Question

Is the dicontinuity in the function of f(x)=abs(x^2-1)/(x-1) removable? if so how do you remove it?

Answer 1

Is the function \[\Large f(x) = \frac{|x^2-1|}{x-1}\] or is it \[\Large f(x) = \left|\frac{x^2-1}{x-1}\right|\]

Answer 2

\[f(x)=| x^2- 1|\div(x-i)\]

Answer 3

oops (x-1) on the bottom

Answer 4

well \[\Large x^2-1=(x-1)(x+1)\]


So \[\Large |x^2-1|=|(x-1)(x+1)|=|x-1||x+1|\]


which means \[\Large |x^2-1|=|x-1||x+1|\]



Now | x - 1 | = x-1 when x is greater than or equal to 1


So \[\Large |x^2-1|=(x-1)|x+1| \ \text{ if x $\ge$ 1}\]


This means that \[\Large |x^2-1|=\frac{(x-1)|x+1|}{x-1} \ \text{ if x $\ge$ 1}\]


From here, cancel the common terms to get \[\Large |x^2-1|=|x+1| \ \text{ where x $\ge$ 1}\]


BUT....


since the original equation has us dividing by x - 1, this means that we must exclude x = 1 from the domain.


So the removable discontinuity is x = 1 and that updates the final equation to be


\[\Large |x^2-1|=|x+1| \ \text{ where x $>$ 1}\]