The function f(x) = x^x is a decreasing funct | vedclass

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(D) To determine the interval where the function $f(x) = x^x$ is decreasing,we find its derivative.Let $y = x^x$.Taking the natural logarithm on both

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$(0, 1/e)$

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(D) To determine the interval where the function $f(x) = x^x$ is decreasing,we find its derivative.Let $y = x^x$.Taking the natural logarithm on both sides,we get $\ln(y) = x \ln(x)$.Differentiating with respect to $x$,we have $\frac{1}{y} \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$.Thus,$f'(x) = \frac{dy}{dx} = y(\ln(x) + 1) = x^x(\ln(x) + 1)$.For the function to be decreasing,we require $f'(x) Since $x^x > 0$ for all $x \in R^{+}$,the condition $f'(x) This gives $\ln(x) Given $x \in R^{+}$,the interval is $(0, 1/e)$.

The function $f(x) = x^x$ is a decreasing function in the interval of . . . . . . . $(x \in R^{+})$

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