Suppose f(x) = [sin(x^2 - 4)]^ -1. Identify any points of discontinuity, and determine (giving reasons) if they are removable, infinite (essential), or jump discontinuities.
Okay, I presume that the [] brackets denote the greatest integer function (int () ). Once I graphed the function on my graphing calculator, it returned a tragically ugly line of dots along y = -1. How can I interpret this and describe it for every single point?
-1 <= sin(x) <= 1 for all x.
So, [sin(x)] is either -1, 0, or 1
This is zero when x^2 - 4 = kπ, or
x = sqrt(kπ + 4) for any integer k. It stays zero till x^2 - 4 = (k + 1/2)π
I'm not sure I understand your answer completely. I now understand that since the function is in greatest integer brackets, that it can only be one of three answers (-1, 0, or 1), but from here, you lose me. Where are the discontinuities located and how can I describe the mass number of what appear to be discontinuities on the graph? Also, would these be considered jump discontinuities since they seem to jump and leaves gaps in the graph? Sorry to have to ask you to explain yorself more, I think I just need a more simplistic description to help me understand.
see my comment on your 8:58 pm repost.
It seems that there might be a misunderstanding regarding the notation used in the function. In this case, the brackets "[ ]" do not denote the greatest integer function, but rather they are used here to enclose the entire function, making it clear where the inverse function begins and ends.
Let's revisit the function f(x) = [sin(x^2 - 4)]^ -1. To identify any points of discontinuity, we need to determine any x-values where the function becomes undefined.
The function is undefined when the denominator of the fraction is equal to zero, as division by zero is not defined. Therefore, we need to find the x-values that make the expression sin(x^2 - 4) equal to zero.
To find these points, we can set the argument of the sine function, x^2 - 4, equal to multiples of π, since the sine function is zero at those values.
So, we have the equation x^2 - 4 = nπ, where n is an integer. Solving for x, we get x = ±√(4 + nπ).
Now, let's analyze the various cases:
1. If n is even, then the expression inside the square root is positive, and the square root yields real values. Thus, we have two distinct x-values when n is even, and these points will not be discontinuities.
2. If n is odd, then the expression inside the square root is negative, and the square root becomes imaginary. Imaginary values are not part of the original function's domain, so we can conclude that these points represent points of discontinuity.
To summarize, the points of discontinuity occur at x = ±√(4 + nπ), where n is an odd integer. These are removable discontinuities since the function can be made continuous at these points by redefining the function value. In this case, since the function is undefined at these points, the limit at these points is also undefined.
In the case of the graph showing a line of dots along y = -1, it may be indicative of a computational limitation or an error in the way the graphing calculator is interpreting the function. It is always important to double-check the input and confirm the correct notation is being used.