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:
The problem has two parts. First, we must identify the function $f(x)$ using a given functional equation and the value of its derivative at $x=0$. Second, we need to construct a new function $g(x)$ and find its global minimum value using calculus.
Step 2: Key Formula or Approach:
1. Solve the functional equation: Differentiating the equation with respect to one variable while holding the other constant can reveal properties of the function's derivative.
2. Find Minimum Value: Once $f(x)$ is known, find the minimum of $g(x)$ by finding its critical points (where $g'(x)=0$) and using the second derivative test to confirm it's a minimum.
Step 3: Detailed Explanation:
Part A: Find the function f(x)
Given the functional equation: $f \left( \frac{x+y}{3} \right) = \frac{f(x)+f(y)}{3}$.
Let's partially differentiate with respect to $x$:
\[ f' \left( \frac{x+y}{3} \right) \cdot \frac{1}{3} = \frac{f'(x)}{3} + 0 \implies f' \left( \frac{x+y}{3} \right) = f'(x) \] This must hold for all $x, y$. This implies that the derivative $f'(x)$ is a constant function.
Let $f'(x) = c$. We are given $f'(0) = 3$, so $c=3$.
Thus, $f'(x) = 3$ for all $x$.
Integrating with respect to $x$ gives $f(x) = 3x + d$, where $d$ is a constant of integration.
To find $d$, substitute this form of $f(x)$ back into the original functional equation:
\[ 3\left(\frac{x+y}{3}\right) + d = \frac{(3x+d) + (3y+d)}{3} \] \[ x+y+d = \frac{3(x+y) + 2d}{3} = (x+y) + \frac{2d}{3} \] \[ d = \frac{2d}{3} \implies 3d = 2d \implies d = 0 \] So, the function is $f(x) = 3x$.
Part B: Find the minimum of g(x)
Now we define the function $g(x) = 3 + e^x f(x) = 3 + 3xe^x$.
To find the minimum value, we first find the derivative, $g'(x)$:
\[ g'(x) = \frac{d}{dx}(3 + 3xe^x) = 0 + 3(1 \cdot e^x + x \cdot e^x) = 3e^x(1+x) \] Set the derivative to zero to find critical points:
$g'(x) = 0 \implies 3e^x(1+x) = 0$. Since $e^x$ is never zero, we must have $1+x=0$, which gives $x=-1$.
To confirm this is a minimum, we use the second derivative test:
$g''(x) = \frac{d}{dx}(3e^x(1+x)) = 3[e^x(1+x) + e^x(1)] = 3e^x(x+2)$.
At $x=-1$, $g''(-1) = 3e^{-1}(-1+2) = 3e^{-1} = \frac{3}{e}>0$. This confirms a local minimum.
The minimum value is $g(-1)$:
\[ g(-1) = 3 + 3(-1)e^{-1} = 3 - \frac{3}{e} = 3\left(1 - \frac{1}{e}\right) = 3\left(\frac{e-1}{e}\right) \] Step 4: Final Answer:
The minimum value of $g(x)$ is $3 \left( \frac{e-1}{e} \right)$.