Question

Range of the function f(x) = x² + 3 / x² + 1 is?

Answer 1

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

Given function is

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

To find the range of the function,

Let assume that

$\rm :\longmapsto\:y = \dfrac{ {x}^{2} + 3}{ {x}^{2} + 1}$

$\rm :\longmapsto\: {yx}^{2} + y = {x}^{2} + 3$

$\rm :\longmapsto\: {yx}^{2} - {x}^{2} = 3 - y$

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

$\rm :\longmapsto\: {x}^{2} = \dfrac{3 - y}{y - 1}$

$\rm :\longmapsto\:x= \sqrt{\dfrac{3 - y}{y - 1} }$

Now, x is defined when

$\rm :\longmapsto\:3 - y \geqslant 0 \: \: \: and \: \: \: y - 1 > 0$

$\rm :\longmapsto\: - y \geqslant - 3 \: \: \: and \: \: \: y > 1$

$\rm :\longmapsto\: y \leqslant 3 \: \: \: and \: \: \: y > 1$

$\bf\implies \:1 < y \leqslant 3$

$\bf\implies \:y \: \in \: (1, \: 3]$

Additional Information :-

If a and b are real numbers, such that a < b, then

$\red{ \boxed{ \sf{ \:(x - a)(x - b) < 0 \: \implies \: a < x < b \: }}}$

$\red{ \boxed{ \sf{ \:(x - a)(x - b) \leqslant 0 \: \implies \: a \leqslant x \leqslant b \: }}}$

$\red{ \boxed{ \sf{ \:(x - a)(x - b) > 0 \: \implies \: x < a \: \: or \: \: x > b \: }}}$

$\red{ \boxed{ \sf{ \:(x - a)(x - b) \geqslant 0 \: \implies \: x \leqslant a \: \: or \: \: x \geqslant b \: }}}$

$\red{ \boxed{ \sf{ \: \frac{x - a}{x - b} < 0 \: \implies \: a < x < b}}}$

$\red{ \boxed{ \sf{ \: \frac{x - a}{x - b} \leqslant 0 \: \implies \: a \leqslant x < b}}}$

$\red{ \boxed{ \sf{ \: \frac{x - a}{x - b} \geqslant 0 \: \implies \: x \leqslant a \: \: \: or \: \: \: x > b}}}$