Let `f(x) = x log x + 3x`. Then

Options

A. f increases in `(e^(-4), oo)`
B. f increases in `(0, oo)`
C. f decreases in `(0, oo)`
D. f decreases in `(0, e^(-2))`

Answer

To solve the problem, we need to analyze the function \( f(x) = x \log x + 3x \) to determine the intervals where it is increasing or decreasing. We will do this by finding the derivative of the function and analyzing its sign.

### Step 1: Find the derivative of \( f(x) \)

The function is given by:
\[ f(x) = x \log x + 3x \]

To find the derivative \( f'(x) \), we will apply the product rule to the term \( x \log x \) and differentiate the constant term \( 3x \).

Using the product rule:
\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]
where \( u = x \) and \( v = \log x \).

Calculating the derivatives:
- \( \frac{du}{dx} = 1 \)
- \( \frac{dv}{dx} = \frac{1}{x} \)

Now applying the product rule:
\[ f'(x) = x \cdot \frac{1}{x} + \log x \cdot 1 + 3 \]
\[ f'(x) = 1 + \log x + 3 \]
\[ f'(x) = \log x + 4 \]

### Step 2: Determine the critical points

To find the critical points, we set the derivative equal to zero:
\[ \log x + 4 = 0 \]
\[ \log x = -4 \]

Exponentiating both sides gives:
\[ x = e^{-4} \]

### Step 3: Analyze the sign of \( f'(x) \)

Now we will analyze the sign of \( f'(x) \) to determine where the function is increasing or decreasing.

- For \( x < e^{-4} \):
\[ \log x < -4 \implies f'(x) < 0 \quad (\text{decreasing}) \]

- For \( x > e^{-4} \):
\[ \log x > -4 \implies f'(x) > 0 \quad (\text{increasing}) \]

### Step 4: Conclusion about the intervals

From the analysis:
- The function \( f(x) \) is decreasing on the interval \( (0, e^{-4}) \).
- The function \( f(x) \) is increasing on the interval \( (e^{-4}, \infty) \).

### Final Result

Thus, the function \( f(x) \) is:
- Decreasing in the interval \( (0, e^{-4}) \)
- Increasing in the interval \( (e^{-4}, \infty) \)