082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

2.4 Review of some common engineering functions and techniques 105
f (x)
a
0
b
x
|a – b|
Figure2.38
The function: f (x) = |x|.
Figure2.39
Distance fromatob=|a −b|.
Example2.22 Find the distance from
(a)x=2tox=9 (b)x=−2tox=9 (c)x=−2tox=−9
Solution (a) Distance = |2 −9| = | −7| = 7
(b) Distance=|−2−9|=|−11|=11
(c) Distance = | −2− (−9)| = |7| = 7
From the definition of the modulus function itfollows:
If |x| =a,then eitherx =aorx = −a.
If|x|<a,then−a<x<a.
If |x| >a, then eitherx >aorx < −a.
For example, if |x| = 4 then eitherx = 4 orx = −4. If |x| < 4 then −4 < x < 4; that
is,xlies between −4 and 4. If |x| > 4, then eitherx > 4 orx < −4; that is,xis either
greater than 4 orless than −4.
Example2.23 Describe the interval on thexaxis defined by
(a) |x| < 2
(b) |x| 3
(c) |x−1|<3
(d) |x+2|>1
Solution (a) |x| < 2 is the same statement as −2 < x < 2; that is,xis numerically less than 2.
Figure2.40illustratesthisregion.Notethattheregionisanopeninterval.Sincethe
pointsx = −2 andx = 2 arenot included, they areshown on the graph as ◦.
(b) If |x| 3theneitherx 3orx −3.ThisisshowninFigure2.41.Therequired
region of thexaxis has two distinct parts. Since the pointsx = 3 andx = −3 are
included inthe interval of interest,they are shown on the graph as •.
(c) |x −1| < 3 isequivalent to −3 <x −1 < 3,thatis −2 <x<4.
(d) |x+2|>1isequivalenttox+2>1orx+2<−1,thatisx>−1orx<−3.