anonymous
  • anonymous
Define f(x) so that ((x^3)-(x^2)-x-2)/((x^2)-4) is continuous at x = 2.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • 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
  • zepdrix
|dw:1348695490464:dw|woops i missed one of the points :x

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

zepdrix
  • 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
  • anonymous
Limits?
anonymous
  • anonymous
So I'll want to find the left and right limits then?
zepdrix
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
Where did you get the x=/= -2 though?
zepdrix
  • 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
  • anonymous
Oh ok then. Thanks a lot.

Looking for something else?

Not the answer you are looking for? Search for more explanations.