# How do you solve and write the following in interval notation: |x - 4| <12?

Apr 3, 2017

$\left(- 8 , 16\right) .$

#### Explanation:

We have to cosider

$2 \text{ Cases : } \left(1\right) : \left(x - 4\right) \ge 0 , \left(2\right) : \left(x - 4\right) < 0.$

This is, because of the Defn. of the Absolute Value Fun. :

 : x to |x| : RR to RR^+uu{0}, where, |x|=x; if x >=0, &, |x|=-x; if x<0.

Case (1) : $\left(x - 4\right) \ge 0.$

(x-4) >=0 rArr |x-4|=x-4, &, because |x-4| < 12, :. x-4 < 12.

$\therefore , \text{ in this Case, } x < 16. \ldots \ldots \left[1\right] .$

Case (2) : $\left(x - 4\right) < 0.$

Here, since, "|x-4|=-(x-4)=4-x, so, |x-4| < 12 rArr 4-x <12.

$\therefore , \text{ in this Case, } - 8 < x \ldots \ldots \ldots \ldots \ldots . \left[2\right] .$

Combining $\left[1\right] \mathmr{and} \left[2\right]$, we get, $- 8 < x < 16.$

This is written, in the Interval Notation, as $\left(- 8 , 16\right) .$

A very quick Soln. can easily be obtained, if we use the

Result :

0 < |x-a| < delta iff a-delta < x < a+delta; x,a in RR, delta in RR^+.#

Enjoy Mths.!