Behavior of Derivative of Function Having Minimum at Specific Point

To determine which of the given options is correct, let us define a new function $g(x)$ as: $g(x) = e^{-x} f(x) \quad \text{for } x \in [0, 1]$

First, we compute the first derivative of $g(x)$ with respect to $x$: $g'(x) = e^{-x} f'(x) - e^{-x} f(x) = e^{-x} \big(f'(x) - f(x)\big)$

Next, we compute the second derivative of $g(x)$: $g''(x) = e^{-x} \big(f''(x) - f'(x)\big) - e^{-x} \big(f'(x) - f(x)\big) = e^{-x} \big(f''(x) - 2f'(x) + f(x)\big)$

We are given in the passage that for all $x \in [0, 1]$: $f''(x) - 2f'(x) + f(x) \ge e^x$

Substituting this inequality into our expression for $g''(x)$: $g''(x) \ge e^{-x} \cdot e^x = 1 > 0 \quad \text{for all } x \in [0, 1]$

Since $g''(x) > 0$ on $[0, 1]$, $g'(x)$ is a strictly increasing function on the interval $[0, 1]$.

We are given that $g(x) = e^{-x} f(x)$ attains its minimum value in the interval $[0, 1]$ at the interior point $x = \frac{1}{4}$. By Fermat's theorem on stationary points for differentiable functions, we must have: $g'\left(\frac{1}{4}\right) = 0$

Since $g'(x)$ is strictly increasing on $[0, 1]$:

1. For $x \in \left(0, \frac{1}{4}\right)$: $g'(x) < g'\left(\frac{1}{4}\right) = 0$

   Substituting $g'(x) = e^{-x} \big(f'(x) - f(x)\big)$: $e^{-x} \big(f'(x) - f(x)\big) < 0$ Since $e^{-x} > 0$ for all $x \in \mathbb{R}$, we get: $f'(x) - f(x) < 0 \implies f'(x) < f(x) \quad \text{for } 0 < x < \frac{1}{4}$

2. For $x \in \left(\frac{1}{4}, 1\right)$: $g'(x) > g'\left(\frac{1}{4}\right) = 0 \implies f'(x) > f(x) \quad \text{for } \frac{1}{4} < x < 1$

Therefore, $f'(x) < f(x)$ holds true for $0 < x < \frac{1}{4}$.

This corresponds to Option (C).