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
Let \( y = f(x) = x^x \).

### Step 2: Take the natural logarithm of both sides
Taking the logarithm, we have:
\[ \log y = \log(x^x) \]
Using the property of logarithms, we can simplify this to:
\[ \log y = x \log x \]

### Step 3: Differentiate both sides
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\log y) = \frac{d}{dx}(x \log x) \]
Using the chain rule on the left side, we get:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \log x) \]
For the right side, we apply the product rule:
\[ \frac{d}{dx}(x \log x) = \log x + 1 \]
Thus, we have:
\[ \frac{1}{y} \frac{dy}{dx} = \log x + 1 \]

### Step 4: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y(\log x + 1) \]
Substituting back \( y = x^x \):
\[ \frac{dy}{dx} = x^x(\log x + 1) \]

### Step 5: Find the stationary points
To find the stationary points, we set \( \frac{dy}{dx} = 0 \):
\[ x^x(\log x + 1) = 0 \]
Since \( x^x \) is never zero for \( x > 0 \), we only need to solve:
\[ \log x + 1 = 0 \]
This simplifies to:
\[ \log x = -1 \]
Exponentiating both sides gives:
\[ x = e^{-1} = \frac{1}{e} \]

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

---