Question

Someone please help me?
Determine if the graph of this equation is continuous for all real numbers, state “no discontinuities” and explain why, using the conditions of continuity.

Answer 1

If there are any discontinuities, please:
Determine what x value the discontinuity occurs
Determine if the discontinuity if removable or non-removable
If removable, identify the value of the hole. If non-removable, show proof using the conditions of continuity.

Answer 2

looks like you have a problem two places, at \(x=-2\) because the domain of \(\frac{x-1}{x+2}\) does not include \(-2\)
also there is a problem at \(x=2\)

Answer 3

so those are the discontinuities?

Answer 4

yes

Answer 5

ok thank you, How do i find all the other information?

Answer 6

neither are removable because the discontinuity at \(x=-2\) is infinite, that is, as \(x\to -2\) you have a vertical asymptote

Answer 7

it is easy to check the discontinuity at \(x=2\) replace \(x\) by 2 in all three expressions and you get 3 different numbers

Answer 8

\[\frac{2-1}{2+2}=\frac{1}{4}\]
\[2-1=1\]
\[2\times 2=4\] and since all three of these numbers are different
\[\lim_{x\to 2}f(x)\] does not exist

Answer 9

Ohh, ok so what does it mean by findint the value of the hole?

Answer 10

there is no hole here

Answer 11

but it says " If removable, identify the value of the hole. If non-removable, show proof using the conditions of continuity."

Answer 12

does that mean its non-removalble?

Answer 13

an example of a hole would be
\[\frac{x^2-4}{x-2}\] because even though this is not defined at \(x=2\) if you factor and cancel you would get
\[\frac{x^2-4}{x-2}=\frac{(x+2)(x-2)}{x-2}=x+2\] to the original expression looks just like \(x+2\) except there is a hole at \((2,4)\)

Answer 14

in your example neither of the discontinuities are removable, because you cannot "remove them"
in my second example you could remove the discontinuity by defining \(f(2)=4\)

Answer 15

oh ok, thank you so much! I get it now!

Answer 16

yw