Find the values of 'x' for which `f(x)=x^(x),x gt 0` is strictly increasing or decreasing.

Answer

To find the values of \( x \) for which the function \( f(x) = x^x \) (where \( x > 0 \)) is strictly increasing or decreasing, we will follow these steps:

### Step 1: Find the derivative \( f'(x) \)

We start with the function:
\[ f(x) = x^x \]

To differentiate \( f(x) \), we take the natural logarithm of both sides:
\[ \ln(f(x)) = \ln(x^x) = x \ln(x) \]

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:
\[ \frac{1}{f(x)} f'(x) = \ln(x) + 1 \]

Now, multiplying both sides by \( f(x) \) gives:
\[ f'(x) = f(x)(\ln(x) + 1) \]

Substituting back \( f(x) = x^x \):
\[ f'(x) = x^x (\ln(x) + 1) \]

### Step 2: Determine when \( f'(x) > 0 \) or \( f'(x) < 0 \)

For \( f(x) \) to be strictly increasing, we need \( f'(x) > 0 \):
\[ x^x (\ln(x) + 1) > 0 \]

Since \( x^x > 0 \) for \( x > 0 \), we only need to analyze the term \( \ln(x) + 1 \):
\[ \ln(x) + 1 > 0 \implies \ln(x) > -1 \implies x > e^{-1} = \frac{1}{e} \]

For \( f(x) \) to be strictly decreasing, we need \( f'(x) < 0 \):
\[ x^x (\ln(x) + 1) < 0 \]
Again, since \( x^x > 0 \), we have:
\[ \ln(x) + 1 < 0 \implies \ln(x) < -1 \implies x < e^{-1} = \frac{1}{e} \]

### Step 3: Conclusion

Thus, we conclude:
- The function \( f(x) = x^x \) is **strictly increasing** for \( x > \frac{1}{e} \).
- The function \( f(x) = x^x \) is **strictly decreasing** for \( 0 < x < \frac{1}{e} \).

### Summary of Results
- **Strictly Increasing**: \( x \in \left(\frac{1}{e}, \infty\right) \)
- **Strictly Decreasing**: \( x \in \left(0, \frac{1}{e}\right) \)