Question

"Find the x-values at which f is not discontinuous in: (x-3)\(x^2-9). Which of the discontinuities are removable?"
Well, I already simplified the problem to 1\(x+3)and I found at nonremovable discontinuity at x=-3... but what does it mean by removable discontinuity? Are there any? If so, how do I find it?

Answer 1

you have already "removed" the discontinuity when you factored and canceled

Answer 2

the original function is undefined both at \(x=3\) and at \(x=-3\) but \(\frac{1}{x+3}\) is defined at 3

Answer 3

in fact if you replace \(x=3\) you get \(\frac{1}{6}\) but if you replace \(x\) by 3 in the original function you would get \(\frac{0}{0}\) which is not a number

Answer 4

therefore your original function has a "removable" discontinuity at \(x=3\) and you can remove it by defining \(f(3)=\frac{1}{6}\) which is what it ought to be