The interval in which the function y = |x^2 - | vedclass

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(C) Let $f(x) = |x^2 - |x| - 2|$.Since $f(x)$ is an even function,we analyze it for $x \ge 0$,where $f(x) = |x^2 - x - 2| = |(x-2)(x+1)|$.For $x \ge 0

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$x \in (0, 2)$

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(C) Let $f(x) = |x^2 - |x| - 2|$.Since $f(x)$ is an even function,we analyze it for $x \ge 0$,where $f(x) = |x^2 - x - 2| = |(x-2)(x+1)|$.For $x \ge 0$,$x+1 > 0$,so $f(x) = |x-2|(x+1)$.If $0 \le x The derivative is $f'(x) = -2x + 1$.Setting $f'(x) = 0$ gives $x = 1/2$.The function increases on $(0, 1/2)$ and decreases on $(1/2, 2)$.Thus,the function is non-monotonic in any interval containing $1/2$,such as $(0, 2)$.Looking at the options,$x \in (0, 2)$ is the correct interval where the function is non-monotonic.

The interval in which the function $y = |x^2 - |x| - 2|$ is non-monotonic is:

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