The function `f (x)= x^x` decreases on the interval

Options

A. `(0,e)`
B. `(0,1)`
C. `(0,1//e)`
D. none of these

Answer

To determine the interval on which the function \( f(x) = x^x \) decreases, we will follow these steps:

### Step 1: Differentiate the function
We start by taking the natural logarithm of both sides:
\[ \log(f(x)) = \log(x^x) \]
Using the property of logarithms, we can simplify the right-hand side:
\[ \log(f(x)) = x \log(x) \]

### Step 2: Differentiate both sides
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:
\[ \frac{1}{f(x)} f'(x) = \log(x) + 1 \]

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

### Step 4: Determine when \( f'(x) < 0 \)
The function \( f(x) \) is decreasing when \( f'(x) < 0 \):
\[ x^x (\log(x) + 1) < 0 \]
Since \( x^x > 0 \) for \( x > 0 \), we can focus on the term \( \log(x) + 1 < 0 \):
\[ \log(x) < -1 \]

### Step 5: Solve the inequality
Exponentiating both sides gives:
\[ x < e^{-1} = \frac{1}{e} \]

### Step 6: Combine the conditions
We also have the condition that \( x > 0 \) (since the logarithm is defined only for positive \( x \)). Therefore, we combine the two inequalities:
\[ 0 < x < \frac{1}{e} \]

### Final Answer
Thus, the function \( f(x) = x^x \) decreases on the interval:
\[ (0, \frac{1}{e}) \]