Relations and Functions — MATHS STD 11 Science — Question

Here the given function is: $\begin{equation} f(x)=\sqrt{9-x^{2}} \end{equation}$
Domain: These are the values of x for which f(x) is defined. From the given f(x) we can say that, f(x) should be real and for that, 9 - x² $\geq$ 0 [since a value less than 0 will give an imaginary value] (3 + x)(3 - x) $\geq$ 0. Now there are two critical points, x = +3 and x = -3. Taking a value less than -3 and putting in the expression we get, (3 - 5)(3 + 5) = -ve value and thus plotting these on number line we have

Since, f(x) is defined for all real numbers that are greater than or equal to -3 and less than or equal to 3, the domain of f(x) is [-3, 3].
Range: The values of f(x) obtained by putting possible values of x. From the f(x) we can see that, the values obtained will only be positive and can be any positive number less than 3. Hence the range of f(x) = [0, 3).