Order between one to one functions and their inverses

No, if the graph of $g(x)$ is above the graph of $f(x)$, then the graph of $g^{-1}(x)$ is to the right of the graph of $f^{-1}(x)$. If these functions are increasing, then being to the right of another function is the same as being below it. However, if they are decreasing, then being to the right of another function is the same as being above it.

For a concrete counterexample, take $f(x)=-x$ and $g(x)=-x+1$. Then $f^{-1}(x)=-x$ and $g^{-1}(x)=-x+1$, so $g^{-1}(x)>f^{-1}(x)$ for all $x$.

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