If `f'(x) gt f(x)" for all "x ge 1 and f(1)=0`, then

Options

A. `e^(x)f(x)` is a decreasing function
B. `e^(-x)f(x)` is an increasing function
C. `f(x) gt0" for all " x in (1, oo)`
D. `f(x) lt 0" for ll " x in [1, oo)`

Answer

To solve the problem, we need to analyze the given conditions and derive the properties of the function \( f(x) \).

### Step-by-Step Solution:

1. **Given Information**:
We know that \( f'(x) > f(x) \) for all \( x \geq 1 \) and that \( f(1) = 0 \).

2. **Analyzing the Inequality**:
Since \( f'(x) > f(x) \), we can rewrite this as:
\[ f'(x) - f(x) > 0 \]
This implies that the function \( g(x) = e^{-x} f(x) \) is increasing. To see this, we can differentiate \( g(x) \):
\[ g'(x) = e^{-x} f'(x) - e^{-x} f(x) = e^{-x} (f'(x) - f(x)) \]
Since \( f'(x) - f(x) > 0 \) for \( x \geq 1 \), it follows that \( g'(x) > 0 \). Thus, \( g(x) \) is an increasing function for \( x \geq 1 \).

3. **Evaluating \( g(1) \)**:
We can evaluate \( g(1) \):
\[ g(1) = e^{-1} f(1) = e^{-1} \cdot 0 = 0 \]
Since \( g(x) \) is increasing and \( g(1) = 0 \), it follows that for \( x > 1 \), \( g(x) > 0 \).

4. **Conclusion about \( f(x) \)**:
Since \( g(x) = e^{-x} f(x) > 0 \) for \( x > 1 \), we can conclude that:
\[ f(x) > 0 \quad \text{for } x > 1 \]

5. **Final Result**:
Therefore, we can conclude that \( f(x) \) is positive for all \( x \geq 1 \).