Which function has a removable discontinuity?
A. p(x)=x/x^2-x-2
B. g(x)=2x-1/x
C. f(x)=5x/1-x^2
D. h(x)=x^2-x-2/x+1

Question

Which function has a removable discontinuity?
A. p(x)=x/x^2-x-2
B. g(x)=2x-1/x
C. f(x)=5x/1-x^2
D. h(x)=x^2-x-2/x+1

shamil98 · Answer

Removable discontinuity means that the top and bottom have a value of that sets the fraction to 0/0.

anonymous · Answer

so how can i figure out which one has a removable discontinuity?

shamil98 · Answer

Find the domain of the function. and Plug in that domain for x.

shamil98 · Answer

the domain of a rational function is found by setting the denominator to equal 0

shamil98 · Answer

h(x)=x^2-x-2/x+1
x + 1 = 0
x= -1


=(-1)^2 + 1 - 2/ -1 + 1

=0/0

anonymous · Answer

so does that mean that D would be the one with a removable discontinuity?

shamil98 · Answer

yeah

shamil98 · Answer

$$[x \in \mathbb{R}: x \ne - 1]$$
is your domain btw.

anonymous · Answer

I see, so for this problem would i do the same thing and set everything to 0?

anonymous · Answer

why is that?

shamil98 · Answer

well actually im not sure..
x-3^x / x^2 
or it might be 2/x ..

anonymous · Answer

:O

anonymous · Answer

is there a way to figure out which one it is for sure?

shamil98 · Answer

Probably, bear with me i'm on three hours of sleep.. one sec..

anonymous · Answer

hahah of course, thank you :)

shamil98 · Answer

Yeah, it's 2/x

anonymous · Answer

thank you soo much :)