sure
ok, separate f(x) = |x+1| into two functions
first is f(x) = x+1
the other is f(x) = -(x+1)
find where they intersect so you know where one stops and the other begins
x+1 = -x - 1
2x = -2
x=-1
and we know that it's essentially linear, so each function is real from (-infinity, infinity)
however, since they intersect at x=-1, we have a point discontinuity at x=-1 (there's a way to prove this, I forget exactly how, but if you just look at the graph you'll see it). There exists no derivative for f(x) at a discontinuity.
so the interval for dy/dx (if we let y=f(x) is (-infinity, -1), (-1, infinity). Now we want to find where f(x) increases, so find the derivative of each function from earlier
dy/dx = 1
dy/dx = -1
f(x) increases when its derivative is positive. 1>0 for all real values, -1 is never >0. So only the first function (f(x) = x+1) increases. |x+1| = x+1 on the interval from (-1, infinity).
f(x) increases on the interval (-1, infinity)
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~Peaf~