.)