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- anonymous

Define f(x) so that ((x^3)-(x^2)-x-2)/((x^2)-4) is continuous at x = 2.

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- anonymous

Define f(x) so that ((x^3)-(x^2)-x-2)/((x^2)-4) is continuous at x = 2.

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- anonymous

I know what it means for a function to be continuous at a certain point and I know that the above function is not continuous at x = 2. But I don't understand what the question means when it asks for one to define the function.

- zepdrix

Hmmm i think they're asking you to do something like this.|dw:1348695324335:dw|
we need to find a value for c that will make it a nice smooth curve
so whats happening is.. you have a HOLE at x=-2, and at x=2.
So we want to go in and patch the hole

- zepdrix

|dw:1348695490464:dw|woops i missed one of the points :x

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- zepdrix

so the way to find the y value of the HOLEs is to approach the hole from the left or right :D
what operation allows us to get closer and closer and closer to a point, without actually touching it?

- anonymous

Limits?

- anonymous

So I'll want to find the left and right limits then?

- zepdrix

mm yes i think so :D let me check first, i dont wanna lead you down the wrong path if i made a mistake.

- zepdrix

Hmm ok this problem is a bit nasty, since you need to factor a term out of a cube. I dont really remember how to do that part, i had to enter it into wolfram :(
|dw:1348696335593:dw|
lemme rewrite the initial function, cause i think i made a tiny notation error :)
so we would have this so far..|dw:1348696501525:dw|

- zepdrix

so f(x)=7/4 will plug the hole at x=2
now we need to plug the hole at x=-2.. so we'll have to do similar to what we did before, except we need to try and factor an (x+2) from that darn cube D:
any of this making sense? XD

- anonymous

I got most of it except why did you choose to get a c1 and c2? Or rather,how did you know that x is not equal to -2? In this case, is c defined as the value that plugs in the gap?

- zepdrix

that's why i corrected the chart, it's not x=c1, its f(x) = c1 , when x=2
the c1 is the Y value that we needed

- anonymous

Where did you get the x=/= -2 though?

- zepdrix

|dw:1348697608343:dw|
lets pretend the graph looks like this
see the holes at -2 and 2?
those are making our function discontinuous.
so to fix that that, we change the function to a piece-wise function and we say
"no no no, when x=-2 or 2, we'll define with a different piece than this function"

- anonymous

Oh ok then. Thanks a lot.

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