Question

Find the domain and range of the real valued function:

(ix)  \[f\left( x \right) = \frac{1}{\sqrt{16 - x^2}}\]

Answer 1

Given:

\[f\left( x \right) = \frac{1}{\sqrt{16 - x^2}}\]

\[(16 - x^2 ) > 0\]
\[ \Rightarrow 16 > x^2 \]
\[ \Rightarrow x \in \left( - 4, 4 \right)\]

\[\frac{1}{\sqrt{16 - x^2}}\] is defined for all real numbers that are greater than  – 4 and less than 4.
Thus, domain of f (x) is {x : – 4 < x < 4} or (– 4, 4).

Range of f :
Let f (x) = y

\[\Rightarrow \frac{1}{\sqrt{16 - x^2}} = y\]
\[ \Rightarrow \frac{1}{16 - x^2} = y^2 \]
\[ \Rightarrow \frac{1}{y^2} = 16 - x^2 \]
\[ \Rightarrow x^2 = 16 - \frac{1}{y^2}\]
\[Since, - 4 < x < 4\]
\[ \Rightarrow 0 \leq x^2 < 16\]
\[ \Rightarrow 0 \leq 16 - \frac{1}{y^2} < 16\]
\[ \Rightarrow - 16 \leq - \frac{1}{y^2} < 0\]
\[ \Rightarrow 16 \geq \frac{1}{y^2} > 0\]
\[ \Rightarrow \frac{1}{16} \leq y^2 < \infty \]
\[ \Rightarrow \frac{1}{4} \leq y < \infty \left( \because y \geq 0 \right)\] 

Hence, range ( f ) = [\[\frac{1}{4}, \infty\] ).