What is the Limit of f(x) as x approaches -3?

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Discussion Overview

The discussion revolves around finding the limit of a piecewise function as x approaches -3, specifically addressing the continuity of the function at that point. Participants explore the implications of the two different definitions of the function on either side of -3 and the necessary conditions for continuity.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants present the piecewise function and express confusion about the limit and continuity at x = -3.
  • One participant suggests that to find the value of 'a' for continuity, the two functions must be set equal at x = -3.
  • Another participant proposes calculating the left-hand limit and right-hand limit as x approaches -3 to ensure they are equal for continuity.
  • There is a discussion about the algebra involved in factoring the numerator and the potential for errors in calculations.
  • Some participants express uncertainty about their calculations and seek confirmation from others regarding their answers.
  • One participant corrects another's algebraic mistake, indicating the importance of careful verification in the problem-solving process.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct value of 'a' or the limit itself, as there are multiple calculations and corrections presented, leading to different interpretations and results.

Contextual Notes

There are unresolved mathematical steps and potential errors in factoring and calculations that contribute to the uncertainty in the answers provided.

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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
 
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asdfsystema said:
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

And you are trying to do WHAT?
 
actually, which limit are you trying to take, a side limit or THE limit,
I too am very confused
 
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
 
you have to set those two functions together and solve for a, you are given x=-3
since we don't 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.
 
thank you for your help . Is there anyone of you math experts out there willing to confirm this answer ? thank you in advance :)
 
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.
 
[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:
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 equal3x^2+3x+A = 4x^2+x+18

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

Is that correct ?
 
  • #10
are u sure u have factored the top correctly? And you are not plugging in 3, but -3.
 
  • #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?
 
  • #12
St. Aegis said:
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.

You did a mistake in here:

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

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