Question

Find the domain and range of the real function $f\left( x \right)=\sqrt{19-{{x}^{2}}}$

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 answer:

The given function is $f\left( x \right)=\sqrt{19-{{x}^{2}}}$
First we will find the domain of f(x),
The value inside the root must be $\ge 0$ for the function to be defined.
Hence we get,
$\begin{align}
  & 19-{{x}^{2}}\ge 0 \\
 & {{x}^{2}}\le 19 \\
 & -\sqrt{19}\le x\le \sqrt{19} \\
\end{align}$
Hence, the domain of the function f(x) is: $\left[ -\sqrt{19},\sqrt{19} \right]$
Now we will find the value of range of f(x),
We know that the value of $\sqrt{x}$ is always $\ge 0$ and hence, the minimum value of $f\left( x \right)=\sqrt{19-{{x}^{2}}}$ can be 0 when the value of x is $\pm \sqrt{19}$.
Now for the maximum value, in $\sqrt{19-{{x}^{2}}}$ the value of ${{x}^{2}}$ is always positive and hence we are subtracting ${{x}^{2}}$ from 19.
So, $\sqrt{19-{{x}^{2}}}$ will always give value $\le \sqrt{19}$, the equal to part is when x = 0.
Hence, the maximum value is $\sqrt{19}$
Hence, the range of $f\left( x \right)=\sqrt{19-{{x}^{2}}}$ will be $\left[ 0,\sqrt{19} \right]$

Note: One can also solve this question by drawing the graph of $y=\sqrt{19-{{x}^{2}}}$, which represents a semicircle of radius $\sqrt{19}$ , and with the help of the graph the possible values in x axis are domain of the function and the possible values in y axis are range of the function.