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**BlackBlaze** $\displaystyle f$ is a one-to-one function with inverse $\displaystyle f^{-1}$.

$\displaystyle p(x) = \frac{1}{1+f(x)}$

Caculate the inverse of the function in terms of $\displaystyle f^{-1}$

So I tried swapping $\displaystyle p(x)$ and $\displaystyle f(x)$.

$\displaystyle f^{-1}(x) = \frac{1}{1+p^{-1}(x)}$

$\displaystyle \frac{1}{f^{-1}(x)} = 1 + p^{-1}(x)$

$\displaystyle \frac{1}{f^{-1}(x)} - 1 = p^{-1}(x)$

The answer in the back of the book however, says this:

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

How do I get to that answer?