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

1 Answer
Dec 21, 2015

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! :-)