Question

Find the domain and range of the real valued function:

(x)  \[f\left( x \right) = \sqrt{x^2 - 16}\]

Answer 1

Given:

\[f\left( x \right) = \sqrt{x^2 - 16}\]

\[( x^2 - 16) \geq 0\]
\[ \Rightarrow x^2 \geq 16\]
\[ \Rightarrow x \in ( - \infty , - 4] \cup [4, \infty )\]

\[\sqrt{x^2 - 16}\]   is defined for all real numbers that are greater than or equal to 4 and less than or equal to –4.
Thus, domain of f (x) is {x : x ≤ – 4 or x ≥ 4} or (–∞, –4] ∪ [4, ∞).

Range of f :
For x ≥ 4, we have:
x²  - 16 ≥ 0

\[\Rightarrow \sqrt{x^2 - 16} \geq 0\]

⇒ f (x) ≥ 0
For x ≤ – 4, we have:
x² -  16 ≥ 0

 \[\Rightarrow \sqrt{x^2 - 16} \geq 0\]

⇒ f (x) ≥ 0

Thus, f (x) takes all real values greater than zero.
Hence, range (f) = [0, ∞).