Find the Domain and Range of \(f(x)=\frac1{\sqrt{16-x^2}}\)
Solution
Domain
Since the square root is in the denominator:
(i) The expression inside the root must be positive.
(ii) The denominator cannot be zero.
Therefore, $$ 16-x^2>0 $$
$$ x^2<16 $$
$$
-4
Hence, the domain is:
$$
(-4,4)
$$
Let
$$
y=\frac1{\sqrt{16-x^2}}
$$
Since
$$
0<16-x^2\le16
$$
Therefore,
$$
0<\sqrt{16-x^2}\le4
$$
Taking reciprocals:
$$
y\ge\frac14
$$
Minimum value occurs at
$$
x=0
$$
$$
f(0)=\frac1{\sqrt{16}}=\frac14
$$
As \(x\to4\) or \(x\to-4\),
$$
\sqrt{16-x^2}\to0
$$
so
$$
f(x)\to\infty
$$
Hence, the range is:
$$
\left[\frac14,\infty\right)
$$
Range