1. This site uses cookies. By continuing to use this site, you are agreeing to our use of cookies. Learn More.

=-=-=-=-=-Calculus question !!!!=-=-=-=-=

Discussion in 'BlackHat Lounge' started by dkdanielkli, Oct 5, 2012.

  1. dkdanielkli

    dkdanielkli Regular Member

    Joined:
    Aug 30, 2011
    Messages:
    291
    Likes Received:
    110
    Location:
    Canada
    I have a quiz tomorrow and I'm pretty much able to solve everything in this chapter. But there is this one question that i'm don't even know how to start and it's probably going to appear on the quiz tomorrow.
    Any help ?
    http://i47.tinypic.com/25ouuzp.png
    PLZ explain how you got the answer.
     
  2. LOL-Blaster

    LOL-Blaster Regular Member

    Joined:
    Aug 29, 2012
    Messages:
    342
    Likes Received:
    706
    Don't you think this is a bad forum to ask this kind of question?
     
  3. huser21

    huser21 Regular Member

    Joined:
    Apr 16, 2009
    Messages:
    219
    Likes Received:
    71
    Location:
    Minny
    Its actually not that hard, only time consuming. And no I wont tell you the answer.
     
  4. directaxcess

    directaxcess Power Member

    Joined:
    Sep 12, 2011
    Messages:
    780
    Likes Received:
    154
    Did you attempt the question? It's pretty easy. Simplify the first expression of the function, plug f(x) into the second equation when x = 2 ... do the same for x = 3 (essentially). I'm too lazy to write this out, so I must say if you just searched this on Google, you would have found the answer :)

    http://answers.yahoo.com/question/index?qid=20100927182406AAqOnL6
     
  5. tacopalypse

    tacopalypse Executive VIP Jr. VIP Premium Member

    Joined:
    Nov 30, 2009
    Messages:
    980
    Likes Received:
    2,485
    Home Page:
    funny how somebody asked almost the exact same question 2 years ago on yahoo answers lol.

    you've got a function that's split up into 3 parts. each part is defined by one of the equations. the parts are joined together at locations x=2 and x=3.

    in order to make the whole function continuous, you need to determine values for a and b such that at the locations x=2 and x=3, the 2 adjoining functions there have the same y-values.

    so you plug in x=2 for the left and middle functions and set them equal to each other, and you'll have an equation in terms of just a and b.

    then you plug in x=3 for the middle and right functions, set then equal to each other, and you have another equation in terms of a and b.

    then just use plain algebra to solve the system of 2 equations with 2 unknowns and you have your answer.