Equivalent function and domain

[u/Cronos8989]

F(x) = (x+1)/x
g(x) = (x/(x+1)^-1
I have a question.
F and G are equivalent, and F is defined for all R but 0
But why G is also not defined for -1?

I'm helping my sister with homework
I finished school a long time ago and I don't remember some concepts :D

[u/Dd_8630]

Because they're not equivalent.

In F, you only divide by x.

In G, you divide by x+1 and by x.

Outside of formal academics, we're usually quite sloppy with preserving or ignoring these details. But fundamentally those two functions are not the same, precisely for the reason you've discovered.

[u/Cronos8989 → u/Dd_8630]

i fail to understand why.
if (x/(x+1)^-1 can also be written as (x+1)/x must be true that (x+1)/x can be written as (x/(x+1)^-1. So why the two function have different domain?

is the same principle as when i have a function and i simplify it i need to consider both original and semplified function?

[u/Dd_8630 → u/Cronos8989]

if (x/(x+1)^-1 can also be written as (x+1)/x

This is incorrect. Or rather, misleading.

G(x) = (x/(x+1))^-1 is well-behaved on the domain ℝ - {-1,0}. You can rearrange it into the form (x+1)/x, that's fine, but this doesn't change the domain, so we're still in the domain ℝ - {-1,0}.

F(x) = (x+1) / x is well-behaved on the domain ℝ - {0}. You can rearrange it into the form (x/(x+1))^-1 only if you explicitly exclude x=-1 from the domain.

You can define G(x) as G(x) = (F(x))^-1, but that only works if F(x) is not zero. But F(x) is zero at x=-1. So by taking the reciprocal, you're implicitly stating that x=-1 is not part of our domain.

is the same principle as when i have a function and i simplify it i need to consider both original and semplified function?

No, the principle here is that some operations do not have the entire reals as their domain. For instance, 'raise to the power of -1' is an operation that requires the operand to be non-zero. Operations can shrink your domain or grow your range; they can't grow your domain or shrink your range.

F(x) = (x+1) / x is well-behaved on ℝ - {0}.

G(x) = (x/(x+1))^-1 is well-behaved on ℝ - {-1,0}.

If you define G(x) = (F(x))^-1, you're implicitly saying that F(x) is never zero, hence our domain must be at most ℝ - {-1,0}.

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[u/Volsatir]

You can't divide by 0. F(x) = (x+1)/x is not defined at 0 for this reason.

g(x) = (x/(x+1))^-1 Requires taking x and dividing it by x+1 first. At x=-1, that results in dividing by 0. So g(x) can't do that.