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

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(C) Given $f(x) = 2x^2 - \ln |x|$.To find the interval where the function is monotonically decreasing,we need $f'(x) < 0$.First,find the derivative: $

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $f(x) = 2x^2 - \ln |x|$.To find the interval where the function is monotonically decreasing,we need $f'(x) First,find the derivative: $f'(x) = 4x - \frac{1}{x}$.Set the derivative to be less than zero: $4x - \frac{1}{x} Simplify the expression: $\frac{4x^2 - 1}{x} Factor the numerator: $\frac{(2x - 1)(2x + 1)}{x} To solve this inequality,find the critical points by setting the numerator and denominator to zero: $x = 1/2, x = -1/2, x = 0$.Test the intervals $(-\infty, -1/2)$,$(-1/2, 0)$,$(0, 1/2)$,and $(1/2, \infty)$.For $x \in (-\infty, -1/2)$,let $x = -1$: $\frac{(2(-1)-1)(2(-1)+1)}{-1} = \frac{(-3)(-1)}{-1} = -3 For $x \in (-1/2, 0)$,let $x = -1/4$: $\frac{(2(-1/4)-1)(2(-1/4)+1)}{-1/4} = \frac{(-1.5)(0.5)}{-0.25} = 3 > 0$ (Not satisfied).For $x \in (0, 1/2)$,let $x = 1/4$: $\frac{(2(1/4)-1)(2(1/4)+1)}{1/4} = \frac{(-0.5)(1.5)}{0.25} = -3 For $x \in (1/2, \infty)$,let $x = 1$: $\frac{(2(1)-1)(2(1)+1)}{1} = \frac{(1)(3)}{1} = 3 > 0$ (Not satisfied).Thus,the function is monotonically decreasing in the interval $(-\infty, -1/2) \cup (0, 1/2)$.

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

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