When looking at a rational fraction f(x) = (x-6)(x+3(x+4) over (x+6) (x-3) (x-4), jamal and angie have two different thoughts. jamal says that the function is defined at x = p-6, x = 3, and x = 4. angie says that the function is undefined at those x values. who is correct? justify your reasoning.

We say a function is undefined when the value gotten isn't present in the domain of the function, like division by a 0 denominator or 0 to the power of 0. That said, we would like to find out whether at each point in the domain, the two points x = 3 and x = 4 exist i.e we don't end up with the indeterminate form. Okay at x =3, we have f(3) = (3-6) (3+3) (3+4)/(3+6) (3-3) (3-4) = (-3) (6) (7) / 0 which is undefined. We don't have to show for x =4 since "4" is negative in the denominator. We would also end up with an undefined function. So, yes, Angie is right.

Just turn the whole gym into pie