If f(x) is a differentiable real valued function satisfying `f''(x)-3f'(x) gt 3 AA x ge 0 and f'(0)=-1,` then `f(x)+x AA x gt 0` is

Options

A. decreasing function of x
B. increasing function of x
C. constant function
D. none of these

Explanation

To solve the problem, we need to analyze the given inequality involving the second derivative of the function \( f(x) \) and its first derivative. Let's break it down step by step.

### Step 1: Understand the given inequality
We are given that:
\[ f''(x) - 3f'(x) > 3 \quad \text{for all } x \geq 0 \]
This can be rearranged to:
\[ f''(x) > 3f'(x) + 3 \]

### Step 2: Rewrite the inequality
We can express this inequality in a more manageable form. We can consider the function:
\[ g(x) = e^{-3x} f'(x) \]
Differentiating \( g(x) \) using the product rule gives:
\[ g'(x) = e^{-3x} f''(x) - 3e^{-3x} f'(x) \]
Substituting the inequality into this expression, we have:
\[ g'(x) > e^{-3x}(3f'(x) + 3) - 3e^{-3x} f'(x) = 3e^{-3x} \]

### Step 3: Analyze the derivative
Since \( g'(x) > 3e^{-3x} \), we can conclude that \( g(x) \) is increasing. This means:
\[ g(x) = e^{-3x} f'(x) \text{ is increasing for } x \geq 0 \]

### Step 4: Evaluate at \( x = 0 \)
We know that:
\[ f'(0) = -1 \]
Thus:
\[ g(0) = e^{0} f'(0) = -1 \]
Since \( g(x) \) is increasing and \( g(0) = -1 \), it follows that for \( x > 0 \):
\[ g(x) > -1 \]
This implies:
\[ e^{-3x} f'(x) > -1 \quad \Rightarrow \quad f'(x) > -e^{3x} \]

### Step 5: Analyze the behavior of \( f'(x) \)
Since \( f'(x) > -e^{3x} \), we can integrate this inequality to find \( f(x) \):
\[ f(x) > -\int e^{3x} \, dx = -\frac{1}{3} e^{3x} + C \]

### Step 6: Evaluate the function \( f(x) + x \)
Now we need to analyze \( f(x) + x \):
\[ f(x) + x > -\frac{1}{3} e^{3x} + C + x \]
The term \( -\frac{1}{3} e^{3x} + C + x \) will dominate as \( x \to \infty \), and since \( e^{3x} \) grows faster than \( x \), we can conclude that \( f(x) + x \) is increasing.

### Conclusion
Thus, we can conclude that \( f(x) + x \) is an increasing function for \( x > 0 \).

### Final Answer
The function \( f(x) + x \) is an **increasing function** for \( x > 0 \).

---