If f is a bijection, which statement correctly describes the graphs of f and f^{-1} and their compositions?

If f is a bijection, which statement correctly describes the graphs of f and f^{-1} and their compositions?

• A. The graphs are identical; compositions equal to identity for all x.
• B. The graph of f^{-1} is the reflection of the graph of f across the line y = x; f(f^{-1}(x)) = x and f^{-1}(f(x)) = x on the appropriate domains.
• C. The graph is rotated 90 degrees; compositions undefined.
• D. The graphs are reflections across y = -x; compositions equal to x.

Answer: (B)

For a bijection, the inverse essentially swaps inputs and outputs, so the graph of the inverse is a mirror of the original graph across the line y = x. If a point (a, b) lies on the graph of f (meaning b = f(a)), then the inverse must pass through (b, a) since a = f^{-1}(b). This symmetry across y = x is what ties the two graphs together. Regarding the compositions, the inverse is designed to undo the mapping, so f composed with its inverse returns the input back on the appropriate domain: f(f^{-1}(x)) equals x for every x in the domain of f^{-1} (which is the range of f). Likewise, f^{-1}(f(x)) equals x for every x in the domain of f. If the domain and codomain are the same set, these identities hold for all elements of that set, giving the identity function. The point is that these compositions are well-defined and act as the identity where they should. The other descriptions don’t fit because the graphs aren’t generally identical unless f happens to be its own inverse, and a 90-degree rotation or a reflection across y = -x does not capture the correct relationship between a function and its inverse.

For a bijection, the inverse essentially swaps inputs and outputs, so the graph of the inverse is a mirror of the original graph across the line y = x. If a point (a, b) lies on the graph of f (meaning b = f(a)), then the inverse must pass through (b, a) since a = f^{-1}(b). This symmetry across y = x is what ties the two graphs together.

Regarding the compositions, the inverse is designed to undo the mapping, so f composed with its inverse returns the input back on the appropriate domain: f(f^{-1}(x)) equals x for every x in the domain of f^{-1} (which is the range of f). Likewise, f^{-1}(f(x)) equals x for every x in the domain of f. If the domain and codomain are the same set, these identities hold for all elements of that set, giving the identity function. The point is that these compositions are well-defined and act as the identity where they should.

The other descriptions don’t fit because the graphs aren’t generally identical unless f happens to be its own inverse, and a 90-degree rotation or a reflection across y = -x does not capture the correct relationship between a function and its inverse.