Question

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 $x, y \in \mathbb{R}$, and $f'(0) = 3$. Then the minimum value of the function $g(x) = 3 + e^x f(x)$, is:

• A. $3 \left( \frac{e+1}{e} \right)$
• B. $3 \left( \frac{e-1}{e} \right)$
• C. $3 - e$
• D. $3e$

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a functional equation and a derivative at a point. We need to find the function $f(x)$, then define $g(x)$ and find its minimum value.
Step 2: Key Formula or Approach:
Identify the nature of $f(x)$ from the functional equation $f \left( \frac{x+y}{3} \right) = \frac{f(x)+f(y)}{3}$.
Step 3: Detailed Explanation:
Given: $f \left( \frac{x+y}{3} \right) = \frac{f(x)+f(y)}{3}$.
Put $y=0$: $f(x/3) = \frac{f(x)+f(0)}{3}$.
Differentiating both sides with respect to $x$:
\[ f'(x/3) \cdot \frac{1}{3} = \frac{f'(x)}{3} \implies f'(x/3) = f'(x) \]
This implies $f'(x)$ is a constant function.
Let $f'(x) = c$. Since $f'(0) = 3$, we have $c = 3$.
Integrating $f'(x) = 3$ gives $f(x) = 3x + d$.
Substitute $f(x) = 3x + d$ back into the original functional equation:
\[ 3 \left( \frac{x+y}{3} \right) + d = \frac{(3x+d) + (3y+d)}{3} \]
\[ (x+y) + d = \frac{3x+3y+2d}{3} = (x+y) + \frac{2d}{3} \]
Equating both sides: $d = \frac{2d}{3} \implies d = 0$.
Thus, $f(x) = 3x$.
Now, $g(x) = 3 + e^x f(x) = 3 + 3x e^x$.
To find the minimum, differentiate $g(x)$:
\[ g'(x) = 3(1 \cdot e^x + x \cdot e^x) = 3e^x(1+x) \]
Setting $g'(x) = 0 \implies 1+x = 0 \implies x = -1$.
Checking the second derivative: $g''(x) = 3e^x(1+x) + 3e^x = 3e^x(x+2)$.
$g''(-1) = 3e^{-1}(1) = \frac{3}{e}>0$. So, $x = -1$ is a point of local minimum.
Minimum value $g(-1) = 3 + 3(-1)e^{-1} = 3 - \frac{3}{e} = 3 \left( 1 - \frac{1}{e} \right) = 3 \left( \frac{e-1}{e} \right)$.
Step 4: Final Answer:
The minimum value of $g(x)$ is $3 \left( \frac{e-1}{e} \right)$.