Question

Find the domain and range of $y\left( x \right)=\sqrt{9-{{x}^{2}}}$.

Answer 1

Hint: For the domain, we have to find the value of $x$ for which the given function is always defined. In this case we will put the value under root $\ge 0$, and then find the values of $x$ for which it is defined. Now for the range of function, we will put the values of $x$ for which it gives maximum and minimum value and that will be the range.

Complete step-by-step solution:
The given function is $f\left( x \right)=\sqrt{9-{{x}^{2}}}$. First, we will find the domain of $f\left( x \right)$. The value inside the root must be $\ge 0$ for the function to be defined. Hence, we get,
$\begin{align}
  & 9-{{x}^{2}}\ge 0 \\
 & \Rightarrow {{x}^{2}}\le 9 \\
 & \Rightarrow x\le \sqrt{9} \\
 & \Rightarrow -3\le x\le 3 \\
\end{align}$
Hence the domain of the function $f\left( x \right)$ is $\left[ -3,3 \right]$.
Now we will find the value of the range of $f\left( x \right)$. We know that the value of $\sqrt{x}$ is always $\ge 0$ and hence, the minimum value of $f\left( x \right)=\sqrt{9-{{x}^{2}}}$ can be 0 when the value of $x$ is $\pm 3$. Now for maximum value, in $\sqrt{9-{{x}^{2}}}$, the value of ${{x}^{2}}$ is always positive and hence we are subtracting ${{x}^{2}}$ from 9. So, $\sqrt{9-{{x}^{2}}}$ will always give value $\le 3$, the equal to part is when $x=0$. Hence the maximum value is $+\sqrt{9}$.
Hence the range of $f\left( x \right)=\sqrt{9-{{x}^{2}}}$ will be $\left[ 0,\sqrt{9} \right]$ or $\left[ 0,+3 \right]$.

Note:One can also solve this question by drawing the graph of $f\left( x \right)=\sqrt{9-{{x}^{2}}}$, which represents a semicircle of radius 3, and with the help of the graph the possible values in x-axis are domain of the function and the possible values in the y-axis are the range of the function.