Question

How do we show that this is true?

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 :)

Answer 1

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)`

Answer 2

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`.)