The function f(x) = 2ln|x| - x|x| is increasi | vedclass

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(A) The function is defined as $f(x) = \begin{cases} 2\ln(x) - x^2 & 	ext{if } x > 0 \ 2\ln(-x) + x^2 & 	ext{if } x < 0 \end{cases}$.To find the in

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$(0, 1)$

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(A) The function is defined as $f(x) = \begin{cases} 2\ln(x) - x^2 & 	ext{if } x > 0 \ 2\ln(-x) + x^2 & 	ext{if } x To find the intervals where the function is increasing,we calculate the derivative $f'(x)$:For $x > 0$,$f'(x) = \frac{2}{x} - 2x = \frac{2(1 - x^2)}{x}$.For $x For the function to be increasing,we require $f'(x) > 0$.Case $1$: $x > 0$. We need $\frac{2(1 - x^2)}{x} > 0$. Since $x > 0$,this implies $1 - x^2 > 0$,which means $x^2 Case $2$: $x  0$. Since $x  0$,the expression $\frac{2(1 + x^2)}{x}$ is always negative for $x Thus,the function is increasing only on the interval $(0, 1)$.

The function $f(x) = 2\ln|x| - x|x|$ is increasing on the interval

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