Question

What is the domain and range of \[y = \dfrac{1}{{x + 1}}\] ?

Answer 1

Hint: A function's domain is the set of numbers that define the result of a function when it is plugged in. A function's range is defined as a set of equation solutions for a given input.

Complete step-by-step answer:
Given that the function is \[y = \dfrac{1}{{x + 1}}\]
We can never divide a fraction by \[0\] when dealing with fractions. We can measure the values that will be excluded in the function by setting the denominator not equal to \[0\] and solving for \[x\] .
That is, \[x + 1 \ne 0\]
 \[ \Rightarrow x \ne - 1\]
Which means \[x\] will take values excluding \[ - 1\] .
To write the notation, when the number is included in the domain, use a bracket; when the number is not included in the domain, use a parenthesis. A union, denoted by the letter \[ \cup \] , connects sections of a domain that are separated by a distance.
So, the domain is \[x \in \left( { - \infty , - 1} \right) \cup \left( { - 1, + \infty } \right)\]
Now we have to calculate the range.
We know that, \[y = \dfrac{1}{{x + 1}}\]
By cross multiplying, we will get,
 \[ \Rightarrow y\left( {x + 1} \right) = 1\]
 \[ \Rightarrow yx + y = 1\]
 \[ \Rightarrow yx = 1 - y\]
 \[ \Rightarrow x = \dfrac{{1 - y}}{y}\]
As we know that the denominator should not equal to \[0\] .
That is, \[y \ne 0\]
Which means \[y\] will take values excluding \[0\] .
So, the range is \[y \in \left( { - \infty ,0} \right) \cup \left( {0, + \infty } \right)\]
The range is written in the same notation as the domain.
That is, \[x \in \left( { - \infty , - 1} \right) \cup \left( { - 1, + \infty } \right)\] is the domain and \[y \in \left( { - \infty ,0} \right) \cup \left( {0, + \infty } \right)\] is the range of the function \[y = \dfrac{1}{{x + 1}}\] .
We are drawing the graph of the function for more clarification:

Note: Keep in mind that the domain of the function is the set of all possible \[x\] values that will satisfy an equation, and the range is a function's output or \[y\] value. If an equation is divided by \[0\] or has a negative square root, it will not produce a valid function.