Question

The range of the function $f\left(x\right)=\sqrt{x^2+\frac{1}{x^2}}$ is equal to.

Answer 1

$(x-\frac{1}{x}{)}^2\ge 0⟹x^2+\frac{1}{x^2}-2\ge 0$

$⟹x^2+\frac{1}{x^2}\ge 2$

$⟹\sqrt{x^2+\frac{1}{x^2}}\in (-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty )$

Range of $f\left(x\right)$ is $(-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty )$