In which interval is the function f(x) = x^x | vedclass

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(C) Let $y = x^x$. Taking the logarithm on both sides,we get $\log y = x \log x$. Differentiating with respect to $x$,we have $\frac{1}{y} \frac{dy}{d

Vedclass · Accepted Answer

$(0, 1/e)$

Vedclass · Answer

(C) Let $y = x^x$. Taking the logarithm on both sides,we get $\log y = x \log x$. Differentiating with respect to $x$,we have $\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. Thus,$\frac{dy}{dx} = x^x(1 + \log x)$. For the function to be decreasing,we must have $\frac{dy}{dx} Since $x^x > 0$ for $x > 0$,the condition $\frac{dy}{dx} This simplifies to $\log x Therefore,$x Since the domain of $x^x$ is $x > 0$,the interval in which the function is decreasing is $(0, 1/e)$.

In which interval is the function $f(x) = x^x$ a decreasing function?

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