# find all possible values of f(x)?

## f(x)=(1-x^2)/(x^2+3

Jun 16, 2018

$f \left(x\right) \in \left(- 1 , + \frac{1}{3}\right]$

#### Explanation:

$f \left(x\right) = \frac{1 - {x}^{2}}{{x}^{2} + 3}$

$f \left(x\right)$ is defined $\forall x \in \mathbb{R} \to$ the domain of $f \left(x\right)$ is $\left(- \infty , + \infty\right)$

Now consider ${\lim}_{x \to \pm \infty} f \left(x\right)$

$f \left(x\right) = \frac{\frac{1}{x} ^ 2 - 1}{1 + \frac{3}{x} ^ 2}$

${\lim}_{x \to + \infty} f \left(x\right) = \frac{0 - 1}{1 + 0} = - 1$

${\lim}_{x \to - \infty} f \left(x\right) = \frac{0 - 1}{1 + 0} = - 1$

Thus, $f {\left(x\right)}_{\min} \to - 1$

Now let's find $f {\left(x\right)}_{\max}$

The maximum value of $f \left(x\right)$ will occur for the value of x where the numerator is a maximum and the denominator is a minimum,

${\left(1 - {x}^{2}\right)}_{\max} \to x = 0$ and ${\left({x}^{2} + 3\right)}_{\min} \to x = 0$

$\to f {\left(x\right)}_{\max} = f \left(0\right) = \frac{1}{3}$

Thus the range (and hence all possible values) of $f \left(x\right)$ is $\left(- 1 , + \frac{1}{3}\right]$

We can observe this result from the graph of $f \left(x\right)$ below.

graph{(1-x^2)/(x^2+3) [-5.54, 5.556, -2.783, 2.764]}