3 Find the range and domain of the
eval function f(x) =√9-x^2
Answers
Answered by
0
Domain:
[
−
3
,
3
]
Range:
[
0
,
3
]
Explanation:
The value under a square root cannot be negative, or else the solution is imaginary.
So, we need
9
−
x
2
≥
0
, or
9
≥
x
2
, so
x
≤
3
and
x
≥
−
3
, or
[
−
3.3
]
.
As
x
takes on these values, we see that the smallest value of the range is
0
, or when
x
=
±
3
(so
√
9
−
9
=
√
0
=
0
), and a max when
x
=
0
, where
y
=
√
9
−
0
=
√
9
=
3
[
−
3
,
3
]
Range:
[
0
,
3
]
Explanation:
The value under a square root cannot be negative, or else the solution is imaginary.
So, we need
9
−
x
2
≥
0
, or
9
≥
x
2
, so
x
≤
3
and
x
≥
−
3
, or
[
−
3.3
]
.
As
x
takes on these values, we see that the smallest value of the range is
0
, or when
x
=
±
3
(so
√
9
−
9
=
√
0
=
0
), and a max when
x
=
0
, where
y
=
√
9
−
0
=
√
9
=
3
Answered by
1
Step-by-step explanation:
f(x) = - x² + √9
—> Domain of the function: (-∞,∞)
As, this function is a parabolic equation,
For, Minimum value of this function,
f'(x) = 0
=> -(2 × x²‐¹) = 0
=> -2x = 0
=> x = 0
f(0) = √9
—>Range of the funtion: (√9,∞)
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