If the domain of a function f: A → N ∪ {0}; f(x) = \(\sqrt{x^2−

1.	If the domain of a function f: A → N ∪ {0}; f(x) = \(\sqrt{x^2−16}\) is A = {4, 5}, then find its range.
Answer» f(x) = \(\sqrt{x^2−16}\); Domain = {4, 5} Range R_f = {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 R_f = {0, 3}

If the domain of a function f: A → N ∪ {0}; f(x) = \(\sqrt{x^2−16}\) is A = {4, 5}, then find its range.

Answer»

f(x) = \(\sqrt{x^2−16}\); Domain = {4, 5}

Range R_f = {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 R_f = {0, 3}