Let #f(x) = 7+|2x-1|#. How do you find all #x# for which #f(x) < 16#?

1 Answer
Douglas K.
Oct 19, 2017

Given: #f(x) = 7 + |2x-1|# and #f(x) < 16#

We can write the inequality:

#7 + |2x-1| < 16#

Subtract 7 from both sides:

#|2x-1| < 9#

Because of the piecewise definition of the absolute value function, #|A| = {(A; A >= 0),(-A; A < 0):}# we can separate the inequality into two inequalities:

#-(2x-1) < 9# and #2x-1 < 9#

Multiply both sides of the first inequality by -1:

#2x-1 > -9# and #2x-1 < 9#

Add 1 to both sides of both inequalities:

#2x > -8# and #2x < 10#

Divide both sides of both inequalities by 2:

#x > -4# and #x < 5#

This can be written as:

#-4 < x < 5#

To check, I will verify that the end points equal 16:

#7 + |2(-4)-1)| = 7 + |-9| = 16#
#7+ |2(5)-1| = 7+|9| = 16#

Both check.

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