Question

What is the domain and range of $\dfrac{1}{{x + 2}}$?

Answer 1

We need to find the domain and range of $\dfrac{1}{{x + 2}}$. Domain of a function is the set of all values which qualify as input in a function. It is basically the set of values which is possible so that the function is defined. Range of a function is the set of all the output we get after substituting the values of the domain.

Complete step by step answer:
We need to find the domain and range of $\dfrac{1}{{x + 2}}$.
Let $y = \dfrac{1}{{x + 2}}$
First of all, let us find the domain of $\dfrac{1}{{x + 2}}$.
For that, we need to find the values of $x$ for which the function $\dfrac{1}{{x + 2}}$ is not defined. A fraction $\dfrac{a}{b}$ is not defined when $b = 0$.
So, let us first find all those values of $x$for which the function $\dfrac{1}{{x + 2}}$ is not defined.
For that, we need to put $x + 2 = 0$
Solving the above equation, we get
$\Rightarrow x + 2 = 0$
$\Rightarrow x = - 2$
Hence, we see that $\dfrac{1}{{x + 2}}$ is not defined for $x = - 2$ only and is defined for all other real values of $x$
So, The domain of $\dfrac{1}{{x + 2}}$ is \mathbb{R} - \left\\{ { - 2} \right\\}, where $\mathbb{R}$ is the set of all real numbers.
Now, we shall find the range of $\dfrac{1}{{x + 2}}$.
We have, $y = \dfrac{1}{{x + 2}}$
Finding the range of $\dfrac{1}{{x + 2}}$ means we need to find all the values that $y$ can take.
To find the range, we need to solve $y = \dfrac{1}{{x + 2}}$ and write $x$ in terms of $y$ so that we can find the values of $y$ for which the function is not defined.
Let us now solve $y = \dfrac{1}{{x + 2}}$
$\Rightarrow y = \dfrac{1}{{x + 2}}$
By cross multiplying, we get
$\Rightarrow y\left( {x + 2} \right) = 1$
Reshuffling the terms, we get
$\Rightarrow \left( {x + 2} \right) = \dfrac{1}{y}$
Now, subtracting $2$ from both the sides, we get
$\Rightarrow x + 2 - 2 = \dfrac{1}{y} - 2$
Solving the terms, we get
$\Rightarrow x = \dfrac{1}{y} - 2$
Here, we see that $y$ can take all the real values except $0$ as if $y = 0$, then $\dfrac{1}{y}$ becomes not defined.
Hence, we say that Range of $\dfrac{1}{{x + 2}}$ is \mathbb{R} - \left\\{ 0 \right\\}, where $\mathbb{R}$ is the set of all real numbers.

Note: We can also do this question by plotting a graph of the given function and then interpreting from the graph the values which are taken by $x$, which constitute the domain of the function. And, the values which are taken by the graph of the function constitute the range of that function. And, while we are solving theoretically, we need to consider all the values for which the function is not defined and not just some points.