Question

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

Answer 1

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.