Question

find the domain of the function f(x)= 2x+1/ xsquare- 3x+2​

Answer 1

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

Given function is

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

can be rewritten as

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

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

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

We know,

Domain of a function f(x) is defined as set of those real values of x for which function f(x) is defined.

$\rm \implies\:f(x) \: is \: defined \: when \: (x - 2)(x - 1) \: \ne \: 0$

$\bf\implies \:x \: \ne \: 1, \: 2$

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

Hence, Domain of

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

is

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

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Additional Information :-

$\boxed{ \tt{ \: x > y \: \: \rm \implies\: - x < - y \: }}$

$\boxed{ \tt{ \: x < y \: \: \rm \implies\: - x > - y \: }}$

$\boxed{ \tt{ \: x < - y \: \: \rm \implies\: - x > y \: }}$

$\boxed{ \tt{ \: x > - y \: \: \rm \implies\: - x < y \: }}$

$\boxed{ \tt{ \: x \geqslant y \: \: \rm \implies\: - x \leqslant - y \: }}$

$\boxed{ \tt{ \: x \leqslant y \: \: \rm \implies\: - x \geqslant - y \: }}$

Answer 2

Answer:

Domain = R - {1,2}

where R is all Real Numbers.