F(x)=(x+1)^-1

f(x)=1/(x+1) so

f(x)/g(x) is

1/[(x+1)(x-2)]

Since division by zero approaches infinity, it is not a real value, and is said to be undefined. Division by zero is "not allowed", which just really means that it does not have a real value. So in this case x cannot equal -1 or 2 so the domain is:

All real numbers except -1 or 2...which can be expressed as:

(-oo,-1),(-1,2),(2,+oo)

f(x)=1/(x+1) so

f(x)/g(x) is

1/[(x+1)(x-2)]

Since division by zero approaches infinity, it is not a real value, and is said to be undefined. Division by zero is "not allowed", which just really means that it does not have a real value. So in this case x cannot equal -1 or 2 so the domain is:

All real numbers except -1 or 2...which can be expressed as:

(-oo,-1),(-1,2),(2,+oo)

F(x)/g(x) is undefined where g(x) = 0. That occurs where x=2.

The domain is all real numbers except 2.

(-∞, 2) ∪ (2, ∞)

The domain is all real numbers except 2.

(-∞, 2) ∪ (2, ∞)

Step-by-step explanation:

Function g(x) = x - 2 is in the denominator. We cannot divide by zero, so the domain of (f/g)(x) is x ≠ 2.

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