The interval in which the function f(x) = 2x^ | vedclass

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(D) Given function is $f(x) = 2x^2 - \log x$ for $x > 0$. To find the interval where the function decreases,we find the derivative $f'(x)$. $f'(x) = \

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$\left(0, \frac{1}{2}\right)$

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(D) Given function is $f(x) = 2x^2 - \log x$ for $x > 0$. To find the interval where the function decreases,we find the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(2x^2 - \log x) = 4x - \frac{1}{x} = \frac{4x^2 - 1}{x}$. For the function to be decreasing,we set $f'(x) $\frac{4x^2 - 1}{x} Since $x > 0$,the denominator is always positive. Thus,we must have $4x^2 - 1 $4x^2 Taking the square root,we get $|x| Given the condition $x > 0$,the interval becomes $0 Therefore,the function decreases in the interval $\left(0, \frac{1}{2}\right)$.

The interval in which the function $f(x) = 2x^2 - \log x$,for $x > 0$ decreases,is

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