Question

How do I explain and show work to this problem

Answer 1

just a quick addition to "jimrgrant1" posting above.

`\cfrac{x^2-1}{x+1}\implies \cfrac{\stackrel{ \textit{difference of squares} }{x^2-1^2}}{x+1}\implies \cfrac{(x-1)(x+1)}{x+1}\implies x-1 ~~ ~\hfill~\x \\\\\\ \textit{why can't x = -1?}\qquad \cfrac{x^2-1}{x+1}\implies \cfrac{(-1)^2-1}{-1+1}\implies \cfrac{0}{0}\implies und efined`

with that constraint, yes, both are the same, otherwise, they differ at the vertical asymptote of x = -1.

Check the picture below.

Answer 2

Yes

Explanation:

given the functions

f(x) = x - 1

and g(x) =
`(x^2-1)/(x+1)` ← ( the numerator is a difference of squares ) , that is

x² - 1 = (x - 1)(x + 1)

Then

g(x) =
`((x-1)(x+1))/(x+1)` ( cancel the common factor (x + 1) , then

g(x) =
`(x-1)/(1)` = x - 1

Thus f(x) = g(x) = x - 1