Question

The domain of the function f(x)=1/(x^2-3x+2) is​

Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = \dfrac{1}{ {x}^{2} - 3x + 2 }$

can be rewritten as

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

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

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

For, f(x) to be defined,

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

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

$\bf\implies \:x \: \in \: R \: - \: \{1,2 \}$

Hence,

Domain of

$\bf :\longmapsto\:f(x) = \dfrac{1}{ {x}^{2} - 3x + 2 }$

is

$\bf\implies \:x \: \in \: R \: - \: \{1,2 \}$

Additional Information :-

Let us take some more examples.

Question :- Find the domain of the following functions :-

$\bf :\longmapsto\:(a) \: \: f(x) = \sqrt{x + 2}$

For, f(x) to be defined,

$\rm :\longmapsto\:x + 2 \geqslant 0$

$\rm :\longmapsto\:x \geqslant - \: 2$

$\bf\implies \:x \: \in \: [ - \: 2, \: \infty )$

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

For, f(x) to be defined,

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

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

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

$\bf\implies \:x \: \in \: ( - \: \infty , \: 1]$

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

For, f(x) to be defined,

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

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

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

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

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

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

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

For, f(x) to be defined,

$\rm :\longmapsto\:3 - x > 0$

$\rm :\longmapsto\: - x > - 3$

$\rm :\longmapsto\: x < 3$

$\bf\implies \:x \: \in \: ( - \: \infty , \: 3)$