Determine the domain and range of the function f(x) = x / x^ - 1

Would appreciate step by step explanation

It looks like the function is supposed to be

f(x) = x / (x ⁻¹)

which can be rewritten as

f(x) = x / (1 / x)

Without any further simplification, notice that x = 0 makes the denominator undefined, so x = 0 is not in the domain. For all other real numbers, we have

x / (1 / x) = x ²

which exists for all real numbers. So the domain (in interval notation) is (-∞, 0) ∪ (0, ∞). If you want this in set notation, it's the set of all real numbers except 0, which can be written as ℝ\ {0} or ℝ- {0}. Or in set-builder notation, {x ∈ ℝ | x ≠ 0}.

For all x ≠ 0, since the function reduces to x ², it will take on all positive real numbers. Since x ² ≥ 0 under normal circumstances, check to see if f(x) also attains this value:

x / (1 / x) = 0

In order for this to be 0, the numerator must be equal to 0. But if x = 0, then the denominator is undefined. So 0 is not in the range, and the range is (0, ∞), or ℝ⁺ \ {0} (where ℝ⁺ denotes the set of positive real numbers), or {x ∈ ℝ | x > 0}.

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol