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It is easily proven that given an everywhere differentiable function $f$ on $\Bbb R$, if $f$ is constant, linear, or a quadratic function, then $$\frac{f(x)-f(y)}{x-y}=\frac{f'(x)+f'(y)}{2}$$ for all distinct $x$ and $y$ in $\Bbb R$. Is it true, conversely that if that particular equation holds, then $f$ is a constant, linear or quadratic function? I thought of this problem after reading a MAA Focus magazine article.

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Suppose that $f$ is a differentiable function, $f'(x)$ is not zero everywhere, and $f$ satisfies $$\frac{f(x)-f(y)}{x-y}=\frac{1}{2}(f'(x)+f'(y)).$$ Then $g(x)=f(x)-f'(0)x-f(0)$ satisfies the equation as well, and we have $g'(0)=0$, $g(0)=0$. Letting $y=0$, we obtain $2g(x)=xg'(x)$, and hence $$\frac{2}{x}=\frac{g'(x)}{g(x)}.$$ Noting that $g(x)$ and $g'(x)$ are not zero everywhere, we integrate to obtain $2\log x+k= \log (g(x))$, and so $$e^k x^2=g(x).$$ Letting $a=e^k$, $b=f'(0)$, $c=f(0)$, it follows that $$f(x)=ax^2+bx+c.$$

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