If `f(x)=int_(0)^(x)(e^(t)+3)dt`, then which of the following is always true `AA x in R`

Options

A. `f(x-3)-f(x-1) le 11`
B. `f(x+1)-f(x-1) lt 4`
C. `f(x+3)-f(x-5) gt 24`
D. `f(x)+f(x+1) lt -21`

Answer

To solve the question, we start with the given function:

\[ f(x) = \int_{0}^{x} (e^{t} + 3) dt \]

### Step 1: Evaluate the integral

We can break the integral into two parts:

\[ f(x) = \int_{0}^{x} e^{t} dt + \int_{0}^{x} 3 dt \]

The first integral can be evaluated as follows:

\[ \int e^{t} dt = e^{t} + C \]

Thus,

\[ \int_{0}^{x} e^{t} dt = e^{x} - e^{0} = e^{x} - 1 \]

The second integral is straightforward:

\[ \int_{0}^{x} 3 dt = 3x \]

Combining these results, we have:

\[ f(x) = (e^{x} - 1) + 3x = e^{x} + 3x - 1 \]

### Step 2: Find the derivative of \( f(x) \)

To analyze the function further, we find its derivative:

\[ f'(x) = \frac{d}{dx}(e^{x} + 3x - 1) = e^{x} + 3 \]

### Step 3: Analyze the derivative

Since \( e^{x} \) is always positive for all \( x \in \mathbb{R} \) and \( 3 \) is a constant positive value, we conclude that:

\[ f'(x) = e^{x} + 3 > 0 \quad \text{for all } x \in \mathbb{R} \]

This means that \( f(x) \) is an increasing function for all real numbers \( x \).

### Step 4: Determine the behavior of \( f(x) \)

Since \( f(x) \) is increasing, it implies that:

1. \( f(x) \) is always greater than \( f(0) \).
2. \( f(0) = e^{0} + 3(0) - 1 = 1 - 1 = 0 \).

Thus, for all \( x \in \mathbb{R} \):

\[ f(x) > 0 \]

### Conclusion

From the analysis above, we can conclude that:

\[ f(x) > 0 \quad \text{for all } x \in \mathbb{R} \]