- #1

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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

- Thread starter asdfsystema
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- #1

- 87

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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

- 1,631

- 4

And you are trying to do WHAT?

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

- #3

- 37

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actually, which limit are you trying to take, a side limit or THE limit,

I too am very confused

I too am very confused

- #4

- 87

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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

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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

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- #7

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[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

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[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.

[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

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(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

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

- #11

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could you check to verify?

- #12

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You did a mistake in here: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.

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

- #13

- 37

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oh good catch, yeah my algebra sucks, sorry people

- #14

- 87

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thanks . made a couple of mistakes haha

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