To determine the interval where the function \( f(x) = x^x \) decreases, we first find the derivative, since a function decreases when its derivative is less than zero.
Rewrite the function as \( f(x) = e^{x \ln x} \). Using the natural logarithm:
\( \ln y = x \ln x \)
Differentiating both sides with respect to \( x \) using implicit differentiation:
\( \frac{dy}{dx} = \frac{d}{dx}(e^{x \ln x}) = e^{x \ln x} \cdot \frac{d}{dx}(x \ln x) \)
Now, differentiate \( x \ln x \):
\( \frac{d}{dx}(x \ln x) = \ln x + 1 \)
Thus, the derivative becomes:
\( \frac{dy}{dx} = x^x (\ln x + 1) \)
To find where \( f(x) \) decreases, solve \( \frac{dy}{dx} < 0 \):
\( x^x (\ln x + 1) < 0 \)
Since \( x^x > 0 \) for all \( x > 0 \), the sign of the derivative depends on \( \ln x + 1 \):
Solve \( \ln x + 1 < 0 \Rightarrow \ln x < -1 \Rightarrow x < \frac{1}{e} \)
However, the domain of \( f(x) = x^x \) is \( x > 0 \), so the decreasing interval is: