Question

Calculus Extra Credit part 1: Limits and Continuity.

Answer 1

OK?

Answer 2

\[\Large f(x) =\frac{ x^3 -3x^2 -x +3}{ x+1} \]

Answer 3

Factor the numerator (you can do it by grouping).

Answer 4

What are the removable discontinuities of f(x) = \[\frac{ x^3 - 3x^2 - x +3 }{ x+1 }\] ? How do you remove them?

Answer 5

\[\Large x^3 -3x^2 -x +3 = x^2(x -3)-(x-3)\]

Answer 6

Can you finish factoring? If you're in calculus, you should be able to :P

Answer 7

Once you find the x-value of the discontinuity, you need to find the limit of the function at that x value.

Answer 8

sorry my computer Is acting up. I got x^3 -3x^2 -x +3 when I factored

Answer 9

\[\Large x^2(x -3)-(x-3) = (x^2-1)(x-3)\]Now finish factoring it.

Answer 10

Use the difference of two squares.

Answer 11

(x+1) and (x-1) = (x^2 -1)

Answer 12

Good, so \[\Large f(x) =\frac{ (x+1)(x-1)(x-3)}{ x+1}\]which factor cancels? That gives your discontinuity. What x value is it?

Answer 13

Hopefully you found it as x=-1.

Now find the limit where x=-1 (it's easy, plug in x=-1)\[\Large \lim_{x \rightarrow -1}(x-1)(x-3)=?\]

Answer 14

sorry my mom needed my help. 10

Answer 15

@agent0smith

Answer 16

All you have to do is plug in -1 into the above, it's not 10...

Answer 17

8

Answer 18

Yep, so when x=-1, you can let the value of the function equal 8.

Answer 19

okay