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Please check Limits problem

  1. Sep 28, 2008 #1
    f(x)= 4 x^3 +13 x^2+ 11 x+ 24 }/ { x + 3 } when x<-3

    f(x)= 3 x^2 + 3 x + a when -3 less than or equal to x

    4x^3+13x^2+11x+24/(x+3 = (x+3)(4x^2+x+8)/ (x+3) = 4x^2+x+8

    The answer is 22? my teacher said it was wrong

    thanks a lot
     
  2. jcsd
  3. Sep 28, 2008 #2
    And you are trying to do WHAT?
     
  4. Sep 28, 2008 #3
    actually, which limit are you trying to take, a side limit or THE limit,
    I too am very confused
     
  5. Sep 28, 2008 #4
    lol i have no idea .. i just showed the first couple of steps that i took.

    here's the entire problem :

    The function f is given by the formula
    f(x)= { 4 x^3 +13 x^2+ 11 x+ 24 }/{ x + 3 }
    when x < -3 and by the formula
    f(x)=3 x^2 + 3 x + a
    when -3 is less than or equal x.
    What value must be chosen for a in order to make this function continuous at -3?


    sorry i just realized i forgot to add the last part
     
  6. Sep 28, 2008 #5
    you have to set those two functions together and solve for a, you are given x=-3
    since we dont care about differentiability,
    { 4 x^3 +13 x^2+ 11 x+ 24 }/{ x + 3 }=3 x^2 + 3 x + a
    4x^2+x+8=3 x^2 + 3 x + a
    3x^2-2x+8=a
    plug in -3
    3(-3)^2-2(-3)+8=27+6+8=41
    but for some reason, i am completley not confident in the answer in this lawl.
     
  7. Sep 28, 2008 #6
    thank you for your help . Is there any one of you math experts out there willing to confirm this answer ? thank you in advance :)
     
  8. Sep 28, 2008 #7
    ok, what you need to do is take two sided limits as x-->-3, that is:

    [tex] \lim_{x\rightarrow -3^+}f(x)[/tex] and [tex]\lim_{x\rightarrow -3^-}f(x)[/tex] , now this function will be continuous at -3, if these two-sided limits are equal. That is take these limits, set them equal to each other, and then solve for a.
     
  9. Sep 28, 2008 #8
    [tex]\lim_{x\rightarrow -3^+}f(x)=\lim_{x\rightarrow -3^+}3 x^2 + 3 x + a[/tex]

    [tex]\lim_{x\rightarrow -3^-}f(x)=\lim_{x\rightarrow -3^-}\frac{4 x^3 +13 x^2+ 11 x+ 24 } { x + 3 }[/tex]

    for the first one , it is pretty straightforward, since -3 does not represent any trouble, but the last one seems to be undefined at -3, so try to factor the numerator, and see if things cancel out, and after that set it equal to the first part. And show us what u tried.
     
    Last edited: Sep 29, 2008
  10. Sep 29, 2008 #9
    i factored out 4x^3+13x^2+11x+24/x+3 to

    (4x^2+x+18) (x+3) / (x+3) canceled them out.

    then set it equal


    3x^2+3x+A = 4x^2+x+18

    i end up plugging in 3 and then getting 51-18 = 33

    Is that correct ?
     
  11. Sep 29, 2008 #10
    are u sure u have factored the top correctly? And you are not plugging in 3, but -3.
     
  12. Sep 29, 2008 #11
    notice how i recommended the same procedure but in one of your steps, your 8 turned into an 18,
    could you check to verify?
     
  13. Sep 29, 2008 #12
    You did a mistake in here:

    a=x^2-2x+8=9+6+8=23
     
  14. Sep 29, 2008 #13
    oh good catch, yeah my algebra sucks, sorry people
     
  15. Sep 29, 2008 #14
    thanks . made a couple of mistakes haha
     
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