Discontinuity Points Of Composite Functions F(g(x))
Hey guys! Today, we're diving into a cool problem involving composite functions and their points of discontinuity. Specifically, we want to find out where the function f(g(x)) is discontinuous within the interval (-1, 1), given the functions f(x) = [2x² + 1] and g(x) defined piecewise. Let's break it down step by step to make sure we understand it perfectly!
Understanding the Functions
First, let’s make sure we're all on the same page about what these functions actually do. Understanding the characteristics of each function is very important for composite functions.
The Function f(x)
The function f(x) = [2x² + 1] involves two key operations: squaring, multiplying by two, adding one, and then applying the greatest integer function (also known as the floor function). The greatest integer function, denoted by [t], returns the largest integer less than or equal to t. For example, [3.14] = 3, [5] = 5, and [-2.7] = -3. Understanding how this function behaves is crucial for identifying discontinuities.
Consider the quadratic part 2x² + 1. This part is continuous everywhere. However, the floor function [t] introduces discontinuities at every integer value. Thus, f(x) will be discontinuous whenever 2x² + 1 is an integer. In other words, we are looking for values of x such that 2x² + 1 = n, where n is an integer. This gives us x² = (n - 1) / 2, so x = ±√((n - 1) / 2). We will be particularly interested in the points where x lies in the range of g(x) when x is in (-1, 1).
The Piecewise Function g(x)
The function g(x) is defined piecewise:
- g(x) = 2x - 3 if x < 0
- g(x) = 2x + 3 if x ≥ 0
This means that for negative values of x, we use the first expression, and for non-negative values, we use the second expression. The function g(x) is continuous everywhere except possibly at x = 0, where the definition changes. Let's check the continuity at x = 0:
- Limit from the left: lim x→0⁻ g(x) = lim x→0⁻ (2x - 3) = -3
- Limit from the right: lim x→0⁺ g(x) = lim x→0⁺ (2x + 3) = 3
Since the left-hand limit and the right-hand limit are not equal, g(x) is discontinuous at x = 0. This discontinuity will play a vital role when we analyze the composite function.
Analyzing the Composite Function f(g(x))
Now, let's dive into the composite function f(g(x)). This means we'll substitute g(x) into f(x), so we get f(g(x)) = [2(g(x))² + 1]. Since g(x) is defined piecewise, we'll have to consider two cases:
Case 1: x < 0
In this case, g(x) = 2x - 3, so f(g(x)) = [2(2x - 3)² + 1] = [2(4x² - 12x + 9) + 1] = [8x² - 24x + 18 + 1] = [8x² - 24x + 19]. We need to find the values of x in the interval (-1, 0) where 8x² - 24x + 19 is an integer. Let's denote h(x) = 8x² - 24x + 19. We are interested in the values of x for which h(x) = n, where n is an integer.
Case 2: x ≥ 0
In this case, g(x) = 2x + 3, so f(g(x)) = [2(2x + 3)² + 1] = [2(4x² + 12x + 9) + 1] = [8x² + 24x + 18 + 1] = [8x² + 24x + 19]. We need to find the values of x in the interval [0, 1) where 8x² + 24x + 19 is an integer. Let's denote k(x) = 8x² + 24x + 19. We are interested in the values of x for which k(x) = n, where n is an integer.
Identifying Discontinuities
Discontinuities can arise from two sources:
- The discontinuity of g(x) at x = 0.
- The points where f(g(x)) jumps because 2(g(x))² + 1 becomes an integer.
Discontinuity at x = 0
We already know that g(x) is discontinuous at x = 0. Let's analyze the behavior of f(g(x)) around x = 0:
- As x approaches 0 from the left, g(x) approaches -3, so f(g(x)) approaches f(-3) = [2(-3)² + 1] = [2(9) + 1] = [19] = 19.
- As x approaches 0 from the right, g(x) approaches 3, so f(g(x)) approaches f(3) = [2(3)² + 1] = [2(9) + 1] = [19] = 19.
Since the left and right limits of f(g(x)) at x = 0 are both equal to 19, f(g(x)) is continuous at x = 0. So, the discontinuity of g(x) at x = 0 does not cause a discontinuity in f(g(x)). This is a key observation!
Discontinuities from Integer Values
Now, let's find the values of x where f(g(x)) jumps. This happens when 2(g(x))² + 1 is an integer. In other words, when 8x² - 24x + 19 or 8x² + 24x + 19 is an integer.
Case 1: x < 0
We need to find x in (-1, 0) such that 8x² - 24x + 19 = n, where n is an integer. Let's analyze the range of h(x) = 8x² - 24x + 19 for x ∈ (-1, 0).
- h(-1) = 8(-1)² - 24(-1) + 19 = 8 + 24 + 19 = 51
- h(0) = 8(0)² - 24(0) + 19 = 19
So, h(x) ranges from 19 to 51 in the interval (-1, 0). The integers in this range are 19, 20, 21, ..., 51. We need to find how many of these integers can be expressed as 8x² - 24x + 19 for some x in (-1, 0). The vertex of the parabola h(x) = 8x² - 24x + 19 occurs at x = -(-24) / (2 * 8) = 24 / 16 = 3 / 2 = 1.5. Since the vertex is outside the interval (-1, 0), h(x) is monotonic (decreasing) on (-1, 0). This means that for each integer value between 19 and 51, there is exactly one x in (-1, 0) such that h(x) = n. Therefore, there are 51 - 19 + 1 = 33 discontinuities in the interval (-1, 0).
Case 2: x ≥ 0
We need to find x in [0, 1) such that 8x² + 24x + 19 = n, where n is an integer. Let's analyze the range of k(x) = 8x² + 24x + 19 for x ∈ [0, 1).
- k(0) = 8(0)² + 24(0) + 19 = 19
- k(1) = 8(1)² + 24(1) + 19 = 8 + 24 + 19 = 51
So, k(x) ranges from 19 to 51 in the interval [0, 1). The integers in this range are 19, 20, 21, ..., 51. We need to find how many of these integers can be expressed as 8x² + 24x + 19 for some x in [0, 1). The vertex of the parabola k(x) = 8x² + 24x + 19 occurs at x = -24 / (2 * 8) = -24 / 16 = -3 / 2 = -1.5. Since the vertex is outside the interval [0, 1), k(x) is monotonic (increasing) on [0, 1). This means that for each integer value between 19 and 51, there is exactly one x in [0, 1) such that k(x) = n. Therefore, there are 51 - 19 + 1 = 33 discontinuities in the interval [0, 1).
Total Number of Discontinuities
We found 33 discontinuities in the interval (-1, 0) and 33 discontinuities in the interval [0, 1). Therefore, the total number of points where f(g(x)) is discontinuous in the open interval (-1, 1) is 33 + 33 = 66. Therefore, m = 66.
So, the final answer is 66. Hope this helps, and happy problem-solving!