Question

Find the domain and range of \[y = \dfrac{3}{{x + 6}}\]?

Answer 1

Hint: We use the definition of domain and range of a function. For domain we find the values where the function is defined, so we deduct the value where the function is undefined. Equating the denominator to 0 we get those values of $x$ for which $y$ is undefined. For range we convert the function in terms of $y$ such that we find the values of $y$ for which $x$ is defined using the same procedure.
Domain of a function \[y = f(x)\] is a set of all those values for which function is defined. So, domain contains all possible values of $x$ for which $y$ exists.
Range of a function contains all the possible values of $y$ for which $x$ exists.

Complete step by step solution:
We are given the function \[y = \dfrac{3}{{x + 6}}\] … (1)
Domain:
Since we have to find the values of x for which y exists, we will exclude those values of x for which function is undefined.
A function will not be defined if its denominator is 0
Equate denominator of the fraction to 0
\[ \Rightarrow x + 6 = 0\]
Shifting constant values to right hand side of the equation
\[ \Rightarrow x = - 6\]
So, all the values except -6 are contained in the domain of the function.
We can write domain of \[y = \dfrac{3}{{x + 6}}\]is \[\left( { - \infty , - 6} \right) \cup \left( { - 6,\infty } \right)\].
Range:
Since we know the range consists of values of y for which x exists, we will convert the function in terms of x to the function in terms of y.
We have the function \[y = \dfrac{3}{{x + 6}}\]
Switch the variables from x to y and vice versa
\[ \Rightarrow x = \dfrac{3}{{y + 6}}\]
Now cross multiply values from both sides of the equation
\[ \Rightarrow y + 6 = \dfrac{3}{x}\]
Shift 6 to right hand side of the equation
\[ \Rightarrow y = \dfrac{3}{x} - 6\]
Take LCM on right hand side
\[ \Rightarrow y = \dfrac{{3 - 6x}}{x}\]
Now this function will only be undefined when the denominator will be 0 i.e. when \[x = 0\]
We can write range of \[y = \dfrac{3}{{x + 6}}\] is \[\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\].
\[\therefore \] The domain of the function \[y = \dfrac{3}{{x + 6}}\] is \[\left( { - \infty , - 6} \right) \cup \left( { - 6,\infty } \right)\] and the range of the function \[y = \dfrac{3}{{x + 6}}\]is \[\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\].

Note: Do not mistake of writing the closed brackets for range and domain here, keep in mind we have to take just the value that occurs before that number where the function is undefined but we have to exclude the value where the function is undefined. So, we use open brackets. Also, do not tend to write intersection of the sets which is wrong as that indicates values existing in both the sets which is null set, so keep in mind we always take union of the sets when writing domain and range as we are trying to combine the possible values.