Find the domain of holeroot 4x-x2
Answers
Step-by-step explanation:
okay we are trying our best
The domain is
x
∈
[
0
,
4
]
The range is
f
(
x
)
∈
[
0
,
2
]
Explanation:
For the domain, what's under the square root sign is
≥
0
Therefore,
4
x
−
x
2
≥
0
x
(
4
−
x
)
≥
0
Let
g
(
x
)
=
√
x
(
4
−
x
)
We can build a sign chart
a
a
a
a
x
a
a
a
a
−
∞
a
a
a
a
a
a
a
0
a
a
a
a
a
a
4
a
a
a
a
a
a
a
+
∞
a
a
a
a
x
a
a
a
a
a
a
a
a
−
a
a
a
a
0
a
a
+
a
a
a
a
a
a
a
+
a
a
a
a
4
−
x
a
a
a
a
a
+
a
a
a
a
a
a
a
+
a
a
0
a
a
a
a
−
a
a
a
a
g
(
x
)
a
a
a
a
a
a
−
a
a
a
a
0
a
a
+
a
a
0
a
a
a
a
−
Therefore
g
(
x
)
≥
0
when
x
∈
[
0
,
4
]
Let,
y
=
√
4
x
−
x
2
hen,
y
2
=
4
x
−
x
2
x
2
−
4
x
+
y
2
=
0
The solutions this quadratic equation is when the discriminant
Δ
≥
0
So,
Δ
=
(
−
4
)
2
−
4
⋅
1
⋅
y
2
16
−
4
y
2
≥
0
4
(
4
−
y
2
)
≥
0
4
(
2
+
y
)
(
2
−
y
)
≥
0
Let
h
(
y
)
=
(
2
+
y
)
(
2
−
y
)
We build the sign chart
a
a
a
a
y
a
a
a
a
−
∞
a
a
a
a
a
−
2
a
a
a
a
#color(white)(aaaaaa)2#
a
a
a
a
a
a
+
∞
a
a
a
a
2
+
y
a
a
a
a
−
a
a
a
a
0
a
a
a
a
+
a
a
a
a
0
a
a
a
a
+
a
a
a
a
2
−
y
a
a
a
a
+
a
a
a
a
0
a
a
a
a
+
a
a
a
a
0
a
a
a
a
−
a
a
a
a
h
(
y
)
a
a
a
a
a
−
a
a
a
a
0
a
a
a
a
+
a
a
a
a
0
a
a
a
a
−
Therefore,
h
(
y
)
≥
0
, when
y
∈
[
−
2
,
2
]
This is not possible for the whole interval, so the range is
y
∈
[
0
,
2
]
graph{sqrt(4x-x^2) [-10, 10, -5, 5]}