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

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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 $f'(x)$. $f'(x) = \frac{d}{dx}(2x^2 - \log x) = 4x - \frac{1}{x}$. The function $f(x)$ is decreasing when $f'(x) $4x - \frac{1}{x} $\Rightarrow \frac{4x^2 - 1}{x} Since $x > 0$,we can multiply by $x$ without changing the inequality sign: $4x^2 - 1 $\Rightarrow 4x^2 $\Rightarrow x^2 $\Rightarrow |x| This gives $x \in (-\frac{1}{2}, \frac{1}{2})$. Given the condition $x > 0$,the interval is $x \in (0, \frac{1}{2})$. Thus,the function decreases in the interval $(0, \frac{1}{2})$.

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

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