f(x) = x^{x} has a stationary point at | vedclass

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(B) Given the function $f(x) = x^{x}$.Taking the natural logarithm on both sides,we get $\ln f(x) = x \ln x$.Differentiating both sides with respect t

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the function $f(x) = x^{x}$.Taking the natural logarithm on both sides,we get $\ln f(x) = x \ln x$.Differentiating both sides with respect to $x$ using the chain rule and product rule:$\frac{1}{f(x)} f'(x) = \ln x + x \cdot \frac{1}{x} = \ln x + 1$.Thus,$f'(x) = f(x)(\ln x + 1) = x^{x}(\ln x + 1)$.$A$ stationary point occurs where $f'(x) = 0$.Since $x^{x} > 0$ for $x > 0$,we set $\ln x + 1 = 0$.$\ln x = -1$.Converting from logarithmic to exponential form,$x = e^{-1} = \frac{1}{e}$.

$f(x) = x^{x}$ has a stationary point at

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