Question

find the domain and range of the following f(x)=1/2x-1​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = \dfrac{1}{2x - 1}$

Domain is defined as set of those real values of x, where function is well defined.

So, in order that f(x) is defined,

$\rm :\longmapsto\:2x - 1 \: \ne \: 0$

$\rm :\longmapsto\:2x \: \ne \: 1$

$\rm :\longmapsto\:x \: \ne \: \dfrac{1}{2}$

Hence,

Domain of

$\bf :\longmapsto\:f(x) = \dfrac{1}{2x - 1} \: is \: x \: \in \: R - \bigg \{\dfrac{1}{2} \bigg \}$

To find the range of f(x)

We first substitute y=f(x) and then solve the equation for x, giving something in the form of x=g(y).

Find the domain of g(y), and this will be the range of f(x).

So,

Let assume that

$\rm :\longmapsto\:y = \dfrac{1}{2x - 1}$

$\rm :\longmapsto\:2x - 1 = \dfrac{1}{y}$

$\rm :\longmapsto\:2x = \dfrac{1}{y} + 1$

$\rm :\longmapsto\:2x = \dfrac{1 + y}{y}$

$\rm :\longmapsto\:x = \dfrac{y + 1}{2y}$

So, x is defined when

$\rm :\longmapsto\:2y \: \ne \: 0$

$\rm :\longmapsto\:y \: \ne \: 0$

So,

$\bf :\longmapsto\: \: y \: \in \: R - \bigg \{0 \bigg \}$

Hence,

Range of

$\bf :\longmapsto\:f(x) = \dfrac{1}{2x - 1} \: is \: y \: \in \: R - \bigg \{0\bigg \}$

Additional Information :-

Let take some more examples :-

Find the domain of the following functions :-

$\bf :\longmapsto\:(1). \: \: f(x) = \sqrt{x - 1}$

Solution

For f(x) to defined,

$\rm :\longmapsto\:x - 1 \geqslant 0$

$\rm :\longmapsto\:x \geqslant 1$

$\bf\implies \:x \: \in \: [1, \: \infty )$

$\bf :\longmapsto\:(2). \: \: f(x) = \sqrt{1 - {x}^{2} }$

For f(x) to be defined,

$\rm :\longmapsto\:1 - {x}^{2} \geqslant 0$

$\rm :\longmapsto\: - ({x}^{2} - 1)\geqslant 0$

$\rm :\longmapsto\: {x}^{2} - 1\leqslant 0$

$\rm :\longmapsto\: (x - 1)(x + 1)\leqslant 0$

$\rm :\longmapsto\: - 1 \leqslant x \leqslant 1$

$\bf\implies \:x \: \in \: [ - 1, \: 1]$

$\bf :\longmapsto\:(3). \: f(x) = \dfrac{1}{ \sqrt{x - 1}}$

For f(x) to be defined,

$\rm :\longmapsto\:x - 1 > 0$

$\rm :\longmapsto\:x > 1$

$\bf\implies \:x \: \in \: (1, \: \infty )$