Question

The range of the function $f\left(x\right)=x+\frac{1}{x}$ is

Answer 1

$f\left(x\right)=x+\frac{1}{x}(x\ne 0)$
Now for range, Let
$y=f\left(x\right)$
$y=x+\frac{1}{x}(x\ne 0)$
$yx=x^2+1$
$x^2-yx+1=0$
$\because$x is real here
So Discriminant D of this quadratic equation is
$D\ge 0$ ,so
$y^2-4\ge 0$
$(y-2)(y+2)\ge 0$

$\Rightarrow y\in (-\infty ,-2]\cup [2,\infty )$
Range of $f\left(x\right)=(-\infty ,-2]\cup [2,\infty )$