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Question: How do you find domain and range of \(f\left( x \right)=\dfrac{1}{2x-4}\) ?...

How do you find domain and range of f(x)=12x−4f\left( x \right)=\dfrac{1}{2x-4} ?

Explanation

Solution

We have been given a fractional expression with linear functions in variable-x in the denominator. The denominator consists of a term in x-variable and a constant term, 4 but the numerator only has constant term, 1. We shall find all values of variable-x for which the given function is defined as domain and then find the inverse of the function to compute the solution set of the range of function.

Complete step by step solution:
Given that f(x)=12x−4f\left( x \right)=\dfrac{1}{2x-4}.
Any fractional function has one condition for which it is defined. The denominator of the fraction must not be equal to zero.
Hence, 2x−4≠02x-4\ne 0.
We shall find the value of x for which the denominator of the fraction is zero, then we shall remove that value of x from the solution set of the domain of the function.
⇒2x−4=0\Rightarrow 2x-4=0
Transposing 4 to the right hand side and dividing both sides by 2, we get
⇒2x=4\Rightarrow 2x=4
⇒x=2\Rightarrow x=2
Thus, the domain of the function is x\in \mathbb{R}-\left\\{ 2 \right\\}.
We shall find the inverse of the given function to find its range.
f(x)=12x−4f\left( x \right)=\dfrac{1}{2x-4}
⇒y=12x−4\Rightarrow y=\dfrac{1}{2x-4}
We shall first cross-multiply the terms and then write them in terms of variable-y.
⇒2yx−4y=1\Rightarrow 2yx-4y=1
⇒2yx=1+4y\Rightarrow 2yx=1+4y
⇒x=1+4y2y\Rightarrow x=\dfrac{1+4y}{2y}
⇒f−1(x)=1+4y2y\Rightarrow {{f}^{-1}}\left( x \right)=\dfrac{1+4y}{2y}
This inverse function is valid for all values of y except y=0y=0.
Therefore, y\in \mathbb{R}-\left\\{ 0 \right\\}.
Hence, the range of f(x)=12x−4f\left( x \right)=\dfrac{1}{2x-4} is \mathbb{R}-\left\\{ 0 \right\\}.
Therefore, for function f(x)=12x−4f\left( x \right)=\dfrac{1}{2x-4}, the domain is x\in \mathbb{R}-\left\\{ 2 \right\\}and the range is f\left( x \right)\in \mathbb{R}-\left\\{ 0 \right\\}.

Note:
In the solution set of the range of a given function, we have excluded 0 from the set of all real numbers because the function is not defined for this value of variable-y. We removed 0 from the solution set of range of function because putting y=0y=0 was making the denominator of the given fraction equal to zero which is not defined for fractional functions.