The function f(x) = x^x is increasing when: | vedclass

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

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

Vedclass · Answer

(A) Let $f(x) = x^x$ for $x > 0$.Taking the natural logarithm on both sides,we get $\ln(f(x)) = x \ln x$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{f(x)} f'(x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.Thus,$f'(x) = x^x(1 + \ln x)$.For the function to be increasing,we require $f'(x) > 0$.Since $x^x > 0$ for all $x > 0$,the condition $f'(x) > 0$ simplifies to $1 + \ln x > 0$.$\ln x > -1$.$\ln x > \ln(e^{-1})$.$x > \frac{1}{e}$.

The function $f(x) = x^x$ is increasing when:

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