Is the Function f(x) = -sin(-x) Invertible on the Interval -π to π?

AI Thread Summary
The function f(x) = -sin(-x) is examined for invertibility on the interval -π to π. The discussion reveals that the function is not monotonic, as it does not consistently increase or decrease across the interval, which is a requirement for invertibility. Graphing the function indicates that it fails to meet this criterion, confirming it is not invertible. Participants also reflect on the importance of understanding monotonicity and the methods used to determine the inverse of functions. Ultimately, the conclusion is that the function does not have an inverse due to its non-monotonic nature.
ZedCar
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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

f^-1(x) = (sin x)^-1
0 = x = 0
 
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Actually, now I'm wondering, is the answer simply that the function is not monotonic and therefore it is not invertible?
 
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
The interval doesn't include \pi or -\pi
ZedCar said:
f^-1(x) = (sin x)^-1
0 = x = 0

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.

What rule or theorem are you using to determine whether a function has an inverse?
 
Mark44 said:
The interval doesn't include \pi or -\pi
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:
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've just input it into http://rechneronline.de/function-graphs/ and it appears that it is not monotonic, and therefore not invertible.

Mark44 said:
What rule or theorem are you using to determine whether a 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).

Below is the method I've been using to determine the inverse functions.

For example,

f(x) = x^2 - 5 0 <= x < ∞
y = x^2 - 5
(y + 5)^0.5 = x
when x = 0, y = -5

therefore f^-1(x) = (x + 5)^0.5
when x = 0, y = -5Is there a specific rule for determining if a function is monotonic?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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