# How do you find the range of f(x)=(2x^2 -1) / ( x^2 +1)?

Jul 23, 2015

You first look for restriction in the domain. In this there are none, since the denominator will always be at least $1$
$x = 0 \to f \left(x\right) = \frac{- 1}{1} = - 1$
The maximum is when $x$ gets really large, the function will be nearing $\frac{2 {x}^{2}}{x} ^ 2 = 2$ without reaching it: ${\lim}_{x \to \infty} f \left(x\right) = 2$
So the range is $- 1 \le f \left(x\right) < 2$