# Twice-differentiable functions proof

Let $f$ be a twice-differentiable function from R to R. Show that if whenever $$f'(c)=\frac{f(b)-f(a)}{b-a}$$ then $c=\frac{a+b}{2}$, it must be true that $f$ is a quadratic polynomial.

This is a question on my homework, and I'm a bit stuck. It's obvious that quadratic polynomials are twice-differentiable. I'm thinking the Mean Value Theorem might be involved, but not sure.

Can anyone give me an tips on how to do this?

Thanks.

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@wj32 That's a good reference, but I see a problem in the proof. In Lemma 2, they say that if $a$ and $b$ are good, then $\frac{a+b}2$ is good, but by their definition, that means that $g'(\frac{a+b}2)=0$ and $g(\frac{a+b}2)=c$. They reference Lemma 1 to claim this, but that only proves that $g'(\frac{a+b}2)=0$, not $g(\frac{a+b}2)=c$. –  Mario Carneiro Dec 11 '12 at 0:17

The first thing to prove is that $f'(\frac{a+b}2)=\frac{f(b)-f(a)}{b-a}$, because this was not actually given as an assumption. However, it follows easily from the Mean Value Theorem, which shows that this is true for some $c\in(a,b)$, and the assumption allows us to conclude that $c=\frac{a+b}2$.

One simple thing to note is that $$f'(0)=\frac{f(x)-f(-x)}{2x},$$ so if $g(x)=f(x)-xf'(0)$, then $$g(x)-g(-x)=f(x)-f(-x)-2f'(0)x=0,$$ so $g(x)$ is even. More generally, for all $a,b,c\in\mathbb R$, $$(c+b)(f(a+c)-f(a+b))=(c-b)(f(a+c+b)-f(a))$$ from a similar argument. Not sure if that helps, though.

Edit: Since $f(x)$ is twice differentiable, we can differentiate $2xf'(c)=f(c+x)-f(c-x)$ w.r.t. $x$ to get

$$2f'(c)=f'(c+x)-f'(c-x)$$ $$0=f''(c+x)-f''(c-x)$$

So if we let $c=x$, we get $f''(2x)=f''(0)$ whence $f''$ is constant and $f$ is quadratic.

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Thank you. For the first part, I'm a little confused as to how you applied the MVT. Because the theorem states that there exists a $c$ in the domain, but how do I know that $c=\frac{a+b}{2}$? –  A A Dec 10 '12 at 6:33
The theorem asserts the existence of some $c$ that satisfies the equation, and your assumption (the problem statement) asserts that if any $c$ exists satisfying the equation, it must be equal to $\frac{a+b}2$. –  Mario Carneiro Dec 10 '12 at 6:36
Thanks, I will work with it and try to figure things out. –  A A Dec 10 '12 at 7:12
@AA see if this latest edit helps you. –  Mario Carneiro Dec 13 '12 at 7:48