Question

Could someone explain this to me?
Find any points of discontinuity for the rational function.

Answer 1

Well here's a hint...what happens when you divide something by 0?

\[\large \frac{1}{0} = ?\]

Answer 2

Nothing happens.. 1 divided by 0 is still 1.

Answer 3

Incorrect:

for an example...lets see what happens here

\[\large \frac{1}{1} = 1\]
\[\large \frac{1}{0.5} = 2\]
\[\large \frac{1}{0.1} = 10\]
\[\large \frac{1}{0.01} = 100\]
\[\large \frac{1}{0.001} = 1000\]

etc... as we get closer and closer to zero, what is happening? we are getting bigger and bigger numbers right?

Answer 4

Oh okay.. I see. So it's like infinite, right?

Answer 5

Correct....\[\large \frac{1}{0} = \infty\]

but what we call that here is "undefined"

So basically, with your rational function,when that is "undefined" is where you will have discontinuities.

Answer 6

So with your question

\[\large \frac{(x + 6)(x + 2)(x + 8)}{(x + 9)(x + 7)}\]

we need that denominator to equal 0 right?

so

when does

\(\large (x + 9)(x + 7) = 0\) ?

Answer 7

Umm... I don't know?

Answer 8

Okay

So what we have is

\[\large (x + 9)(x + 7) = 0\]

Tel me what would happen if we make x = -9...

Answer 9

Well 9+(-9) is zero...

Answer 10

Exactly, but lets just do everything out just to see something

\[\large (x + 9)(x + 7) = 0\]
when x = -9
\[\large ((-9) + 9)((-9) + 7) = 0\]
\[\large (0)(-2) = 0\]

and what we see from that...is that when x = -9...the first set of parenthesis does = 0....and since that first set of parenthesis is being MULTIPLIED to the second set, it doesn't matter what is in that second set right? since anything times 0 = 0

Knowing that

\[\large (x + 9)(x + 7) = 0\]

where else would this equal 0?

Answer 11

So would -7 work as well? (Sorry, math is extremely confusing to me.)

Answer 12

Math may be somewhat confusing at some points, but apparently not to you because that is correct :)

Since -7 + 7 would make that second set of parenthesis = 0.....anything being multiplied to that would also = 0

So there we have it, there are 2 points of discontinuity in this rational function

and they occur at x = -9 and at x = -7

Answer 13

Thank you so much for your help!