find the
domain of the function f(x)
x2 +32+ 5/
x2 - 5 + 4
Answers
Answer:
Domain :
R
−
{
−
1
,
3
}
Range:
R
Explanation:
To Find the domain
Equate the denominator(
x
2
−
2
x
−
3
) to zero, then solve the equation for
x
→
x
2
−
2
x
−
3
=
0
→
x
=
−
(
−
2
)
±
√
(
−
2
)
2
−
4
⋅
(
1
)
⋅
(
−
3
)
2
⋅
1
→
x
=
1
±
2
⇒
x
=
−
1
and
x
=
3
This means that, when
x
=
−
1
or
3
, we have the
x
2
−
2
x
−
3
=
0
Implying that
f
(
x
)
=
x
0
which is undefined.
Hence, the domain is all real numbers except
−
1
and
3
Also written as
D
f
=
(
−
∞
,
−
1
)
∪
(
−
1
,
3
)
∪
(
3
,
+
∞
)
To Find the Range
Step 1
say
f
(
x
)
=
y
and rearrange the function as a quadratic equation
y
=
x
x
2
−
2
x
−
3
→
y
(
x
2
−
2
x
−
3
)
=
x
→
y
x
2
−
2
y
x
−
3
y
−
x
=
0
→
y
x
2
+
(
−
2
y
−
1
)
x
−
3
y
=
0
Step 2
We know from the quadratic formula,
x
=
−
b
±
√
b
2
−
4
a
c
2
a
that the solutions of
x
are real when
b
2
−
4
a
c
≥
0
So likewise, we say,
(
−
2
y
−
1
)
2
−
4
⋅
(
y
)
⋅
(
−
3
y
)
≥
0
b
a
c
Step 3
We solve the inequality for the values set of values of
y
⇒
4
y
2
+
4
y
+
1
+
12
y
2
≥
0
→
16
y
2
+
4
y
+
1
≥
0
→
16
[
y
2
+
1
4
y
+
1
16
]
≥
0
→
16
[
(
y
+
1
8
)
2
−
1
64
+
1
16
]
≥
0
→
16
[
(
y
+
1
8
)
2
+
3
64
]
≥
0
Notice that for all values of
y
the left hand side of the inequality be greater than (but not equal) to zero.
We then conclude that,
y
can take all real values.
y
∈
R
⇔
f
(
x
)
∈
R
So the Range is
R