Find range of the real function
[tex]f = \left \{ \bigg(x, \: \dfrac{ {x}^{2} }{ {x}^{2} + 1} \bigg) : x \: \in \: R \right \}[/tex]
from R into R​.

Respuesta :

Step-by-step explanation:

[tex] \large\underline{\sf{Solution-}}[/tex]

Given function is

[tex]\rm \longmapsto\:f = \bigg(x, \: \dfrac{ {x}^{2} }{ {x}^{2} + 1} \bigg) : x \: \in \: R[/tex]

To find the range of the function f, Let assume that

[tex]\rm \longmapsto\:y = \dfrac{ {x}^{2} }{ {x}^{2} + 1} [/tex]

[tex]\rm \longmapsto\: {yx}^{2} + y = {x}^{2} [/tex]

[tex]\rm \longmapsto\: {yx}^{2} - {x}^{2} = - y[/tex]

[tex]\rm \longmapsto\: - {x}^{2} (1 - y) = - y[/tex]

[tex]\rm \longmapsto\: {x}^{2} (1 - y) = y[/tex]

[tex]\rm \longmapsto\: {x}^{2} = \dfrac{y}{1 - y} [/tex]

[tex]\rm\implies \:x = \sqrt{\dfrac{y}{1 - y} } [/tex]

For x to be defined,

[tex]\rm \longmapsto\:y \geqslant 0 \: \: and \: \: 1 - y > 0[/tex]

[tex]\rm \longmapsto\:y \geqslant 0 \: \: and \: \: - y > - 1[/tex]

[tex]\rm \longmapsto\:y \geqslant 0 \: \: and \: \: y < 1[/tex]

[tex]\bf\implies \:y \: \in \: [0, \: 1)[/tex]

Hence,

[tex]\bf\implies \:Range \: of \: f \: \in \: [0, \: 1)[/tex]