Find the range of the following function.

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

$(0,\frac{1}{2}]$

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

Above function is not defined for those values of x for which $x^2+2=0$. But there are no real values of x for which $x^2 = -2$. So, we can take any real value of x. Therefore, domain is set R.

For range let $y=\frac{1}{x^2+2}$

So, $x^2+2=\frac{1}{y}$ or $x^2=\frac{1}{y}-2=\frac{1-2y}{y}$

Now $x^2≥0$ for all real values of x.

$∴\frac{1-2y}{y}≥0$

$⇒\frac{2y-1}{y}≤0⇒0<y≤1/2$

Therefore, range is $(0,\frac{1}{2}]$