1. Sep 28, 2008

### asdfsystema

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. Sep 28, 2008

### sutupidmath

And you are trying to do WHAT?

3. Sep 28, 2008

### St. Aegis

actually, which limit are you trying to take, a side limit or THE limit,
I too am very confused

4. Sep 28, 2008

### asdfsystema

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. Sep 28, 2008

### St. Aegis

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. Sep 28, 2008

### asdfsystema

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. Sep 28, 2008

### sutupidmath

ok, what you need to do is take two sided limits as x-->-3, that is:

$$\lim_{x\rightarrow -3^+}f(x)$$ and $$\lim_{x\rightarrow -3^-}f(x)$$ , 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. Sep 28, 2008

### sutupidmath

$$\lim_{x\rightarrow -3^+}f(x)=\lim_{x\rightarrow -3^+}3 x^2 + 3 x + a$$

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

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
9. Sep 29, 2008

### asdfsystema

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. Sep 29, 2008

### sutupidmath

are u sure u have factored the top correctly? And you are not plugging in 3, but -3.

11. Sep 29, 2008

### St. Aegis

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. Sep 29, 2008

### sutupidmath

You did a mistake in here:

a=x^2-2x+8=9+6+8=23

13. Sep 29, 2008

### St. Aegis

oh good catch, yeah my algebra sucks, sorry people

14. Sep 29, 2008

### asdfsystema

thanks . made a couple of mistakes haha