Question

Find the domain and range of a real function $f(x)=\sqrt{9-{{x}^{2}}}$.

Answer 1

To find the domain we have to find the possible values of x for that putting $9-{{x}^{2}}\ge 0$. Gives $-3\le x\le 3$ now its turn for finding range we now have to find max and minimum values of f(x) or y so for that after squaring expression will look like ${{y}^{2}}=9-{{x}^{2}}$ which gives $x=\sqrt{9-{{y}^{2}}}$ on shifting , clearly x is defined when $9-{{y}^{2}}\ge 0$, because the expression inside root cannot be negative.
Which gives $9\le {{y}^{2}}$ which gives range as $-3\le y\le 3$ but we know one thing that is the root of any expression is always positive ($y=\sqrt{9-{{x}^{2}}}$) so y should be greater than or equals to 0. So, the range will be $0\le y\le 3$

Complete step-by-step solution:
Given a function $f(x)=\sqrt{9-{{x}^{2}}}$and we have to find domain and range of this function so lets consider an example $\sqrt{x}$ , domain of a function means possible values of x which satisfies the function so here the domain is $x\ge 0$, similarly domain of function $\sqrt{9-{{x}^{2}}}$ is $9-{{x}^{2}}\ge 0$ which simplifies to $9\ge {{x}^{2}}$ so from here we got domain as $3\ge x\ge -3$
After solving domain now we have to find the range of this function $y=\sqrt{9-{{x}^{2}}}$ , we know property that is root of any expression is always positive so y will be greater or equals to 0,now solving the function and finding the possible range of values of y and for that we can simplify it to ${{y}^{2}}=9-{{x}^{2}}$
Which on solving ${{x}^{2}}=9-{{y}^{2}}$ so we got x as $x=\sqrt{9-{{y}^{2}}}$ now looking at this we can say that values inside root can never be negative so we can write this condition as $9-{{y}^{2}}\ge 0$
Which on further calculation gives $-3\le y\le 3$ but we know that y will be greater or equals to 0 $y\ge 0$ so taking intersection of both ranges we got our result as $0\le y\le 3$
Hence domain of function is $-3\le x\le 3$ and range is $0\le y\le 3$.

Note: Sometimes if x is given in denominator and we are asked to find domain than we have to consider one more condition that is $x\ne 0$ for example $f(x)=\sqrt{\dfrac{9}{x}-{{x}^{2}}}$ in this function while calculating domain from the method mentioned above we will also consider $x\ne 0$, then after taking intersection of the results we get final domain.