Show that f(x) = x^2 - 2x (x is (negative infinity, 1) is 1:1 and find its inverse. How should I approach this question? what should I do?
$\displaystyle f(x) = x^2 - 2x$
$\displaystyle = x^2 - 2x + (-1)^2 - (-1)^2$
$\displaystyle = (x - 1)^2 - 1$.
So the turning point is at $\displaystyle (1, -1)$.
Since the function is cut off at $\displaystyle x = 1$, that means you don't have any repeating $\displaystyle y$ values. Therefore $\displaystyle f(x)$ is one-to-one.
Now, to find the inverse, swap the $\displaystyle x$ and $\displaystyle y$...
$\displaystyle x = y^2 - 2y$
$\displaystyle x = y^2 - 2y + (-1)^2 - (-1)^2$
$\displaystyle x = (y - 1)^2 - 1$
$\displaystyle x + 1 = (y - 1)^2$
$\displaystyle -\sqrt{x + 1} = y - 1$ (since we are only accepting the lower half)
$\displaystyle y = 1 - \sqrt{x + 1}$.
So $\displaystyle f^{-1}(x) = 1 - \sqrt{x + 1}$.