How do you graph and solve #5 + |1-x/2| >=8#?

1 Answer
Dec 21, 2015

Answer:

#x >= 8 " or " x <= -4#

Explanation:

1) Simplifying

First of all, bring #5# to the other side. You can do so by subtracting #5# on both sides of the inequality:

#5 + abs(1 - x/2) >= 8#

#<=> abs(1 - x/2) >= 3 #

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2) Evaluating the absolute value function

To evaluate the absolute value function, we need to find out when #1 - x/2 >= 0# holds and when # 1 - x/2 < 0 # holds.

To do this, let's find the point where #1 - x/2 = 0#:

# 1 - x/2 = 0 " " <=> " " x/2 = 1 " " <=> " " x = 2#

Plugging #x = 1# and #x = 3# gives:

# 1 - x/2 >= 0 color(white)(xxx) "for " x <= 2#

# 1 - x/2 < 0 color(white)(xxx) "for " x > 2#

Now, you can evaluate the absolute value function:

# abs(1 - x/2) = { (color(white)(xx) 1 - x/2, color(white)(xxx) "for " 1 - x/2 >= 0 ), (-(1 - x/2), color(white)(xxx) "for " 1 - x/2 < 0) :}#

# color(white)(xxxxx) = { (color(white)(x) 1 - x/2, color(white)(xxxxx) "for " x <= 2 ), (-1 + x/2, color(white)(xxxxx) "for " x > 2) :}#

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3) Solving the two cases

3a) Let #x <= 2#.

This means that #1 - x/2 >= 0# and #abs(1 - x/2) >= 1 - x/2#.

# => 1 - x/2 >= 3#

... subtract # 1# from both sides of the inequality...

#<=> - x/2 >= 2#

... multiply both sides with #-2#.
Be careful: if multiplying with a negative number or dividing by a negative number, you need to flip the inequality sign!

#<=> x <= - 4#

Now, we need to combine the condition #x <= 2# with the solution #x <= -4#.
As #x <= -4# is the more restrictive condition, this is the solution for this case.

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3b) Let #x > 2#.

This means that #1 - x/2 < 0# and #abs(1 - x/2) = - 1 + x/2#.

# => - 1 + x/2 >= 3#

... add #1# to both sides of the inequality...

# <=> x/2 >= 4#

... multiply both sides of the inequality with #2#...

#<=> x >= 8#

Between the two conditions, #x >= 8# and #x > 2#, the former is the more restrictive condition.

Thus, this is the solution for the second case.

In total, the solution is #x >= 8 " or " x <= -4#

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4) Graphing

You can graph the absolute value function #abs(1 - x/2)# using the "elbow" of the absolute value function and the slope:

  • the "elbow" is the point of the function where #1 - x/2 = 0# holds which is #x = 2#. Thus, the elbow is #(2; 0)#.
  • The slope is the factor of #x/2#, so it's #1/2#.

Thus the absolute function looks as follows:

graph{abs(1 - x/2) [-10, 10, -5, 5]}

The graph of

#abs(1 - x/2) >= 3#

is the part of the graph that is above the horizontal line at #y = 3#:

graph{(y - abs(1 - x/2))(y - 3) = 0 [-15, 15, -5, 10]}

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Hope that this helped! :-)