The discussion revolves around the invertibility of the function f(x) = -sin(-x) over the interval -π to π. Participants explore whether the function is monotonic and thus invertible, considering the implications of the interval's endpoints.
The discussion is ongoing, with participants sharing their thoughts on the function's monotonicity and exploring different methods for determining invertibility. Some guidance regarding graphing the function has been offered, but no consensus has been reached on the function's invertibility.
Participants note the specific interval does not include the endpoints -π and π, which raises questions about the implications for the function's behavior and invertibility. There is also mention of a lack of specific rules being applied to determine monotonicity in this context.
The interval doesn't include [itex]\pi[/itex] or [itex]-\pi[/itex]ZedCar said:Homework Statement
Determine if the following function is invertible.
If it is, find the inverse function.
f (x) = -sin(-x)
-∏ < x < ∏
The Attempt at a Solution
f(x) = -sin(-x)
y = -sin(-x)
-(sin y)^-1 = -x
(sin y)^-1 = x
when x = -∏, y = 0
when x = ∏, y = 0
ZedCar said:f^-1(x) = (sin x)^-1
0 = x = 0
Yes, I wasn't sure what to do about that. On other questions I've tried it usually states, for example, -∏ <= x <= ∏, but in this example it didn't have an equals in the interval.Mark44 said:The interval doesn't include [itex]\pi[/itex] or [itex]-\pi[/itex]
Mark44 said:Have you sketched a graph of this function? If you know the graph of y = f(x), you can get the graph of y = -f(-x) by doing a couple of reflections. Having a graph should give you a good idea of whether your function has an inverse.
I wasn't really using any specific rule for determining if the function is monotonic, as so far the functions I've been trying aren't too difficult eg (X^2 - 5) or (3x + 3).Mark44 said:What rule or theorem are you using to determine whether a function has an inverse?