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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.
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
|dw:1348695490464:dw|woops i missed one of the points :x
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?
So I'll want to find the left and right limits then?
mm yes i think so :D let me check first, i dont wanna lead you down the wrong path if i made a mistake.
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|
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
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?
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
Where did you get the x=/= -2 though?
|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"
Oh ok then. Thanks a lot.