A one-to-one function has an inverse. The inverse is another function that undoes the action of the first one, so if we evaluate a function [tex]f[/tex] at some point [tex]x[/tex] to get the number [tex]f(x)[/tex], evaluating the inverse at [tex]f(x)[/tex] will recover the original input [tex]x[/tex]. In other words,
[tex]f^{-1}(f(x)) = x[/tex]
The process works in the opposite direction, too:
[tex]f\left(f^{-1}(x)\right) = x[/tex]
From the given definition of [tex]g[/tex], we have [tex]g(-4) = 3[/tex], so taking inverses on both sides, we find
[tex]g(-4) = 3 \implies g^{-1}(g(-4)) = g^{-1}(3) \implies \boxed{g^{-1}(3) = -4}[/tex]
Given [tex]h(x)=2x-13[/tex], evaluating [tex]h[/tex] at its inverse will recover [tex]x[/tex], so that
[tex]h\left(h^{-1}(x)\right) = x \implies 2h^{-1}(x) - 13 = x \implies \boxed{h^{-1}(x) = \dfrac{x+13}2}[/tex]
[tex](h\circ h^{-1})(x)[/tex] is another way of writing the compound function [tex]h\left(h^{-1}(x)\right)[/tex]. As already discussed, this reduces to [tex]x[/tex], so
[tex]\boxed{\left(h\circ h^{-1}\right)(-9) = -9}[/tex]
