Question

Find the range $x^2+\frac{1}{x^2+2}$

Answer 1

$f\left(x\right)=x^2+\frac{1}{x^2+2}$

$f\left(x\right)=\{-\infty ,\infty \}$

$x=0f\left(x\right)=x^2+\frac{1}{x^2+2}$

$=0+\frac{1}{0+2}=\frac{1}{2}$

So the range of $f\left(x\right)=(\frac{1}{2},\infty )$