Minimum Value of Function Defined by Functional Equation

To find the minimum value of the function $g(x) = 3 + e^x f(x)$, we first need to determine the explicit expression for the function $f(x)$.

Step 1: Determine $f(0)$

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

Substitute $x = 0$ and $y = 0$: $f(0) = \frac{f(0) + f(0)}{3}$ $3 f(0) = 2 f(0) \implies f(0) = 0$

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Step 2: Determine $f'(x)$ and $f(x)$

Differentiating both sides of the functional equation partially with respect to $x$: $\frac{d}{dx} \left[ f\left(\frac{x+y}{3}\right) \right] = \frac{d}{dx} \left[ \frac{f(x)+f(y)}{3} \right]$ $\frac{1}{3} f'\left(\frac{x+y}{3}\right) = \frac{1}{3} f'(x) \implies f'\left(\frac{x+y}{3}\right) = f'(x)$

Since this relation holds for all $x, y \in \mathbb{R}$, we can substitute $y = -x$: $f'(0) = f'(x)$

We are given that $f'(0) = 3$, which implies: $f'(x) = 3 \quad \text{for all } x \in \mathbb{R}$

Integrating $f'(x) = 3$ with respect to $x$: $f(x) = 3x + C$

Using the initial condition $f(0) = 0$: $f(0) = 3(0) + C = 0 \implies C = 0$

Thus, the function is: $f(x) = 3x$

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Step 3: Analyze $g(x)$ for Extrema

Substitute $f(x) = 3x$ into the definition of $g(x)$: $g(x) = 3 + 3x e^x$

To find the critical points of $g(x)$, compute its derivative $g'(x)$: $g'(x) = \frac{d}{dx} \left( 3 + 3x e^x \right)$ $g'(x) = 3 e^x + 3x e^x = 3 e^x (1 + x)$

Set $g'(x) = 0$ to find the critical points: $3 e^x (1 + x) = 0$

Since $e^x > 0$ for all real $x$: $1 + x = 0 \implies x = -1$

Now, analyze the sign of $g'(x)$:

• For $x < -1$, $g'(x) < 0$, so $g(x)$ is strictly decreasing.
• For $x > -1$, $g'(x) > 0$, so $g(x)$ is strictly increasing.

Hence, $g(x)$ attains its global minimum at $x = -1$.

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Step 4: Calculate the Minimum Value

Substitute $x = -1$ into $g(x)$: $g(-1) = 3 + 3(-1) e^{-1} = 3 - \frac{3}{e} = 3 \left( 1 - \frac{1}{e} \right) = 3 \left( \frac{e - 1}{e} \right)$

Thus, the minimum value of $g(x)$ is $3\left(\frac{e-1}{e}\right)$.

Correct Answer: Option B