Question

How do you find the domain and range of $\sqrt {x - 3} - \sqrt {x + 3}$?

Answer 1

In this question we are asked to find the domain and range of the function, this can be done by the definition of the domain and range of the function, The domain of a function $f\left( x \right)$ is the set of all values for which the function is defined, and the range of the function is the set of all values that $f$ takes, by using the definitions we will get the required result.

Complete answer:
Given function is $\sqrt {x - 3} - \sqrt {x + 3}$,
In order to find the domain and range, we need to find the values of $x$ for which the function is defined.
Since we have square roots, we know that they both need to be greater than or equal to 0. Since the first square root is a subtraction, it is more restrictive so it will control the domain, i.e., Set the radicand in $x - 3$ greater than or equal to $0$ to find where the expression is defined.
$\Rightarrow x - 3 \geqslant 0$,
Now add 3 to both sides of the equation we get,
$\Rightarrow x - 3 + 3 \geqslant 0 + 3$,
Now simplifying we get.
$\Rightarrow x \geqslant 3$,
So, the domain of the function is all the real numbers greater than or equal to 3, which can be written as, $\left[ {3,\infty } \right)$.
Now to find the range,
$\sqrt {x - 3} - \sqrt {x + 3} \geqslant 0$,
$\Rightarrow \sqrt {x - 3} \geqslant \sqrt {x + 3}$,
By looking, we know that this subtraction will always be negative because $\sqrt {x + 3} > \sqrt {x - 3}$,
Now take some values for $x$,

$x$	$\sqrt {x - 3} - \sqrt {x + 3}$
6	$\sqrt 3 - \sqrt 6$, which is negative
5	$\sqrt 2 - \sqrt 5$, which is negative
4	$1 - \sqrt 7$which is negative
3	$- \sqrt 6$which is negative
2	$\sqrt { - 1} - \sqrt 5$which is not a real number
1	$\sqrt { - 2} - \sqrt 4$which is not a real number

Therefore taking the smallest valid value of $x$ gives us $- \sqrt 6$. Thus, our function will always be greater than or equal to $- \sqrt 6$, and however, we know that $\sqrt {x - 3} \ne \sqrt {x + 3}$, so the function will never touch 0. Thus 0 is the limiting value of function.
So, the range of the given function is $\left( { - \infty , - \sqrt 6 } \right]$.
So, the domain of the function is all real numbers greater than or equal to 3 which is given by interval notation $\left[ {3,\infty } \right)$ and the range of the function is all real numbers greater than or equal to $- \sqrt 6$which is given by interval notation $\left( { - \infty , - \sqrt 6 } \right]$.

Final Answer:
$\therefore$ The domain of the given function $\sqrt {x - 3} - \sqrt {x + 3}$ is given by interval notation $\left[ {3,\infty } \right)$ and the range which is given by interval notation $\left( { - \infty , - \sqrt 6 } \right]$.

Note:
A function is a relation between domain and range such that each value in the domain corresponds to only one value in the range. Relations that are not functions violate this definition. They feature at least one value in the domain that corresponds to two or more values in the range.