Asked by bob
is f(x))=-x^{3}-1 one-to-one, onto, or neither
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GPT-5 mini
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Assuming f: R → R with f(x) = −x^3 − 1:
- One-to-one (injective): If f(a) = f(b) then −a^3 − 1 = −b^3 − 1, so a^3 = b^3 and hence a = b. Equivalently, f is strictly decreasing, so it is injective.
- Onto (surjective): For any y ∈ R solve y = −x^3 − 1 ⇒ x^3 = −(y + 1) ⇒ x = ³√(−(y + 1)), which is a real number for every y. Thus every real y has a preimage.
Therefore f is both one-to-one and onto (a bijection).
- One-to-one (injective): If f(a) = f(b) then −a^3 − 1 = −b^3 − 1, so a^3 = b^3 and hence a = b. Equivalently, f is strictly decreasing, so it is injective.
- Onto (surjective): For any y ∈ R solve y = −x^3 − 1 ⇒ x^3 = −(y + 1) ⇒ x = ³√(−(y + 1)), which is a real number for every y. Thus every real y has a preimage.
Therefore f is both one-to-one and onto (a bijection).
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