Question

The given function is $\dfrac{\left| x-4 \right|}{x-4}$. Find the range of the function.

Answer 1

We need to find the form of $\left| x \right|$. The given function is of the form $\dfrac{\left| y \right|}{y}$. So, we find the values that it can achieve for all values of y. As the values are single values the set of range will be countable points.

Complete step by step answer:
Let’s consider $f\left( x \right)=\dfrac{\left| x-4 \right|}{x-4}$, $x\in \mathbb{R}$. Its range set is R.
In general, the range defines the set where the function maps its points of domain.
Range is also called the image of domain.
So, for $f\left( x \right)=\dfrac{\left| x-4 \right|}{x-4}$, $\mathbb{R}$ is the domain where $x\in \mathbb{R}$ and the range of the function is the set R where $R=f\left( x \right)$.
We know that the modulus function acts in a manner such that for function $\left| y \right|$ it will be
\begin{aligned} & \left| y \right|=\left\\{ y,\forall \right.y\ge 0 \\\ & \left| y \right|=\left\\{ -y,\forall \right.y<0 \\\ \end{aligned}
We break the given function in two functions. One being all x where $x\ge 4$ and the other part where $x<4$.
Now, for $x\ge 4$ we get the value of $x-4\ge 0$. So, we treat $\left| x-4 \right|$ as $x-4$ for all $x\ge 4$.
And for $x<4$ we get the value of $x-4<0$. So, we treat $\left| x-4 \right|$ as $-\left( x-4 \right)$ for all $x<4$.
So, the function becomes $f\left( x \right)=\dfrac{\left| x-4 \right|}{x-4}=\dfrac{x-4}{x-4}=1$ for all values of x where $x\ge 4$.
And $f\left( x \right)=\dfrac{\left| x-4 \right|}{x-4}=\dfrac{-\left( x-4 \right)}{x-4}=-1$ for all values of x where $x<4$.
So, in total the given function can be arranged in a way such that
$\begin{aligned} & f\left( x \right)=1,\forall x\in (-\infty ,4] \\\ & f\left( x \right)=-1,\forall x\in (-4,\infty ) \\\ \end{aligned}$
So, the function only takes values of 1 and -1 for all real values of x

The range of the function $f\left( x \right)=\dfrac{\left| x-4 \right|}{x-4}$ is R=\left\\{ -1,1 \right\\}.

Note: We need to remember that the range set always doesn’t have to be a span of points. It can be a singleton point or points sometime. Here we cannot formulate the given function in its simplest form until we have the modulus function transformed into its normal form. That’s why we broke the domain of the function into two parts.