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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 $g(\frac{a+b}2)=c$.2012-12-11

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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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    @AA see if this latest edit helps you.2012-12-13