`x^x` has a stationary point at

Options

A. x = e
B. x = 1/e
C. x = 1
D. `x = sqrte`

Answer

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

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

### Step 2: Take the natural logarithm
To differentiate \( f(x) \), we first take the natural logarithm of both sides:
\[ \ln(f(x)) = \ln(x^x) \]
Using the logarithmic identity \( \ln(a^b) = b \ln(a) \), we can rewrite this as:
\[ \ln(f(x)) = x \ln(x) \]

### Step 3: Differentiate using the chain rule
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:
\[ \frac{1}{f(x)} \cdot f'(x) = \ln(x) + 1 \]

### Step 4: Solve for \( f'(x) \)
Now, we can solve for \( f'(x) \):
\[ f'(x) = f(x) \cdot (\ln(x) + 1) \]
Substituting back \( f(x) = x^x \):
\[ f'(x) = x^x (\ln(x) + 1) \]

### Step 5: Set the derivative to zero
To find the stationary points, we set \( f'(x) = 0 \):
\[ x^x (\ln(x) + 1) = 0 \]
Since \( x^x \) is never zero for \( x > 0 \), we focus on the term \( \ln(x) + 1 = 0 \):
\[ \ln(x) = -1 \]

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

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