Originally Posted by

**topsquark** Worst comes to worst, pick a bunch of x values and find the g(x)'s.

$\displaystyle g(x) = |x^2 - 1| - |x^2 - 4|$

$\displaystyle g(x) = |(x + 1)(x - 1)| - |(x + 2)(x - 2)|$

Critical points are where the arguments of the absolute value bars is zero, so I have critical points at $\displaystyle x = -2, -1, 1, 2$.

For $\displaystyle (-\infty, -2)$ $\displaystyle (x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $\displaystyle (x + 2)(x - 2) > 0 \implies |(x + 2)(x - 2)| = (x + 2)(x - 2)$.

So on $\displaystyle (-\infty, -2)$ $\displaystyle g(x) = (x + 1)(x - 1) - (x + 2)(x - 2) = 3$

For $\displaystyle (-2, -1)$ $\displaystyle (x + 1)(x - 1) > 0 \implies |(x + 1)(x - 1)| = (x + 1)(x - 1)$ and $\displaystyle (x + 2)(x - 2) < 0 \implies |(x + 2)(x - 2)| = -(x + 2)(x - 2)$.

So on $\displaystyle (-2, -1)$ $\displaystyle g(x) = (x + 1)(x - 1) + (x + 2)(x - 2) = 2x^2 - 5$

So far we have:

$\displaystyle g(x) = \begin{cases} 3, \text{ for }-\infty \leq x \\ 2x^2 - 5, \text{ for } -2 < x \leq -1 \end{cases} $

I'll let you figure the other two intervals.

-Dan