Question

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

Answer 1

`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! :-)