Question

Given the function f(x) = (x-2)/(x^3-8), what value should be assigned to f(x) to make the function continuous at x=2?

Please walk me through this. I have the answer but do not know how to get it.

Answer 1

When you look at that, you see that when x = 2, both the top and the bottom are zero.
Therefore (x-2) is a factor of the bottom.
Divide (x^3-8) by (x-2) to get the rest of the denominator
I get the denominator is actually factored
(x-2)(x^2 + 2x +4)
so now we have

f(x) = (x-2) / [ (x-2)(x^2 +2x + 4)]

well, lo and behold we can write our original f(x) as

f(x) = 1 / (x^2 + 2x +4)
which is just fine when x = 2

Answer 2

How do you factor the (x-2) opt, I do not understand the steps to do this?

Answer 3

Ok, well I can see that (x-2) is a factor of the bottom because when x = 2 the bottom x^3- 8 = 8-8 = 0
so
I must be able to divide (x^3-8) by (x-2) with no remainder.
Write (x^3-8) as (x^3 + 0x^2 + 0x -8)
and do a long division
first: put x^2 up top
x^2 (x-2)
then x^3 + 0x^2 +0x - 8
subtract x^3 - 2 x^2

2 x^2 +0 x - 8 is remainder so far
then +2x
+2x (x-2) gives
2x^2 - 4x
subtract again
4 x - 8 is now remainder on the bottom
then +4
+4 (x-2) gives
4x -8
subtract that and remainder is zero
so the result is
x^2
+ 2x
+4