Find interval in which `f(x) = x^x`, `x > 0` is strictly increasing

Options

A. `(1/e, infty)`
B. `(2, infty)`
C. `(1/4, infty)`
D. `(1/(16), infty)`

Answer

To find the interval in which the function \( f(x) = x^x \) (for \( x > 0 \)) is strictly increasing, we need to follow these steps:

### Step 1: Find the derivative of \( f(x) \)
To determine where the function is increasing, we first need to find its derivative. We can express \( f(x) = x^x \) in a more manageable form using logarithms.

Let \( y = x^x \). Taking the natural logarithm of both sides, we get:
\[ \ln y = x \ln x \]

Now, we differentiate both sides with respect to \( x \):
\[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \]

Multiplying both sides by \( y \) gives us:
\[ \frac{dy}{dx} = y (\ln x + 1) = x^x (\ln x + 1) \]

### Step 2: Set the derivative equal to zero
To find the critical points, we set the derivative equal to zero:
\[ x^x (\ln x + 1) = 0 \]

Since \( x^x > 0 \) for \( x > 0 \), we can simplify this to:
\[ \ln x + 1 = 0 \]

### Step 3: Solve for \( x \)
Solving for \( x \) gives:
\[ \ln x = -1 \implies x = e^{-1} = \frac{1}{e} \]

### Step 4: Determine the intervals
Now, we need to analyze the sign of the derivative \( \frac{dy}{dx} = x^x (\ln x + 1) \) around the critical point \( x = \frac{1}{e} \).

1. **For \( 0 < x < \frac{1}{e} \)**:
- Here, \( \ln x < -1 \), so \( \ln x + 1 < 0 \).
- Therefore, \( \frac{dy}{dx} < 0 \) (the function is decreasing).

2. **For \( x > \frac{1}{e} \)**:
- Here, \( \ln x > -1 \), so \( \ln x + 1 > 0 \).
- Therefore, \( \frac{dy}{dx} > 0 \) (the function is increasing).

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