Question

Sir pls tell me what is the range of function x+1/x

Answer 1

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

Given function is

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

Let assume that,

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

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

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

$\rm :\longmapsto\: {x}^{2} + 1 = xy$

$\rm :\longmapsto\: {x}^{2} + 1 - xy = 0$

$\rm :\longmapsto\: {x}^{2} - xy + 1 = 0$

Its a quadratic equation, so roots of this equation is evaluated by using Quadratic formula, given as

$\red{ \boxed{ \sf{ \:x = \frac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}}}}$

Here,

$\red{\rm :\longmapsto\:a = 1 \: \: \: \: } \\ \red{\rm :\longmapsto\:b = - y} \\ \red{\rm :\longmapsto\:c = 1 \: \: \: \: }$

On substituting the values, we get

$\rm :\longmapsto\:x \: = \: \dfrac{y \: \pm \: \sqrt{ {y}^{2} - 4 \times 1 \times 1} }{2 \times 1}$

$\rm :\longmapsto\:x \: = \: \dfrac{y \: \pm \: \sqrt{ {y}^{2} - 4} }{2}$

Now, for x to be real,

$\rm :\longmapsto\: {y}^{2} - 4 \geqslant 0$

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

$\rm :\longmapsto\:(y - 2)(y + 2) \geqslant 0$

$\bf\implies \:y \leqslant - 2 \: \: or \: \: y \geqslant 2$

$\bf\implies \:y \: \in \: ( - \infty , \: - \: 2\bigg] \: \cup \: \bigg[2, \: \infty )$

Additional Information :-

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

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

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

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

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