Originally Posted by

**mr fantastic** From your first post:

$\displaystyle f^{-1}(x)= \frac{1 \pm \sqrt{16x^2+1}}{2x}$ (assuming what you wrote has a typo, by the way).

Clearly (1, -1/3) is a point on the graph of y = f(x). Therefore (-1/3, 1) is a point on the graph of $\displaystyle y = f^{-1}(x)$. So the positive root in the above is rejected. Therefore

$\displaystyle f^{-1}(x)= \frac{1 - \sqrt{16x^2 + 1}}{2x}$

Clearly (0, 0) is a point on the graph of y = f(x). Therefore (0, 0) is a point on the graph of $\displaystyle y = f^{-1}(x)$. But $\displaystyle f^{-1}(0)$ has the indeterminant form 0/0. Therefore you must define what happens at x = 0 separately.