A function is increasing when its derivative is positive. We need to find the derivative of this function, but since it involves an absolute value, we'll need to rewrite it in piecewise form first.

Absolute value functions can be rewritten in a piecewise form based on when the expression inside the absolute value bars is positive or negative. In this case:

#f(x)={(x^2-4", when "x^2-4>=0),(-(x^2-4)", when "x^2-4<0):}#

So our function changes "pieces" when #x^2-4=0#. A little quick algebra shows this to be at #x=+-2#. It's probably not that hard to see that when #x in (-2, 2)#, the value of #x^2-4# is negative. (Simply plugging any value from this interval into #x^2-4# will confirm this.) So our final piecewise form of #f(x)# is:

#f(x)={(x^2-4", "x<"-"2uux>2),(-(x^2-4)", -"2<=x<=2):}#

Now that #f(x)# is in piecewise form, we can take its derivative:

#f'(x)={(2x", "x<"-"2", "x>2),(-2x", -"2< x <2):}#

Notice the change from #<=# to #<# signs in the 2nd piece. That's because when a piecewise function #f# changes pieces at a point #c#, #f'(x)# is *undefined at #c#* when #lim_(x->c^-)f'(x)!=lim_(x->c^+)f'(x)#. (In other words, if the function "pieces" on both sides of #c# disagree on what they think the slope should be at #c#, then there is no slope at that point.)

That is certainly true here at both #x="-"2# and #x=2#, since #2("-"2) !=-2("-"2)# and #2(2) !=-2(2)#. Anyway, the last thing to do is solve for #f'(x)>0#:

#ul("When "x<"-"2uux>2):# (the first piece)

#f'(x)>0=>2x>0=>x>0#

#:.f(x)" increasing on "x>0nn(x<"-"2uux>2)#

#=>f(x)" increasing on "x>2#

#ul("When ""-"2< x <2: )# (the second piece)

#f'(x)>0=>-2x>0=>x<0#

#:.f(x)" increasing on "x<0nn("-"2< x <2)#

#=>f(x)" increasing on -"2< x < 0#

So our final answer is the union of these two intervals:

#f(x)# is increasing on #("-"2,0)uu(2,oo)#.

Similar work can be done to determine when #f(x)# is decreasing, except we set #f'(x) <0# (instead of #> 0#) and solve for #x#.

Here is a graph of #f(x)=abs(x^2-4)#:

graph{abs(x^2-4) [-10, 10, -1.96, 8.04]}