Question

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

Answer 1

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.