1. ## Continuous Functions

I'm having some trouble with this question:

$\displaystyle Consider\ the\ function$

$\displaystyle f(x)=\frac{x^3-6x^2+11x-6}{x-a}$

$\displaystyle This\ is\ not\ continuous\ at\ x=a.$ $\displaystyle For\ which\ values\ of\ a\ is\ the\ disconuity\ removable?$

2. Hi

The discontinuity is removable if a is a solution of $\displaystyle x^3 - 6x^2 + 11x - 6 = 0$ because in this case you can write

$\displaystyle x^3 - 6x^2 + 11x - 6 = (x-a) P(x)$ where P is a second degree polynomial

And then $\displaystyle f(x) = \frac{x^3 - 6x^2 + 11x - 6}{x-a} = \frac{(x-a) P(x)}{x-a} = P(x)$ is continuous at a

3. Originally Posted by acevipa
I'm having some trouble with this question:

$\displaystyle Consider\ the\ function$

$\displaystyle f(x)=\frac{x^3-6x^2+11x-6}{x-a}$

$\displaystyle This\ is\ not\ continuous\ at\ x=a.$ $\displaystyle For\ which\ values\ of\ a\ is\ the\ disconuity\ removable?$
first is $\displaystyle x^3-6x^2+11x-6$ composite it is

$\displaystyle x^3-6x^2+11x-6 = (x-2)(x^2-4x+3)$

is $\displaystyle x^2-4x+3)$ composite

so the first value of a is 2

you can continue