Increasing/decreasing intervals and local extrema for f(x) = |x^2 - 4|

Since $x^2 - 4 \ge 0$ for $|x|\ge 2$ and $x^2-4<0$ for $|x|<2$, we can write $f(x) = x^2-4$ for $|x|\ge 2$ and $f(x) = 4-x^2$ for $-2<x<2$. Differentiating each piece gives $f'(x) = 2x$ for $|x|>2$ and $f'(x) = -2x$ for $-2<x<2$; the derivative fails to exist at $x=\pm2$ due to the corners, so $x = -2, 0, 2$ are the critical numbers.

Checking signs: on $(-\infty,-2)$, $f'(x)=2x<0$ (decreasing); on $(-2,0)$, $f'(x)=-2x>0$ (increasing); on $(0,2)$, $f'(x)=-2x<0$ (decreasing); on $(2,\infty)$, $f'(x)=2x>0$ (increasing). By the First Derivative Test, $f'$ changes negative-to-positive at $x=-2$ and $x=2$, giving local minima there (value $0$), and changes positive-to-negative at $x=0$, giving a local maximum there (value $4$).

Therefore $f$ is decreasing on $(-\infty,-2)$ and $(0,2)$, increasing on $(-2,0)$ and $(2,\infty)$, with local minima at $x=-2$ and $x=2$, and a local maximum at $x=0$.