Let f: ℝ arrow ℝ be a differentiable function such that...

Given $f\left(\dfrac{x+y}{3}\right) = \dfrac{f(x) + f(y)}{3}$

Substituting $x = 0 and$y = 0$, we get:<br><br>$f(0) = \dfrac{2f(0)}{3} \Rightarrow f(0) = 0$<br><br>Differentiating the given equation partially with respect to$x$, treating$y$as a constant:<br><br>$f'\left(\dfrac{x+y}{3}\right) \cdot \dfrac{1}{3} = \dfrac{f'(x)}{3}$<br><br>$\Rightarrow f'\left(\dfrac{x+y}{3}\right) = f'(x)$<br><br>Substituting$x = 0$, we get:<br><br>$f'\left(\dfrac{y}{3}\right) = f'(0) = 3$<br><br>Since this is true for all$y \in \mathbb{R}$,$f'(x) = 3$for all$x \in \mathbb{R}$.<br><br>Integrating both sides with respect to$x$:<br><br>$f(x) = 3x + C$<br><br>Using$f(0) = 0$, we get$C = 0$. Thus,$f(x) = 3x$.<br><br>Now, the function$g(x)$is given by:<br><br>$g(x) = 3 + e^x f(x) = 3 + 3x e^x$<br><br>To find the minimum value, we differentiate$g(x)$with respect to$x$:<br><br>$g'(x) = 3(e^x + x e^x) = 3e^x(1 + x)$<br><br>Setting$g'(x) = 0$gives$x = -1$.<br><br>For$x < -1$,$g'(x) < 0$and for$x > -1$,$g'(x) > 0$. Therefore,$x = -1$is a point of global minimum.<br><br>The minimum value of$g(x)$is:<br><br>$g(-1) = 3 + 3(-1)e^{-1} = 3 - \dfrac{3}{e} = 3\left(\dfrac{e-1}{e}\right)$<br><br>Answer:$3\left(\dfrac{e-1}{e}\right)