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Question: How do you find domain and range of \(f\left( x \right)=\dfrac{1}{x-5}\) ?...

How do you find domain and range of f(x)=1x−5f\left( x \right)=\dfrac{1}{x-5} ?

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, 5 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 answer:
Given that f(x)=1x−5f\left( x \right)=\dfrac{1}{x-5}.
Any fractional function has one condition for which it is defined. The denominator of the fraction must not be equal to zero.
Hence, x−5≠0x-5\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.
⇒x−5=0\Rightarrow x-5=0
Transposing 5 to the right hand side, we get
⇒x=5\Rightarrow x=5
Thus, the domain of the function is x\in \mathbb{R}-\left\\{ 5 \right\\}.
We shall find the inverse of the given function to find its range.
f(x)=1x−5f\left( x \right)=\dfrac{1}{x-5}
⇒y=1x−5\Rightarrow y=\dfrac{1}{x-5}
We shall first cross-multiply the terms and then write them in terms of variable-y.
⇒yx−5y=1\Rightarrow yx-5y=1
⇒yx=5y+1 ⇒x=5y+1y \begin{aligned} & \Rightarrow yx=5y+1 \\\ & \Rightarrow x=\dfrac{5y+1}{y} \\\ \end{aligned}
⇒f−1(x)=5y+1y\Rightarrow {{f}^{-1}}\left( x \right)=\dfrac{5y+1}{y}
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)=2x−5f\left( x \right)=\dfrac{2}{x-5} is \mathbb{R}-\left\\{ 0 \right\\}.
Therefore, for function f(x)=2x−5f\left( x \right)=\dfrac{2}{x-5}, the domain is x\in \mathbb{R}-\left\\{ 5 \right\\}and the range is f\left( x \right)\in \mathbb{R}-\left\\{ 0 \right\\}

Note:
A function, ff which is a mapping from variable-x to variable-y is invertible if and only if there exists a function f−1{{f}^{-1}} which is a mapping from variable-x to variable-y, such that the composition of f−1{{f}^{-1}} with ff is equal to 1.