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factor the denominator, then set it equal to zero and solve
I have a graph D:
if one of the factors cancels with a factor in the numerator, then it is "removable" because you can remove it
oh then it is even easier, you do it with your eyeballs instead of algebra
course it would be much easier to help if we could see the picture :D
I'm trying. give me a moment.
okay sorry it looks bad. the single point is at (-1,1) and the open circle is at (-1,3) and the other line runs between -2 and -3 on the x-axis. @satellite73
so the one with the hole is "removable" the other one isn't
Can you explain why?
|dw:1441327899031:dw| The empty dot could have been a removable discontinuity, because we could have added a definition of the function for f(-1)=3. However, since f(-1) has been defined as -1, this discontinuity becomes non-removable. x=-2 is a vertical asymptote, which is a non-removable discontinuity.
I am so confused
the one with the hole is removable cause you can remove it by filling in the hole
the one with an "infinite discontinuity" can't be adjusted to make it continuous, so it is not removable
which one is that? I'm sorry for so many questions but I don't understand
Not everyone may agree with me, but according to the definition, a removable discontinuity is when a point (a hole) is undefined, in which case we can fill the hole and the function becomes continuous. (a hole is where the limit of the function from the left and right are defined and identical) However, in your case, the value of f(-1) has been defined (f(-1)=1), so we can no longer fill the hole (to make it continuous), so it is no longer removable.
yes, I learned what it is and I'm looking at my notes right now and it just isn't making any sense at all.
when it says to identify the x-values, is it asking for the domain?
Please explain what bothers you.
Just the question in general... what it's asking me to do
There are different kinds of discontinuities - removable discontinuities (where a hole interrupts an otherwise continuous curve) - essential discontinuities, which subdivides into - vertical asymptotes, and (as in x=-3) - step discontinuities (a jump in y-value) See: http://www.mathwords.com/e/essential_discontinuity.htm and http://www.mathwords.com/r/removable_discontinuity.htm The question is asking you to locate and identify the type of discontinuity in the given example.
I only learned continuous, removable discontinuity, jump discontinuity, and infinite discontinuity.
Well these are the same, but with a different name. continuous is not a discontinuity jump => step vertical asymptote =>infinite removable is where you can plug the "hole". In the given example, since f(-1)=1, you can no longer plug the "hole" (-1,3), so in my opinion, it becomes a jump discontinuity.