#### Question

Find the domain and range of the real valued function:

(viii) \[f\left( x \right) = \sqrt{9 - x^2}\]

#### Solution

Given:

\[f\left( x \right) = \sqrt{9 - x^2}\]

\[(9 - x^2 ) \geq 0\]

\[ \Rightarrow 9 \geq x^2 \]

\[ \Rightarrow x \in \left[ - 3, 3 \right]\]

\[ \Rightarrow 9 \geq x^2 \]

\[ \Rightarrow x \in \left[ - 3, 3 \right]\]

\[\sqrt{9 - x^2}\] is defined for all real numbers that are greater than or equal to – 3 and less than or equal to 3.

Thus, domain of

For any value of

Hence, the range of

Thus, domain of

*f*(*x*) is {*x*: – 3 ≤*x*≤ 3} or [– 3, 3].For any value of

*x*such that – 3 ≤*x*≤ 3, the value of*f*(*x*) will lie between 0 and 3.Hence, the range of

*f*(*x*) is {*x*: 0 ≤*x*≤ 3} or [0, 3].

Is there an error in this question or solution?

Solution Find the Domain and Range of the Real Valued Function: (Viii) F ( X ) = √ 9 − X 2 Concept: Cartesian Product of Sets.