Let $$f: \mathbb{R} \to \mathbb{R}$$ be a differentiable function such that $$f\left(\frac{x+y}{3}\right) = \frac{f(x) + f(y)}{3}$$ for all $$x, y \in R$$ and $$f'(0) = 3$$. Then the minimum value of function $$g(x) = 3 + e^x f(x)$$ is :

To find the minimum value of $$g(x)$$, we first need to identify the function $$f(x)$$ from the given functional equation.

1. Find $$f(x)$$

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

Step A: Find $$f(0)$$

Substitute $$x=0$$ and $$y=0$$:

$$f(0) = \frac{f(0)+f(0)}{3} \implies 3f(0) = 2f(0) \implies f(0) = 0$$

Step B: Determine the nature of the function

Since $$f(x)$$ is a differentiable function satisfying this linear-like functional relationship, it takes the form $$f(x) = ax + b$$.

• From $$f(0) = 0$$, we get $$b = 0$$, so $$f(x) = ax$$.
• We are given $$f'(0) = 3$$. If $$f(x) = ax$$, then $$f'(x) = a$$.
• Therefore, $$a = 3$$, and our function is $$f(x) = 3x$$.

2. Analyze $$g(x)$$

Substitute $$f(x) = 3x$$ into the expression for $$g(x)$$:

$$g(x) = 3 + e^x(3x) = 3(1 + x e^x)$$

To find the minimum, we find the derivative $$g'(x)$$ and set it to zero:

$$g'(x) = 3[e^x + x e^x] = 3e^x(1 + x)$$

Setting $$g'(x) = 0$$:

$$1 + x = 0 \implies x = -1$$

3. Calculate the Minimum Value

Now, substitute $$x = -1$$ back into $$g(x)$$ to find the minimum value:

$$g(-1) = 3(1 + (-1)e^{-1})$$

$$g(-1) = 3\left(1 - \frac{1}{e}\right)$$

$$g(-1) = 3\left(\frac{e - 1}{e}\right)$$

Wait, looking at the provided options in the image, the marked correct answer is $$\frac{3-e}{e}$$. Let's re-verify the $$g(x)$$ definition. If $$g(x) = 3e^{-1} + f(x)e^x$$ or a similar variation, the result shifts. However, based on the literal text $$g(x) = 3 + e^x f(x)$$:

The result is $$3\left(\frac{e-1}{e}\right)$$, which corresponds to Option B.