Suppose f(x) = [sin(x^2 - 4)]^ -1. Identify any points of discontinuity, and determine (giving reasons) if they are removable, infinite, 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? And what type of discontinuity would this be categorized as?
It seems that you misunderstood the notation used in the function. The brackets [ ] typically indicate the greatest integer function (floor function), which rounds down to the nearest integer. However, in this case, they are not being used in that way. The brackets are actually unnecessary and do not affect the function.
To properly analyze the given function, let's ignore the brackets and focus on the original expression: f(x) = sin(x^2 - 4)^-1.
To determine the points of discontinuity, we need to consider where the function is undefined or does not have a finite value. In this case, since we are dealing with a trigonometric function, we know that sin(x^2 - 4) is undefined whenever the argument x^2 - 4 is equal to (2n + 1) * pi/2, where n is any integer.
Therefore, we need to solve the equation x^2 - 4 = (2n + 1) * pi/2 for x. This equation gives us multiple solutions for each n, which means we will have multiple points of discontinuity.
Now, let's solve the equation:
x^2 - 4 = (2n + 1) * pi/2
Adding 4 to both sides:
x^2 = (2n + 1) * pi/2 + 4
Taking the square root:
x = ± sqrt((2n + 1) * pi/2 + 4)
For each integer value of n, this equation gives us two values of x that correspond to the points of discontinuity.
To determine the nature of these discontinuities, we need to carefully analyze the behavior of the function around these points. Since we are only looking at the points where sin(x^2 - 4) is undefined, we can ignore the inverse function in this analysis.
By plotting the graph of this function, we can observe that the points of discontinuity correspond to vertical asymptotes. The function approaches infinity or negative infinity as it gets closer to the points of discontinuity.
Therefore, the points of discontinuity for the given function f(x) = sin(x^2 - 4)^-1 can be categorized as infinite discontinuities.