f(x) =√(4-x^2) then range is
Answers
Answered by
0
Step-by-step explanation:
Explanation:
As we are looking at
f
(
x
)
=
√
4
−
x
2
+
√
x
2
−
1
we cannot have
|
x
|
<
1
or
|
x
|
>
2
,
for if
|
x
|
<
1
,
√
x
2
−
1
is not defined and if
|
x
|
>
2
, then
√
4
−
x
2
is not defined.
Hence domain of
x
is the interval
[
−
2
,
−
1
]
∪
[
1
,
2
]
Note that
f
(
−
2
)
=
f
(
−
1
)
=
f
(
1
)
=
f
(
2
)
=
√
3
Further
f
'
(
x
)
=
1
2
√
4
−
x
2
⋅
(
−
2
x
)
+
1
2
√
x
2
−
1
⋅
2
x
=
x
(
1
√
x
2
−
1
−
1
√
4
−
x
2
)
Thisis
0
when
√
x
2
−
1
=
√
4
−
x
2
or
2
x
2
=
5
or
x
=
±
√
2.5
and where
f
(
x
)
=
2
√
1.5
=
√
6
Hence, range is
[
√
3
,
√
6
]
graph{ sqrt(4-x^2) + sqrt(x^2-1) [-2.24, 2.76, 0.46,
Mark me as branliest and follow me and message me also
Similar questions