Let f(x)=|x|with a domain of (-∞, 0]

Question

a. What is the range of f?

b. Is this function one-to-one?

c. What is the domain and range of f^-1(x)?

d. What is the rule for f^1(x)?

e.d. is the domain and range of is the domain and range of

Answer 1

a. With a domain of (-∞,0], f(x)=|x| is the same as f(x)=-x. So the range would be [0,∞).

b. The question of one-to-one is asking if no two distinct x-values exist such that each returns the same output under f. Or graphically, a function is one-to-one if it passes the horizontal-line test, as f does. So yes, f is one-to-one.

c. To find the domain and range of f^-1, we need only look at the range and domain of f. So the domain of f^-1 is [0,∞) and the range is (-∞,0].

d. The rule of f^-1 (I assume you intended a negative there), if I understand it correctly (this is not a terminology I have ever heard used) is found by reversing the rule of f(x): x=|f^-1(x)|. Now to reverse the absolute value, we add a ± sign to x: ±x=f^-1(x). Now this is not a function (it fails the vertical-line test), so we refer back to our domain and range, and we see that there are no negative values in the domain or positive values in the range. Thus f^-1(x)=-x as a rule fits the definition.