Question

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

Answer 1

Hint: 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.