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How do you find the domain and range of $\dfrac{1}{{1 + x}}$?

Last updated date: 20th Jun 2024
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Hint: To find the domain and range of the given function, we have to form a general equation for $x$ and $y$ . Then we need to check whether every element in $x$ has its image or not. And we need to find in what category the values of $x$ and $y$ come under.

Complete step-by-step solution:
Let us consider the given equation,
$f(x) = y = \dfrac{1}{{1 + x}}$
$x,y \in \mathbb{R}$ , for any value of $x$ , we have an image in $y$ , except when $x = - 1$ .
To find the general equation for $x$ , we solve the above equation and we get,
$\Rightarrow y(1 + x) = 1 \\ \Rightarrow y + xy = 1 \\ \Rightarrow xy = 1 - y \\ \Rightarrow x = \dfrac{{1 - y}}{y} \\$
$x,y \in \mathbb{R}$ , for any value of $y$ , we have pre image $x$ , except $y = 0$
This is the required equation for $x$ which is the preimage of $y$ . For $x = - 1$ , we don’t have an image in $y$ because when we substitute $x = - 1$ in $y$ we get, $\dfrac{1}{0}$ which is undefined.
And also, for the image $y = 0$ , the value of $x$ will also be undefined. And hence the domain and range of the function will be the real numbers.

Additional information: There are different types of function they are one-one function, into function, onto function and bijective function. These types define the nature of the function with the help of domain, range and its co-domain.

Note: Let us consider a function $f(x) = y = {x^2}$ , when we put $x = 1$ , we get $y = 1$ . Here the value of $x$ is considered as a domain and the value $y = 1$ is considered as a range of the domain $x = 1$ . If any of the domain $x$ is present without the image in $y$ , then it is not a function.