We know that
f
(
x
)
is used to represent a function of one variable: the input variable is
x
and the output is the value
f
(
x
)
. Here
x
is the
independent
variable and
y
=
f
(
x
)
is the
dependent variable
.
Suppose we consider a function with
two independent
input variables
x
and
y
, for example
f
(
x
,
y
)
=
x
+
2
y
+
3.
If we specify values for
x
and
y
then we have a
single
value
f
(
x
,
y
)
. For example, if
x
=
3
and
y
=
1
then
f
(
x
,
y
)
=
3
+
2
+
3
=
8
. We write
f
(
3
,
1
)
=
8
.
Find the values of
f
(
2
,
1
)
,
f
(
−
1
,
−
3
)
and
f
(
0
,
0
)
for the following functions.
f
(
x
,
y
)
=
x
2
+
y
2
+
1
f
(
x
,
y
)
=
2
x
+
x
y
+
y
3
Answer
f
(
2
,
1
)
=
2
2
+
1
2
+
1
=
6
;
f
(
−
1
,
−
3
)
=
(
−
1
)
2
+
(
−
3
)
2
+
1
=
11
;
f
(
0
,
0
)
=
1
f
(
2
,
1
)
=
4
+
2
+
1
=
7
;
f
(
−
1
,
−
3
)
=
−
2
+
3
−
27
=
−
26
;
f
(
0
,
0
)
=
0
In a similar way we can define a function of three independent variables. Let these variables be
x
,
y
and
u
and the function
f
(
x
,
y
,
u
)
.
Given
f
(
x
,
y
,
u
)
=
x
2
+
y
u
+
2
, find
f
(
0
,
1
,
0
)
,
f
(
−
1
,
−
1
,
2
)
.
f
(
0
,
1
,
0
)
=
0
2
+
1
×
0
+
2
=
2
;
f
(
−
1
,
−
1
,
2
)
=
1
−
2
+
2
=
1
Find
f
(
2
,
−
1
,
1
)
for
f
(
x
,
y
,
u
)
=
x
y
+
y
u
+
u
x
.
Evaluate
f
(
x
,
y
,
u
,
t
)
=
x
2
−
y
2
−
u
2
−
2
t
when
x
=
1
,
y
=
−
2
,
u
=
3
,
t
=
1.
Answer
f
(
2
,
−
1
,
1
)
=
2
×
(
−
1
)
+
(
−
1
)
×
1
+
1
×
2
=
−
1
f
(
1
,
−
2
,
3
,
1
)
=
1
2
−
(
−
2
)
2
−
3
2
−
2
×
1
=
−
14
(this is a function of 4 independent variables).