Question

Math: One to One function...help!

Answer 1

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 $f$ at some point $x$ to get the number $f(x)$, evaluating the inverse at $f(x)$ will recover the original input $x$. In other words,

$f^{-1}(f(x)) = x$

The process works in the opposite direction, too:

$f\left(f^{-1}(x)\right) = x$

From the given definition of $g$, we have $g(-4) = 3$, so taking inverses on both sides, we find

$g(-4) = 3 \implies g^{-1}(g(-4)) = g^{-1}(3) \implies \boxed{g^{-1}(3) = -4}$

Given $h(x)=2x-13$, evaluating $h$ at its inverse will recover $x$, so that

$h\left(h^{-1}(x)\right) = x \implies 2h^{-1}(x) - 13 = x \implies \boxed{h^{-1}(x) = \dfrac{x+13}2}$

$(h\circ h^{-1})(x)$ is another way of writing the compound function $h\left(h^{-1}(x)\right)$. As already discussed, this reduces to $x$, so

$\boxed{\left(h\circ h^{-1}\right)(-9) = -9}$