1. ## 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?

2. I think you require

$\displaystyle \frac{x^2+4x-5}{x^2-1}= \frac{(x+5)(x-1)}{(x+1)(x-1)}= \frac{(x+5)}{(x+1)}$

3. 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.

4. how did you type the answer in that format. Which app or website did you use.

how did you type the answer in that format. Which app or website did you use.

6. Originally Posted by suchgreatheights
f(x)= ((x^2)+4x-5)/((x^2)-1)
What would be the formula for the extended function that's continuous at x=1?
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.

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