Determine the values of x for which f(x) = xₓ, x > 0 is increasing or decreasing.

To determine the values of x for which the function f(x) = x^x is increasing or decreasing, we need to analyze its derivative.

Finding the Derivative

First, we can express the function in a more manageable form using logarithmic differentiation:

• Let y = x^x.
• Taking the natural logarithm of both sides gives us: ln(y) = x ln(x).
• Now, differentiate both sides with respect to x:

Applying the Product Rule

Using the product rule on the right side:

• 1/y * dy/dx = ln(x) + 1.
• Thus, dy/dx = y(ln(x) + 1) = x^x(ln(x) + 1).

Analyzing the Derivative

Now, we need to find when this derivative is positive or negative:

• f'(x) = x^x(ln(x) + 1).
• Since x^x is always positive for x > 0, the sign of f'(x) depends on the term (ln(x) + 1).

Setting the Derivative to Zero

To find critical points, set ln(x) + 1 = 0:

• ln(x) = -1.
• This implies x = e^(-1) = 1/e.

Determining Intervals of Increase and Decrease

Now, we can analyze the sign of ln(x) + 1 in the intervals:

• For x < 1/e, ln(x) < -1, so f'(x) < 0 (decreasing).
• For x > 1/e, ln(x) > -1, so f'(x) > 0 (increasing).

Summary of Findings

In conclusion:

• f(x) = x^x is decreasing for 0 < x < 1/e.
• f(x) = x^x is increasing for x > 1/e.