f(x) = \(\sqrt{x^2−16}\); Domain = {4, 5}
Range Rf = {f(x) | x ∈ Domain}
= {f(4), f(5)}
Now, f(x) =\(\sqrt{x^2−16}\)
∴ f( 4) = \(\sqrt{(4)^2−16}\)
= \(\sqrt{16-16}\) = 0
f(5) = \(\sqrt{(5)^2−16}\)
= \(\sqrt{25-16}\) = \(\sqrt{9}\) = 3 (∵ x is positive.}
Hence, Range Rf = {0, 3}