The function `f(x) = x^(x) , x to 0` , has a stationary point at

Options

A. x = e
B. ` x = (1)/( e)`
C. x = 1
D. ` x = sqrt(e)`

Answer

To find the stationary point of the function \( f(x) = x^x \) as \( x \) approaches 0, we need to follow these steps:

### Step 1: Rewrite the function using logarithms
We start by taking the natural logarithm of both sides:
\[ \ln(f(x)) = \ln(x^x) = x \ln(x) \]

### Step 2: Differentiate both sides
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\ln(f(x))) = \frac{d}{dx}(x \ln(x)) \]
Using the chain rule on the left side and the product rule on the right side, we get:
\[ \frac{1}{f(x)} \frac{df}{dx} = \ln(x) + 1 \]

### Step 3: Solve for \( f'(x) \)
Now, we can solve for \( f'(x) \):
\[ \frac{df}{dx} = f(x)(\ln(x) + 1) \]
Substituting back \( f(x) = x^x \):
\[ f'(x) = x^x (\ln(x) + 1) \]

### Step 4: Set the derivative to zero
To find the stationary points, we set the derivative equal to zero:
\[ x^x (\ln(x) + 1) = 0 \]
Since \( x^x \) is never zero for \( x > 0 \), we focus on the other factor:
\[ \ln(x) + 1 = 0 \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ \ln(x) = -1 \]
Taking the exponential of both sides:
\[ x = e^{-1} = \frac{1}{e} \]

### Conclusion
Thus, the stationary point of the function \( f(x) = x^x \) as \( x \) approaches 0 is:
\[ x = \frac{1}{e} \]