Ask your own question, for FREE!
Mathematics 9 Online
OpenStudy (anonymous):

Determine if each function is continuous. If the function is not continuous, find the x-axis location of and classify each discontinuity.

OpenStudy (anonymous):

\[f(x)=\frac{ x+1 }{ x^2-x-2 }\]

OpenStudy (anonymous):

P.D.: 2,-1

terenzreignz (terenzreignz):

This is where factoring the denominator might come in handy.

OpenStudy (anonymous):

the (x+1) cancel out right?

terenzreignz (terenzreignz):

Okay, you seem to have gotten it... now to classify... If a certain point of discontinuity makes the denominator and ONLY the denominator zero, then it is an infinite discontinuity. On the other hand, if a certain point of discontinuity makes BOTH the numerator and denominator zero, then it is a removable discontinuity (or a hole).

OpenStudy (anonymous):

I think there is a hole at x=1?

terenzreignz (terenzreignz):

x = 1 is not even a point of discontinuity... check the P.D's you gave me :)

OpenStudy (anonymous):

the hole is at x=-1? but then the equation would be 0/(x-2)

terenzreignz (terenzreignz):

hole is at x = -1 because at x = -1, both the numerator and denominator are zero :)

OpenStudy (anonymous):

So it's a removable discontinuity.

terenzreignz (terenzreignz):

Yup. And the other one, x = 2 ?

OpenStudy (anonymous):

? umm they're both removable?

OpenStudy (anonymous):

or is that the x-axis location?

terenzreignz (terenzreignz):

No... I said... a P.D. would be removable if it makes BOTH the numerator and denominator zero ... so check

OpenStudy (anonymous):

x=2 only makes the denominator 0

terenzreignz (terenzreignz):

So... is it a removable discontinuity?

OpenStudy (anonymous):

No

terenzreignz (terenzreignz):

Then what is it...?

OpenStudy (anonymous):

So it's an infinite discontinuity right?

terenzreignz (terenzreignz):

Yup.

OpenStudy (anonymous):

when they say x-axis location, would that be x=2?

terenzreignz (terenzreignz):

Yes.

OpenStudy (anonymous):

alright once again thank you, that was my last question, i can finish the rest xD

terenzreignz (terenzreignz):

Awesome :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!