<h2>Domain and Range of a Function</h2><h3>
Answer:</h3>
Domain: 
Range: 
<h3>
Step-by-step explanation:</h3>
Domain:
In finding the Domain of a function, the values for the input of the function should not make the output of the function <em>undefined</em> or <em>complex</em>. Because of this, we can think of the values for the input that make the output of the function <em>undefined</em> or <em>complex</em> so that we will <u>not include them in our Domain</u>. We can only make the output <em>undefined</em> if the input makes the denominator
. In
, there's no value for
that makes the denominator
as it is constant,
(Note: All expressions implicitly have
as their denominator even though it's not written). We can only make the output of the function <em>complex</em> if the value of
makes the function take the
root of a negative number where
. There's no radical sign in
so we shouldn't worry about the output of
being <em>complex</em>. Because there's no value for the input,
, that can make the output of
<em>undefined</em> or <em>complex</em>, its Domain can be any number.
Domain: 
Range:
In finding the Range, it is actually the same logic as finding the Domain but first, we'll have to do a bit of rewriting for the given function.
Let
so
.
First, we need to make
the subject of the equation.
Making
the subject:
.
In the Domain, we'll have to think of the value for the input,
, that makes the output <em>undefined</em> or <em>complex</em>. In the Range, the same logic for the Domain, we'll have to think for the value of
, that makes the
<em>undefined</em> or <em>complex</em>. That's why we made
in the equation the subject. In our rewritten equation, we can see that
is under the <em><u>square</u></em> root. Which means if
is negative,
will be <em>complex</em>. So we have to make
be greater than or equal to
(
) so that
won't be complex.
Solving for the inequality, 

Range: 