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Let us see in detail building process of membership functions
μ
x
and
1
()
()
μ
x
μ
x
. The left boundary of membership function
unambiguously defines
2
3
()
μ
x
the right boundary of membership function
. It is necessary to construct
2
()
()
μ
x
membership function
μ
x
which will explicitly define the left boundary
,
2
1
()
μ
x
or to construct the left boundary of membership function
with which we
2
()
μ
explicitly.
Let us construct membership function
x
define
1
()
()
μ
x
μ
x
. Let right boundary
look like
1
2
(
)
ab
. Let us consider the objects allocated by an expert to
level
X
. Having made paired comparisons of these objects and subsequent
buildings similar to those above, we obtain linear function
y
=
1
−
a
x
−
b
−
/
>
0
,
3
3
3
3
, which is
the right boundary of required function. If this function satisfies to two conditions:
()
y
=
a
x
+
b
1
1
(
)
y
−
b
/
a
<
0
y
0
>
1
;
, then
3
3
⎧
1
−
b
1
0
≤
x
≤
1
;
⎪
a
⎪
⎪
1
1
−
b
b
()
μ
x
=
a
x
+
b
,
1
<
x
≤
−
1
;
⎨
1
1
1
a
a
⎪
1
1
b
⎪
0
−
1
<
x
≤
1
⎪
a
⎩
1
1
−
b
⎧
0
0
≤
x
≤
1
;
⎪
a
⎪
1
1
−
b
b
⎪
1
1
1
−
a
x
−
b
,
<
x
≤
−
;
1
1
⎪
a
a
1
1
⎪
⎪
b
b
()
1
−
1
<
x
≤
−
3
;
μ
x
=
⎨
2
a
a
⎪
1
3
b
1
−
b
⎪
1
−
a
x
−
b
,
−
3
<
x
≤
3
;
⎪
3
3
a
a
⎪
3
3
1
−
b
⎪
3
0
<
x
≤
1
⎪
a
⎩
3
()
(
)
y
=
a
x
+
b
If function
satisfies the conditions:
y
0
>
1
;
y
−
b
/
a
≥
0
,
1
1
3
3
()
membership
function
μ
x
is
defined
under
the
condition
1
(
)
(
)
by
. In this case one of coefficients still remains
unknown, it is determined from a normal equation. We obtain the membership
functions
−
/
a
=
−
a
b
/
a
+
b
=
0
3
3
1
3
3
1
