Respuesta :
Consider f, such that f(x)=y, where x is in the Domain of f, and y is in the Range of f.
Let f(a)=b. So, (a, b) is in the graph of f.
The inverse function of f, [tex]f^{-1}[/tex] (if exists) is such that if f(x)=y, then [tex]f^{-1}(y)=x[/tex].
So if (a, b) is a particular point in the graph of f, then (b, a) is a point in the graph of [tex]f^{-1}[/tex].
A point where they intersect is a point where (a, b)=(b, a), that is a=b. Among our choices, only (2, 2) satisfies this condition.
Answer: (2, 2)
Let f(a)=b. So, (a, b) is in the graph of f.
The inverse function of f, [tex]f^{-1}[/tex] (if exists) is such that if f(x)=y, then [tex]f^{-1}(y)=x[/tex].
So if (a, b) is a particular point in the graph of f, then (b, a) is a point in the graph of [tex]f^{-1}[/tex].
A point where they intersect is a point where (a, b)=(b, a), that is a=b. Among our choices, only (2, 2) satisfies this condition.
Answer: (2, 2)
Answer:
C. (2, 2)
Step-by-step explanation:
this is the correct answer on ed-genuity, hope this helps! :)