Question

Find the domain and range of $f\left( x \right) = \sqrt {\left( {x - 1} \right)\left( {3 - x} \right)} $

Answer 1

Hint: For finding the domain and the range of the function we need to obtain the zeroes of it. And from this, we will have the domain and zeroes. So the domain will be any input given to the solution and the range will be the output of the solution.

Complete answer:
So we have the function given as $f\left( x \right) = \sqrt {\left( {x - 1} \right)\left( {3 - x} \right)} $
Now for finding the domain and range we will equate the function or we can say the polynomial to zero. And as we know that under the square root the polynomial must be positive.
So the domain will be obtained by solving $\left( {x - 1} \right)\left( {3 - x} \right) \geqslant 0$
So from this, we can obtain the zeroes of the polynomial, we have
$ \Rightarrow \left( {x - 1} \right)\left( {3 - x} \right) = 0$
If $\left( {x - 1} \right) = 0$ , then
$ \Rightarrow x = 1$ and
If $\left( {3 - x} \right) = 0$ , then
$ \Rightarrow x = 3$
So we have the two values of $x$ which is $1,3$
Hence, the polynomial will be solved for the external intervals which are $x > 1,x > 3$
And it can be written as $x \in \left( {1, + \propto } \right)$
Now since the function $f\left( x \right)$ is positive due to the presence of the square root. Therefore, the range will include all the positive real numbers and mathematically it can be written as $x \geqslant 0$ .

Hence, the domain of function will be $x \in \left( {1, + \propto } \right)$ and the range of function will be $x \geqslant 0$

Note: So in a simpler term to understand the domain and range we can say that the domain means the number we feed to the function and the range will be the number it gives back to us. For greater understanding, if we would think of the domain as all the possible inputs and the range as all the possible outputs then we will have the right idea to understand it. And in this simple way, we can understand it.