The function `f(x) = x^(x) , x gt 0` , is increasing on the interval

Options

A. (0, e]
B. `(0, (1)/(e))`
C. `[(1)/(e),infty)`
D. none of these

Answer

To determine the interval on which the function \( f(x) = x^x \) is increasing for \( x > 0 \), we need to find the derivative of the function and analyze its sign.

### Step 1: Define the function
Let \( f(x) = x^x \).

### Step 2: Take the natural logarithm
To differentiate \( f(x) \), we can first take the natural logarithm of both sides:
\[ \ln f(x) = \ln(x^x) = x \ln x \]

### Step 3: Differentiate using implicit differentiation
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\ln f(x)) = \frac{1}{f(x)} \cdot f'(x) \]
Using the product rule on the right side:
\[ \frac{d}{dx}(x \ln x) = \ln x + 1 \]
Thus, we have:
\[ \frac{1}{f(x)} f'(x) = \ln x + 1 \]

### Step 4: Solve for \( f'(x) \)
Now, multiply both sides by \( f(x) \):
\[ f'(x) = f(x)(\ln x + 1) \]
Substituting back \( f(x) = x^x \):
\[ f'(x) = x^x (\ln x + 1) \]

### Step 5: Determine when \( f'(x) \) is greater than 0
For the function to be increasing, we need:
\[ f'(x) > 0 \implies x^x (\ln x + 1) > 0 \]
Since \( x^x > 0 \) for \( x > 0 \), we only need to consider:
\[ \ln x + 1 > 0 \]
This simplifies to:
\[ \ln x > -1 \]
Exponentiating both sides gives:
\[ x > e^{-1} = \frac{1}{e} \]

### Step 6: Conclusion
Thus, the function \( f(x) = x^x \) is increasing on the interval:
\[ \left(\frac{1}{e}, \infty\right) \]