# Thread: Inverse function, in terms with another inverse function

1. ## Inverse function, in terms with another inverse function

$\displaystyle f$ 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?

2. Originally Posted by 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?
It helps to apply the procedure correctly.

Let $\displaystyle y = \frac{1}{1 + f(x)}$. Swap y and x:

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

Your job is to make y the subject:

$\displaystyle \frac{1}{x} = 1 + f(y) \Rightarrow \frac{1}{x} - 1 = f(y)$

and it should be clear what the last step is.

3. Right at the second step, I saw what a gross error I made. Don't I feel silly.

Thanks for the speedy response, as well as the formatting.

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