Question

Sol. [Key Concept: Put $x = y = 0$ to get $f(0) = 0$; differentiating gives $f'(x) = 3$, so $f(x) = 3x$, then minimise $g$ via $g' = 0$.]

$f\left( \frac{x + y}{3} \right) = \frac{f(x) + f(y)}{3}$, put $x = y = 0$

$$f(0) = \frac{2f(0)}{3} $$$\Rightarrow f(0) = 0$.

$$\begin{array}{r} f^{'}\left( \frac{x + y}{3} \right) \cdot \frac{1}{3} = \frac{1}{3}f^{'}(x)\#(1) \end{array}$$

put $x = 0$

f' $\left( \frac{y}{3} \right)\frac{1}{3} = \frac{1}{3} \times 3$

$$f^{'}\left( \frac{y}{3} \right) = 3 $$put $y = 3x$

$$f^{'}(x) = 3 $$integrate both sides

$$f(x) = 3x + c $$from eq. (1)

$${f(0) = 0 }{\Rightarrow f(x) = 3x }$$Now

$${g(x) = 3 + e^{x} \cdot 3x }{g^{'}(x) = 3\left\lbrack e^{x} + x \cdot e^{x} \right\rbrack }{= 3e^{x}(x + 1) }$$$g^{'}(x) = 0$, at $x = - 1$

$${(g(x))_{\text{min}\text{~}} = 3 + e^{- 1}( - 3) }{= 3\left\lbrack 1 - \frac{1}{e} \right\rbrack = \frac{3(e - 1)}{e} }$$