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

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(C) Given the function $f(x) = 2x^2 - \ln |x|$.For the function to be monotonically decreasing,we must have $f'(x) < 0$.First,find the derivative: $f'

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the function $f(x) = 2x^2 - \ln |x|$.For the function to be monotonically decreasing,we must have $f'(x) First,find the derivative: $f'(x) = 4x - \frac{1}{x}$.Set the derivative to be less than zero: $4x - \frac{1}{x} This simplifies to $\frac{4x^2 - 1}{x} To solve this inequality,we find the critical points where the numerator or denominator is zero: $x = -1/2, 0, 1/2$.Testing the intervals:For $x \in (-\infty, -1/2)$,$f'(x) For $x \in (-1/2, 0)$,$f'(x) > 0$.For $x \in (0, 1/2)$,$f'(x) For $x \in (1/2, \infty)$,$f'(x) > 0$.Thus,the function is monotonically decreasing in the interval $(-\infty, -1/2) \cup (0, 1/2)$.

In which interval is the function $f(x) = 2x^2 - \ln |x|$ $(x \ne 0)$ monotonically decreasing?

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