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# Absolute Value Inequalities

## Key Questions

• There are many facets to this question. I will address a simply one and let the questioner expand to a specific example he/she may have.

Assume: $\left\mid f \left(x\right) \right\mid < a$ $\left[a \in \mathbb{R}\right]$

Then, either $f \left(x\right) < a \mathmr{and} - f \left(x\right) < a$

Applying the rules of inequalities, either $f \left(x\right) < a \mathmr{and} f \left(x\right) > - a$

Which leads to the compound inequality: $- a < f \left(x\right) < a$

Therefore in solving absolute value inequalities of this and similar forms simply consider both positive and negative possibilities of the function and solve for each.

Example: $\left\mid x - 3 \right\mid < 5$

Either $\left(x - 3\right) < 5 \to x < 8$

Or $- \left(x - 3\right) < 5 \to \left(x - 3\right) > - 5 \to x > - 2$

Thus: $- 2 < x < 8$

Which can be expressed in interval notation as: $x \in \left(- 2 , 8\right)$

• Example

The solution set of $| x | > - 1$ is the set of all real numbers.

I hope that this was helpful.

• If all the terms of the inequality are absolute values, then Yes, but if the inequality contains a term which is not force to an absolute value then No.

If all terms of an inequality are absolute values then the only way either side could be negative is if the collection of terms on that side contained a subtraction. For example:
$| a | - | b | < | c |$
But such an inequality could always be written without the subtraction by adding an amount equal to that being subtracted to each side
$\left(| a | - | b | < | c |\right) \to \left(| a | < | c | + | b |\right)$

Since this is the case one side of the inequality must have an absolute value that is less than (or less than or equal to) some positive number. Lets call this positive number $K$.

The minimal side of the inequality must be $< \left(K\right)$ AND $> \left(- K\right)$ (a compound relationship).

Note however, if any of the terms are not absolute values, this does not apply. For example:
$a < | b |$
is not a compound relationship; $a$ is not restricted except by an upper limit of $| b |$.

• This key question hasn't been answered yet.
• This key question hasn't been answered yet.

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