f(x)= ((x^2)+4x-5)/((x^2)-1)
What would be the formula for the extended function that's continuous at x=1?
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The reason that is not continuous at x= 1, as it stands, is that the denominator is 0 at x= 1. The reason it is possible to "make it" continuous (it has a "removable" discontinuity) is that the numerator is also 0 at x= 1. And the fact that the numerator is 0 at x= 1 tells you that x- 1 is a factor. You should have thought immediately of $\displaystyle x^2+ 4x- 5= (x- 1)(x+ b)$. Multiply out the right side and it should be easy to see what b must equal.