The function ` g (x) = x^(x)` has a critical point at

Options

A. x = e
B. x = 1
C. ` x = (1)/( e )`
D. ` x = sqrt(6)`

Answer

To find the critical points of the function \( g(x) = x^x \), we will follow these steps:

### Step 1: Find the derivative \( g'(x) \)
To find the critical points, we first need to compute the derivative of \( g(x) \). Since \( g(x) = x^x \) is a variable raised to a variable, we can use logarithmic differentiation.

1. Take the natural logarithm of both sides:
\[ \log(g(x)) = \log(x^x) \]

2. Using the property of logarithms, we can simplify the right-hand side:
\[ \log(g(x)) = x \log(x) \]

### Step 2: Differentiate both sides
Now, we differentiate both sides with respect to \( x \):
1. The left-hand side using the chain rule:
\[ \frac{1}{g(x)} g'(x) \]
2. The right-hand side using the product rule:
\[ \frac{d}{dx}(x \log(x)) = \log(x) + 1 \]

Putting it together, we have:
\[ \frac{1}{g(x)} g'(x) = \log(x) + 1 \]

### Step 3: Solve for \( g'(x) \)
Now, we can solve for \( g'(x) \):
\[ g'(x) = g(x) (\log(x) + 1) \]
Substituting \( g(x) = x^x \):
\[ g'(x) = x^x (\log(x) + 1) \]

### Step 4: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ x^x (\log(x) + 1) = 0 \]
Since \( x^x \) is never zero for \( x > 0 \), we can focus on the term \( \log(x) + 1 = 0 \):
\[ \log(x) + 1 = 0 \implies \log(x) = -1 \]

### Step 5: Solve for \( x \)
To solve for \( x \), we exponentiate both sides:
\[ x = e^{-1} = \frac{1}{e} \]

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