A function f: R rightarrow R satisfies fleft( | vedclass

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(A) Given the functional equation $f\left(\frac{x+y}{3}\right) = \frac{f(x)+f(y)+f(0)}{3}$.Setting $x=0$ and $y=0$, we get $f(0) = \frac{f(0)+f(0)+f(0

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(A) Given the functional equation $f\left(\frac{x+y}{3}ight) = \frac{f(x)+f(y)+f(0)}{3}$.Setting $x=0$ and $y=0$, we get $f(0) = \frac{f(0)+f(0)+f(0)}{3} = f(0)$, which is always true.Let $f(0) = c$. Then $f\left(\frac{x+y}{3}ight) = \frac{f(x)+f(y)+c}{3}$.Setting $y=0$, we get $f\left(\frac{x}{3}ight) = \frac{f(x)+f(0)+c}{3} = \frac{f(x)+2c}{3}$.This implies $f(x) = 3f\left(\frac{x}{3}ight) - 2c$.Since $f$ is differentiable at $x=0$, let $f'(0) = a$.Using the definition of the derivative, $f'(x) = \lim_{h 	o 0} \frac{f(x+h)-f(x)}{h}$.By differentiating the functional equation with respect to $x$, we find that $f(x)$ must be a linear function of the form $f(x) = ax+c$.Substituting $f(x) = ax+c$ into the original equation: $a\left(\frac{x+y}{3}ight)+c = \frac{(ax+c)+(ay+c)+c}{3} = \frac{a(x+y)+3c}{3} = a\left(\frac{x+y}{3}ight)+c$.This holds for all $x, y$. Thus, $f(x)$ is a linear function.

$A$ function $f: R \rightarrow R$ satisfies $f\left(\frac{x+y}{3}\right) = \frac{f(x)+f(y)+f(0)}{3}$ for all $x, y \in R$. If the function $f$ is differentiable at $x=0$, then $f$ is:

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