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J 1 B (
), μ S (
)) =
(
)
x
y
x
1
y
2
y 2
2 .
J 2 VS (
), μ VB (
)) =
((
)
,
) =[
(
,
) ]
x
y
min
1
x
min
1
x
y
Hence,
μ Q 1 (
x
) =
J 1 (
0
.
4
,
1
y
) =
0
.
4
· (
1
y
)
,
0
.
36
,
if
y
0
.
6
2
y 2
2
μ Q 2 (
x
) =
J 2 ((
1
0
.
4
)
,
) = (
min
(
0
.
6
,
y
))
=
y 2
,
if
y
<
0
.
6
,
whose graphics are,
µ Q*
µ Q2*
µ Q1*
Finally,
0
.
4
(
1
y
),
if 0
y
0
.
463
y 2
μ Q (
y
) =
max
Q 1 (
y
), μ Q 2 (
y
)) =
,
if 0
.
463
y
0
.
6
0
.
36
,
if 0
.
6
y
1
.
Example 3.4.13 Let's find the function CRI: X
Y , in the case with X
=[
0
,
1
]
,
Y
=[
0
,
1
]
, and
=
r1: If x is small, then y
9
=
r2: If x is big, then y
2,
1
x
,
if
y
=
9
it follows
μ Q 1 (
y
) = (
1
x
{ 9 } (
y
) =
0
,
if
y
=
9
,
x
,
if
y
=
2
μ Q 2 (
y
) =
x
μ { 2 } (
y
) =
0
,
if
y
=
2
,
and
x
,
if y
=
2
μ Q (
y
) =
max
Q 1 (
y
), μ Q 2 (
y
)) =
1
x
,
if y
=
9
that gives,
0
,
otherwise
2 x
+
9
(
1
x
)
CRI
(
x
) =
=
9
7 x
,
x
+
1
x
 
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