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(B) Let $y = x^x$. Taking the natural logarithm on both sides,we get $\log y = x \log x$ for $x > 0$.Differentiating both sides with respect to $x$,we

Vedclass · Accepted Answer

$x = \frac{1}{e}$

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(B) Let $y = x^x$. Taking the natural logarithm on both sides,we get $\log y = x \log x$ for $x > 0$.Differentiating both sides 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 a stationary point,we set $\frac{dy}{dx} = 0$.Since $x^x > 0$,we must have $1 + \log x = 0$,which implies $\log x = -1$.Therefore,$x = e^{-1} = \frac{1}{e}$.To verify,we find the second derivative $\frac{d^2y}{dx^2} = x^x(1 + \log x)^2 + x^x \cdot \frac{1}{x}$.At $x = \frac{1}{e}$,$\frac{d^2y}{dx^2} = (\frac{1}{e})^{1/e}(0)^2 + (\frac{1}{e})^{1/e} \cdot e = e \cdot (\frac{1}{e})^{1/e} > 0$.Since the second derivative is positive,$y$ has a minimum at $x = \frac{1}{e}$.

$x^x$ has a stationary point at

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