Let `f:R to R` be a differentiable function such that `f(x)=x^(2)+int_(0)^(x)e^(-t)f(x-t)dt`. `f(x)` increases for

Options

A. `xgt1`
B. `xlt-2`
C. `xgt2`
D. none of these

Answer

To solve the problem, we need to analyze the given function \( f(x) \) and determine the intervals where it is increasing. The function is defined as:

\[ f(x) = x^2 + \int_0^x e^{-t} f(x - t) \, dt \]

### Step 1: Differentiate \( f(x) \)

We will differentiate both sides of the equation with respect to \( x \):

\[ f'(x) = \frac{d}{dx} \left( x^2 \right) + \frac{d}{dx} \left( \int_0^x e^{-t} f(x - t) \, dt \right) \]

The derivative of \( x^2 \) is \( 2x \).

For the integral, we apply Leibniz's rule:

\[ \frac{d}{dx} \left( \int_0^x e^{-t} f(x - t) \, dt \right) = e^{-x} f(0) + \int_0^x e^{-t} \frac{d}{dx} f(x - t) \, dt \]

Using the chain rule, we have:

\[ \frac{d}{dx} f(x - t) = f'(x - t) \]

Thus, we can rewrite the derivative as:

\[ f'(x) = 2x + e^{-x} f(0) + \int_0^x e^{-t} f'(x - t) \, dt \]

### Step 2: Substitute \( f(0) \)

To evaluate \( f(0) \), we substitute \( x = 0 \) into the original function:

\[ f(0) = 0^2 + \int_0^0 e^{-t} f(0 - t) \, dt = 0 \]

So, \( f(0) = 0 \).

### Step 3: Simplify \( f'(x) \)

Now substituting \( f(0) = 0 \) into the derivative:

\[ f'(x) = 2x + \int_0^x e^{-t} f'(x - t) \, dt \]

### Step 4: Analyze the expression for \( f'(x) \)

We want to find when \( f'(x) > 0 \):

\[ f'(x) = 2x + \int_0^x e^{-t} f'(x - t) \, dt > 0 \]

### Step 5: Find critical points

The term \( 2x \) is positive when \( x > 0 \) and negative when \( x < 0 \). The integral term is always non-negative since \( e^{-t} > 0 \) for all \( t \).

### Step 6: Determine intervals for \( f'(x) \)

1. For \( x < 0 \):
- \( 2x < 0 \)
- The integral term is non-negative, but it cannot outweigh \( 2x \). Thus, \( f'(x) < 0 \).

2. For \( x = 0 \):
- \( f'(0) = 0 \).

3. For \( x > 0 \):
- \( 2x > 0 \).
- The integral term is also non-negative, making \( f'(x) > 0 \).

### Conclusion

Thus, \( f(x) \) is increasing for \( x \in (0, \infty) \).

### Final Answer

The function \( f(x) \) increases for \( x \in (0, \infty) \).