In which interval does the function f(x) = 2x | vedclass

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(D) The function is $f(x) = 2x^2 - \log |x|$.To find the interval of monotonic increase,we find the derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2x^2 - \l

Vedclass · Accepted Answer

$(-1/2, 0) \cup (1/2, \infty)$

Vedclass · Answer

(D) The function is $f(x) = 2x^2 - \log |x|$.To find the interval of monotonic increase,we find the derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2x^2 - \log |x|) = 4x - \frac{1}{x}$.For the function to be monotonically increasing,we require $f'(x) > 0$.$4x - \frac{1}{x} > 0 \implies \frac{4x^2 - 1}{x} > 0$.This inequality holds when the numerator and denominator have the same sign.Case $1$: $x > 0$. Then $4x^2 - 1 > 0 \implies x^2 > 1/4 \implies x > 1/2$.Case $2$: $x Combining these,the function increases in the interval $(-1/2, 0) \cup (1/2, \infty)$.

In which interval does the function $f(x) = 2x^2 - \log |x|$ $(x \neq 0)$ monotonically increase?

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