# Extended Function Formula

• Jan 10th 2010, 07:49 PM
suchgreatheights
Extended Function Formula
f(x)= ((x^2)+4x-5)/((x^2)-1)
What would be the formula for the extended function that's continuous at x=1?
• Jan 10th 2010, 08:00 PM
pickslides
I think you require

$\displaystyle \frac{x^2+4x-5}{x^2-1}= \frac{(x+5)(x-1)}{(x+1)(x-1)}= \frac{(x+5)}{(x+1)}$
• Jan 10th 2010, 08:15 PM
suchgreatheights
I don't know what an extended function is and I have no idea how to find a formula for one, so I'll just hope this is what the question is asking for. Thanks.
• Jan 10th 2010, 09:26 PM
how did you type the answer in that format. Which app or website did you use.
• Jan 10th 2010, 10:15 PM
pickslides
Quote:

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.