Originally Posted by

**A Beautiful Mind** Okay. I got this function.

$\displaystyle f(x)= x/9x^2-4$

The answer is zero, but I don't understand how it's zero because I worked it out like this.

I took the bottom half and realized it was $\displaystyle (3x+2)(3x-2)$.

$\displaystyle 3x+2=0 $

$\displaystyle 3x=-2$

$\displaystyle x= -2/3$

$\displaystyle 3x-2=0$

$\displaystyle 3x=2$

$\displaystyle x = 2/3$

So what does that mean?

$\displaystyle 2/3 $

----

$\displaystyle

9 (2/3)^2-4$

The bottom half would be zero after the computation, am I right? And you can't divide by zero. But how does that make x zero?