How do you decide whether the relation f(x)= x/ [(x+2)(x-2)] defines a function?

May 5, 2016

This formula does define a function since it defines a unique value for any $x$ in the (implicit) domain.

Explanation:

For any Real value of $x$ except $x = \pm 2$, the formula:

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

uniquely defines a unique Real value.

When $x = \pm 2$, the denominator of $f \left(x\right)$ is zero, so $f \left(x\right)$ is undefined and these values of $x$ are not in the domain.

So the given formula defines a function on the (implicit) domain:

$\left(- \infty , - 2\right) \cup \left(- 2 , 2\right) \cup \left(2 , \infty\right)$

The graph of $f \left(x\right)$ satisfies the vertical line test.

graph{x/((x+2)(x-2)) [-10, 10, -5, 5]}

Any vertical line only intersects $f \left(x\right)$ at one point at most.