How do you graph and solve #abs(x+1)<=11#?

1 Answer
Jun 7, 2016

Answer:

Solution:
#-12 <= x <= 10#

Explanation:

Start from a definition of the absolute value.
For non-negative real #x# the value of #abs(x)# is #x#.
For negative real #x# the value of #abs(x)# is #-x#.

In short:
if #x>=0#, #|x| = x#
if #x<0#, #|x| = -x#

Using this definition, let's divide the domain of the #|x+1|# in two intervals:
1. #x+1 >= 0#, where #|x+1| = x+1#
2. #x+1 < 0#, where #|x+1| = -(x+1)#

Case 1. Looking for solutions in interval #x+1 >= 0# (or, equivalently, #x >= -1#)
Since in this interval #|x+1| = x+1#, our inequality looks like
#x+1 <= 11# and the solution is #x <= 10#.
Together with a boundary of the interval, we have come to a solution:
#-1 <= x <= 10#

Case 2. Looking for solutions in interval #x+1 < 0# (or, equivalently, #x < -1#)
Since in this interval #|x+1| = -(x+1)#, our inequality looks like
#-(x+1) <= 11# and the solution is #x >= -12#.
Together with a boundary of the interval, we have come to a solution:
#-12 <= x < -1#

So, two intervals are the solution to an original inequality:
#-1 <= x <= 10# and
#-12 <= x < -1#

These can be combined into
#-12 <= x <= 10#

Here is a graph of #y=|x+1|#

graph{|x+1| [-20, 20, -15, 15]}
It illustrates the solution