Question
The range of the real-valued function $$f\left( x \right) = \sqrt {9 - {x^2}} $$ is :
A.
$$\left[ {0,\,3} \right]$$
B.
$$\left[ { - 3,\,3} \right]$$
C.
$$\left[ { - 3,\,0} \right]$$
D.
none of these
Answer :
$$\left[ {0,\,3} \right]$$
Solution :
$$9 - {x^2} \geqslant 0\,\,\, \Rightarrow - 3 \leqslant x \leqslant 3$$
As $$9 - {x^2}$$ is an even function, the value of $$f\left( x \right)$$ changes in $$0 \leqslant x \leqslant 3$$
Therefore,
$$\max \,f\left( x \right) = \sqrt {9 - 0} = 3{\text{ and }}\min \,f\left( x \right) = \sqrt {9 - 9} = 0$$