Some functions that aren't invertible can be made invertible by restricting their domains. For example, the function x^2 is invertible if we restrict x to the interval [0inf), or to any subset of that interval. In that case, the inverse function is sqrt(x). (We could also restrict to the domain (-inf,0], in which case the inverse function would be -sqrt(x).) Similarly, by restricting the domain of the function f(x) = 2x^2 - 4x - 9 to an interval, we can make it invertible. What is the largest such interval that includes the point x = 0?
Find the turning point (vertex)
It occurs where \(\frac{-b}{2a}\) = \(\frac{-(-4)}{2(-9)}\)= \(\frac{4}{-18}=\frac{2}{-9}\).
Since \(x=\) \(0\) is to the left of \(x = 1\) the larger interval is \((-\infty,\frac{2}{-9})\)
(Not sure if this is the correct answer, but it's what I got.)
