Question

Find the domain and range of the following real function $ f\left( x \right) = \sqrt {9 - {x^2}} $ .
A.Domain = $ \left[ { - 2,2} \right] $ and range $ = \left[ {0,2} \right] $
B.Domain $ = \left[ {0,2} \right] $ and range $ = \left[ { - 2,2} \right] $
C.Domain $ \left[ { - 3,3} \right] $ and range $ = \left[ {0,3} \right] $
D.Domain $ \left[ {0,2} \right] $ and range $ = \left[ { - 2,2} \right] $

Answer 1

Hint: The domain is the set of all possible and legitimate values of $ x $ which can be put in a function without making it indeterminate while range is the set of all the possible values of $ f\left( x \right) $ which can be obtained from the function. The term present in the square root is always greater than 0.

Complete step-by-step answer:
The given function is
 $ y = \sqrt {9 - {x^2}} $
The critical points are considered by observing the terms in the square root
 $ \sqrt {9 - {x^2}} $
Always, $ 9 - {x^2} \geqslant 0 $
It implies that $ x \leqslant \pm 3 $
The domain of the function is $ \left[ { - 3,3} \right] $ any value which lies between $ - 3 $ and $ 3 $ is the legitimate value of x.
The greatest value of the function can be obtained by substituting which lies in the domain of the function
 $ \Rightarrow f\left( 0 \right) = \sqrt {9 - {0^2}} = 3 $
The lowest value that can be obtained by the function is obtained by putting $ x = \pm 3 $ in the function
 $ \Rightarrow f\left( { \pm 3} \right) = \sqrt {9 - {{\left( { \pm 3} \right)}^2}} = 0 $
Thus, the range of the function is $ \left[ {0,3} \right] $ .
So, the correct answer is “Option C”.

Note: The important concept is that of a domain and properties of a square root function should be clear in mind before solving such problems.
For a square root function to be valid, the value of the function present inside the square root should always be positive i.e., greater than 0.
Domain is the set of legitimate values of function, which does not make the function as indeterminate.
For instance, for the function $ y = \dfrac{1}{{1 - x}} $ , the domain set includes all the real values of $ x $ except $ + 1 $ . Because at $ x = 1 $ the denominator tends to zero and the function goes to infinity. So, domain of the function is $ R - \left\{ 1 \right\} $