Proof of (f'(b)+f'(a))/2+f(a) = f(b) for |b-a| = 1

[computerex]

We were given a problem:
"The slope of f(x) at point x is twice the x value. f(2) = 3. Find f(3)."

I did this the conventional way:

f(x) = $$\int$$ 2x dx
= x^2 + c
Solving with the initial condition of f(2) = 3 gives me f(x) = x^2 - 1. So f(3) = 8.

My class did it a different way. They found f'(3) and f'(2), took the average and added that to the initial value 3, to get eight.

So can someone prove:
(f'(b)+f'(a))/2+f(a) = f(b) given |b-a| = 1 ?

 

[Mentallic]

Well, no, because it's not true in general. Try it for say, y=x³.

It can be proven for all quadratics though. Use the fact that if |b-a|=1 then your 2 function values are of the form f(a) and b=a+1 so f(b)=f(a+1).

So just prove for all quadratics f(x)=ax²+bx+c

$$f'(x+1)+f'(x)=2(f(x+1)-f(x))$$

 

[physicsman2]

Well, no, because it's not true in general. Try it for say, y=x³.

It can be proven for all quadratics though. Use the fact that if |b-a|=1 then your 2 function values are of the form f(a) and b=a+1 so f(b)=f(a+1).

So just prove for all quadratics f(x)=ax²+bx+c

$$f'(x+1)+f'(x)=2(f(x+1)-f(x))$$

That's what I thought, but I wasn't sure.

I would go the way you did, OP. I would just find the antiderivative and go from there, which is what you exactly did.