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

1 Answer
Apr 3, 2017

Answer:

#(-8,16).#

Explanation:

We have to cosider

#2" Cases : "(1) : (x-4) >=0, (2) : (x-4) < 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) : #(x-4) >=0.#

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

#:.," in this Case, "x < 16.......[1].#

Case (2) : #(x-4) < 0.#

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

#:.," in this Case, "-8 < x................[2].#

Combining #[1] and [2]#, we get, #-8 < x < 16.#

This is written, in the Interval Notation, as #(-8,16).#

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.!