Please check Limits problem

  • #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
 

Answers and Replies

  • #2
1,631
4
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?
 
  • #3
37
0
actually, which limit are you trying to take, a side limit or THE limit,
I too am very confused
 
  • #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
 
  • #5
37
0
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.
 
  • #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 :)
 
  • #7
1,631
4
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.
 
  • #8
1,631
4
[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:
  • #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 ?
 
  • #10
1,631
4
are u sure u have factored the top correctly? And you are not plugging in 3, but -3.
 
  • #11
37
0
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
1,631
4
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
37
0
oh good catch, yeah my algebra sucks, sorry people
 
  • #14
thanks . made a couple of mistakes haha
 

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