If f(x)=x^x,then the interval in which f(x) d | vedclass

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(A) Given $f(x) = x^x$. Taking the natural logarithm on both sides,we get $\ln(f(x)) = x \ln(x)$.Differentiating both sides with respect to $x$,we hav

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$\left[0, \frac{1}{e}\right]$

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(A) Given $f(x) = x^x$. Taking the natural logarithm on both sides,we get $\ln(f(x)) = x \ln(x)$.Differentiating both sides with respect to $x$,we have $\frac{f'(x)}{f(x)} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$.Thus,$f'(x) = x^x(1 + \ln(x))$.For $f(x)$ to be a decreasing function,we must have $f'(x) \leq 0$.Since $x^x > 0$ for all $x > 0$,the condition $f'(x) \leq 0$ implies $1 + \ln(x) \leq 0$.This gives $\ln(x) \leq -1$,which means $x \leq e^{-1} = \frac{1}{e}$.Since the domain of $f(x) = x^x$ is $x > 0$,the interval where $f(x)$ decreases is $\left(0, \frac{1}{e}\right]$.Comparing with the given options,the correct interval is $\left[0, \frac{1}{e}\right]$.

If $f(x)=x^x$,then the interval in which $f(x)$ decreases is

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