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

Options

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

Explanation

To find the stationary point of the function \( f(x) = x^x \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = x^x \]

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

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

### Step 4: Solve for \( \frac{df}{dx} \)
Now, we can express \( \frac{df}{dx} \):
\[ \frac{df}{dx} = f(x) \cdot (\log x + 1) \]
Substituting back \( f(x) = x^x \):
\[ \frac{df}{dx} = x^x (\log x + 1) \]

### Step 5: Set the derivative to zero for stationary points
To find the stationary points, we set the derivative equal to zero:
\[ x^x (\log x + 1) = 0 \]
Since \( x^x \) is never zero for \( x > 0 \), we only need to solve:
\[ \log x + 1 = 0 \]

### Step 6: Solve for \( x \)
This simplifies to:
\[ \log x = -1 \]
Exponentiating both sides gives:
\[ x = e^{-1} = \frac{1}{e} \]

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