Asked by htht
The function f(x) = ln(x) has a domain of all real numbers greater than zero and a range of all real numbers. The inverse of this function is f–1(x) = ex. Which conclusion can be drawn by comparing the two functions?
The domain of f–1(x) is all real numbers and the range is all real numbers.
The domain of f–1(x) is all real numbers greater than 0 and the range is all real numbers.
The domain of f–1(x) is all real numbers and the range is all real numbers greater than 0.
The domain of f–1(x) is all real numbers greater than 0 and the range is all real numbers greater than 0.
All Answers 1
Answered by
GPT-5 mini
AI
The correct conclusion is:
The domain of f–1(x) is all real numbers and the range is all real numbers greater than 0.
Reason: f(x)=ln x has domain (0, ∞) and range (−∞, ∞). The inverse swaps domain and range, so f⁻¹(x)=e^x has domain (−∞, ∞) and range (0, ∞).
The domain of f–1(x) is all real numbers and the range is all real numbers greater than 0.
Reason: f(x)=ln x has domain (0, ∞) and range (−∞, ∞). The inverse swaps domain and range, so f⁻¹(x)=e^x has domain (−∞, ∞) and range (0, ∞).
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.