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?
Actually, in spite of what I wrote in an earlier post, it's possible that the brackets are used just to avoid having nested parentheses. In that case, f(x) has discontinuities wherever sin(x^2-4) is zero.
That is, where x^2 - 4 is a multiple of pi.
If x^2 - 4 = k*pi,
Then f is discontinuous at x = sqrt(k*pi + 4) for all integer k
Ah, the tragically ugly line of dots, a classic in the world of mathematics. In this case, it seems like there might be some confusion with the notation. The brackets [ ] are usually used to indicate the greatest integer function, but in this context, they might just be ordinary brackets.
Let's assume that the brackets are indeed ordinary brackets and not the greatest integer function. In that case, the function f(x) = [sin(x^2 - 4)]^-1 simplifies to f(x) = 1/sin(x^2 - 4).
Now, let's talk about the points of discontinuity. For the sine function, we know that it is not defined for certain values of its argument, namely when the argument is an odd multiple of π/2. In this case, the argument is x^2 - 4.
To find the points of discontinuity, we need to solve the equation x^2 - 4 = nπ/2, where n is an odd integer. This gives us x = √(4 + nπ/2) and x = -√(4 + nπ/2).
Since the argument of the sine function is squared, we will have two points of discontinuity for each value of n. So, we have an infinite number of points of discontinuity (quite the party, I must say).
But wait, there's more! Is this a removable, essential, or jump discontinuity? Well, if the function can be defined or redefined to make it continuous at a particular point, then it's a removable discontinuity. If the function approaches different values from the left and right sides of the point, then it's a jump discontinuity. And if the function behaves chaotically or oscillates around the point, then it's an essential discontinuity.
In our case, since the function is 1/sin(x^2 - 4) and the sine function oscillates between -1 and 1, it means that the function will approach different values from the left and right sides of each discontinuity point. So, we can say that all the points of discontinuity are jump discontinuities. They're jumping around like a bunch of excited kangaroos!
Well, that's one interpretation of the tragically ugly line of dots. I hope it brings a smile to your face, or at least a chuckle.
It seems there is a misunderstanding regarding the notation used in the function. The brackets, [ ], typically signify the greatest integer function, also known as the floor function, which takes a real number and returns the greatest integer less than or equal to that number. However, in this context, the brackets do not denote the greatest integer function.
Let's clarify the notation and examine the function f(x) = [sin(x^2 - 4)]^ -1 without the brackets. The function is actually the reciprocal of the sine function of the quantity (x^2 - 4).
To identify the points of discontinuity, we need to consider the points where the function is not defined or where the function changes abruptly.
The sine function is defined for all real numbers, so the only points of potential discontinuity will occur when the denominator, sin(x^2 - 4), is equal to zero.
To find the points where sin(x^2 - 4) = 0, we can solve the equation x^2 - 4 = nπ, where n is an integer.
x^2 - 4 = nπ
x^2 = nπ + 4
x = ±√(nπ + 4)
So, the points of potential discontinuity are x = ±√(nπ + 4), where n is an integer.
To determine if these points are removable, infinite (essential), or jump discontinuities, we need more information. Specifically, we need to examine the behavior of the function around these points. This can be done by analyzing the limits of the function as x approaches these points.
If the limit exists and is finite as x approaches a point of potential discontinuity, it is a removable discontinuity.
If the limit does not exist or is infinite as x approaches a point of potential discontinuity, it is an infinite (essential) discontinuity.
If the limit exists, but is different from the function value at that point, it is a jump discontinuity.
Without further information or the complete equation, we cannot determine the exact nature of the discontinuities at x = ±√(nπ + 4). Additional analysis would be required to determine the behavior of the function in these regions.
It seems there might be some confusion in your question. The brackets [ ] typically denote the greatest integer function, but in the function you provided, the brackets are not needed to determine the points of discontinuity or analyze the function.
Let's analyze the function f(x) = [sin(x^2 - 4)]^ -1, assuming that the brackets are not related to the greatest integer function.
To find the points of discontinuity, we need to look for values of x for which the function is undefined. In this case, the function f(x) = [sin(x^2 - 4)]^ -1 is undefined when the denominator, [sin(x^2 - 4)], is equal to zero.
To determine when the denominator is equal to zero, we can solve the equation sin(x^2 - 4) = 0.
First, let's consider the equation x^2 - 4 = 0.
The solutions to this equation are x = -2 and x = 2.
Now, let's find the values of x for which sin(x^2 - 4) = 0 by considering the sine function.
The sine of an angle is equal to zero at multiples of pi, i.e., sin(n * pi) = 0, where n is any integer.
Therefore, we need to solve the equation x^2 - 4 = n * pi for x.
For n = 0, x^2 - 4 = 0, which gives us x = -2 and x = 2 (the same solutions as before).
For n = 1, x^2 - 4 = pi, which has no real solutions.
For n = -1, x^2 - 4 = -pi, which also has no real solutions.
So, the points of discontinuity for f(x) occur at x = -2 and x = 2.
To determine if these points of discontinuity are removable or essential, we need to investigate the behavior of the function as x approaches these points.
If the limit of f(x) as x approaches a point of discontinuity exists and is finite, the point of discontinuity is removable. If the limit is infinite or does not exist, the point of discontinuity is essential.
In this case, when we take the limit of f(x) as x approaches -2 or 2, we find that the limit does not exist. Therefore, the discontinuities at x = -2 and x = 2 are essential (also known as infinite discontinuities).
In summary, the function f(x) = [sin(x^2 - 4)]^ -1 has essential (infinite) discontinuities at x = -2 and x = 2.