The inverse of a function is a function if and only if the original function is one-to-one, meaning that each input corresponds to exactly one output.
In this case, the original function 𝑓(𝑥) = √𝑥 + 3 is not one-to-one, since multiple inputs can result in the same output. For example, both √4 + 3 and (-√4) + 3 result in an output of 5.
To ensure that the inverse relation is a function, we can restrict the domain of the original function. One way to do this is to restrict the original function to only non-negative values for 𝑥, since the square root only gives non-negative outputs.
Therefore, by restricting the domain of 𝑓(𝑥) = √𝑥 + 3 to 𝑥 ≥ 0, the inverse relation would be a function.
Explain why the inverse of 𝑓(𝑥) = √𝑥 + 3 is or is not a function. If not, how would you restrict the domain of the original function to guarantee that the inverse relation is a function?
3 answers
wrong. √4 = 2, not ±2
(-√4) + 3 = -2+3 = 1, not 5!
So f(x) is one-to-one, with a domain of all real x ≥0
(-√4) + 3 = -2+3 = 1, not 5!
So f(x) is one-to-one, with a domain of all real x ≥0
You are correct, thank you for pointing out the mistake in my previous response.
Since √4 = 2 and not ±2, and (-√4) + 3 = -2 + 3 = 1 and not 5, the function 𝑓(𝑥) = √𝑥 + 3 is one-to-one when restricted to non-negative values of 𝑥.
Therefore, the inverse of 𝑓(𝑥) = √𝑥 + 3 is a function without any further domain restrictions, as the original function is already one-to-one with a domain of all real numbers greater than or equal to 0. Thank you for the clarification.
Since √4 = 2 and not ±2, and (-√4) + 3 = -2 + 3 = 1 and not 5, the function 𝑓(𝑥) = √𝑥 + 3 is one-to-one when restricted to non-negative values of 𝑥.
Therefore, the inverse of 𝑓(𝑥) = √𝑥 + 3 is a function without any further domain restrictions, as the original function is already one-to-one with a domain of all real numbers greater than or equal to 0. Thank you for the clarification.