Question

Find the domain and range of $\dfrac{1}{{x + 6}}$ ?

Answer 1

Use the definition of domain 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.
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.

Complete step by step solution:
We are given the function $f(x) = \dfrac{1}{{x + 6}}$ … (1)
Domain of the function:
From the definition of domain of function, we know that domain consists of all those values of ‘x’ for which the function is defined.
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 in equation (1) 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 $f(x) = \dfrac{1}{{x + 6}}$ is $\left( { - \infty , - 6} \right) \cup \left( { - 6,\infty } \right)$ .
$\therefore$ The domain of the function $\dfrac{1}{{x + 6}}$ is $\left( { - \infty , - 6} \right) \cup \left( { - 6,\infty } \right)$

Note: Do not make mistake of writing the closed brackets for 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, many 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 as we are trying to combine the possible values.