# How would you “remove the discontinuity” of f ? In otherwords, how would you define f(x) in order to make continuous f(x) = (x^2 - x - 2)/(x - 2) at 2?

*print*Print*list*Cite

The value of the function `f(x) = (x^2 - x - 2)/(x - 2)` is defined for all values of x other than x = 2.

If we try to determine the value of f(x) at x = 2, we get `lim_(x->2) (x^2 - x - 2)/(x - 2) = (4 - 2 - 2)/(2 - 2) = 0/0` which is not defined.

When the graph of `f(x) = (x^2 - x - 2)/(x - 2)` is plotted, there is a break at x = 2. The discontinuity here is removable as f(x) needs to be redefined only for a single point.

To remove this discontinuity, we can define the function such that `f(x) = (x^2 - x - 2)/(x - 2)` for all `x!=2` and `f(x) = (x^2 - x - 2)/(x - 2) = C` , where C is a constant for x = 2

To remove the discontinuity of `f(x) = (x^2 - x - 2)/(x - 2)` , we could redefine it as:

`f(x) = (x^2 - x - 2)/(x - 2), x!= 2`

`f(x) = 3, x = 2`

To remove the discontinuity, factor the numerator x^2 - x - 2 into (x - 2)(x + 1). The factor (x - 2) in the numerator will cancel with the factor (x - 2) in the denominator leaving the function f(x) = x + 1 which will give the value of 3 when x = 2.

Discontinuity Removed.