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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- Jan 10th 2010, 07:49 PMsuchgreatheightsExtended 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 PMpickslides
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 PMsuchgreatheights
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 PMbhaskar2
how did you type the answer in that format. Which app or website did you use.

- Jan 10th 2010, 10:15 PMpickslides
Read the tutroial here: http://www.mathhelpforum.com/math-help/latex-help/

- Jan 11th 2010, 04:26 AMHallsofIvy
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.