Question

what is the point of discontinuity of
f(x) = x^2-64/x-8

Could someone please explain this to me? Thanks.

Answer 1

Let me see if I can.

Basically, the question is since

\[\frac{x^2 - 64}{x - 8} = x + 8\]

However they are only equal if \(x \ne 8\)

Answer 2

if denominator becomes zero, the function will become undefined so the discontinuity of the function will be the zero of the denominator.

Answer 3

In other words, if you graphed

\[\frac{x^2 - 64}{x - 8}\]

There would be an empty circle in the graph at x = 8

Answer 4

so that would be considered the point of discontinuity?

Answer 5

Or would it be 8,16?

Answer 6

however,if you factor out the numerator the denominator will cancel out, so the function will have a hole at x=8...

Answer 7

so the point of disconitunity is where there is a hole?

Answer 8

and that hole is at 8?

Answer 9

yes you are right..

Answer 10

thank you!

Answer 11

no problem