Recall the definition of absolute value:

|N| = { N if N>=0,

-n if N<0

}

Defines function y = f(x) = |x+2|-|x-1|-x

We are interested in intervals of x where f(x)<-3/2

Breaks the number line into pieces.....

x < -2

------------

then x+2 < 0 and x-1 <0,

so |x+2| = -(x+2) and |x-1|=-(x-1)

Then function becomes:

-(x+2) - -(x-1) - x =

-(x+2) + (x-1) - x =

-x + -2 + x - 1 -x =

-3 -x

so -3-x < -3/2

3+x > 3/2

x> 3/2 - 3

x>-3/2

But x<-2, so there are no solutions in this interval

becuase x>-3/2 and x<-2 cannot happen at the same time

=============================================================

-2 < x < -1

------------------

then 0 < x+2 < 1 and -3 < x-1 < -2

so then x+2>0 and x-1 < 0

that means |x+2| = x+2 and |x-1| = -(x+1)

then function becomes:

x+2 - -(x-1) - x =

x+2 + x-1 - x =

x + 1

x+1 < -3/2

x < -5/2

x < -5/2

But -2 < x < -1, so there are no solutions in this interval

becuase -2 < x < -1 and x<-5/2 cannot happen at the same time

==================================================================

-1 < x < 0

-------------------

then 1 < x+2 < 2 and -2 < x-1 < -1

SO then x+2 > 0 and x-1 < 0

That means |x+2|=x+2 and |x-1| = -(x-1)

the function becomes

x+2 - -(x-1) - x =

x+2 + (x-1) - x =

x + 1

x+1 < -3/2

x < -5/2

But -1 < x < 0 , so there are no solutions in this interval

because -1 < x < 0

=================================================================

0 <x < 1

------------------

2 < x+2 < 3 or -1 < x-1 < 0

So then x+2>0 and x-1 < 0

that means |x+2| = x+2 and |x-1| = -(x-1)

the function becomes

x+2 - -(x-1) - x =

x+2 + (x-1) -x =

x + 1

As previously seen, twice before xc < -5/2

But 0 < x < 1, so there are not solutions in

this interval becuase x < -5/2 and 0 < x < 1

cannot happen at the same time

===============================================

1 < x < 2

----------------

then 3 < x+2 < 4 and 0 < x-1 < 3

This time both quantities are positive,

so |x+2| = x+2 and |x-1| = x-1

the function becomes

x+2 -(x-1) - x =

x +2 -x + 1 - x =

-x + 3

-x + 3 < -3/2

-x < -3/2 - 3

-x < -3/2 - 6/2

-x < -9/2

x>9/2

But 1 < x < 2, so there are no solutions

in this interval because x>9/2 and 1 <x<2

cannot happen at the same time

========================================================

x>2

then x+2 > 4 and x-1 > 1

Again both quantities are positive,

so |x+2| = x+2 and |x-1| = x-1

x+2 -(x-1) - x =

x+2 -x + 1 - x =

-x + 3

per the previous case, x>9/2

so x>9/2 IS a solution

If you examine the graph of y = f(x) = |x+2|-|x-1|-x

you will see that the graph drops below -3/2 when x>9/2.

But there are the analytics