Question

Range of $f\left(x\right)=$$x^2+\frac{1}{x^2+4}$is

Answer 1

Since the function has only terms of $x^2$ so any positive or negative value of x will give same value of f(x)

So the smallest value of x that we can input in f(x) is 0

$f\left(0\right)=0^2+\frac{1}{0^2+4}=\frac{1}{4}$

If we take $x\to \infty$ then $f\left(x\right)\to \infty$

So range of $f\left(x\right)=[\frac{1}{4},\infty )$