Mean Value Theorem Application On Arcsin(x) Over [0, 1/2]
In this article, we delve into the application of the Mean Value Theorem (MVT) to the function f(x) = arcsin(x), also denoted as sin⁻¹(x), over the interval [0, 1/2]. The Mean Value Theorem is a cornerstone of calculus, providing a powerful connection between the average rate of change of a function and its instantaneous rate of change. Understanding its application is crucial for grasping many advanced concepts in mathematical analysis. This article will not only determine if the Mean Value Theorem can be applied to the given function and interval but also find the point(s) guaranteed by the theorem, providing a comprehensive understanding of this concept. Let's embark on this mathematical journey together!
H2: Understanding the Mean Value Theorem
Before we apply the Mean Value Theorem to our specific function, let's first establish a firm understanding of the theorem itself. The Mean Value Theorem is a fundamental result in calculus that relates the average rate of change of a function over an interval to its instantaneous rate of change at some point within that interval. In simpler terms, it states that if a function is continuous on a closed interval and differentiable on the open interval, there exists at least one point within the interval where the tangent to the function's graph is parallel to the secant line connecting the endpoints of the interval.
The formal statement of the Mean Value Theorem is as follows:
If a function f(x) satisfies the following two conditions:
- f(x) is continuous on the closed interval [a, b].
- f(x) is differentiable on the open interval (a, b).
Then, there exists at least one point c in the open interval (a, b) such that:
f'(c) = (f(b) - f(a)) / (b - a)
Where:
- f'(c) represents the derivative of the function f(x) evaluated at the point c, which is the instantaneous rate of change at c.
- (f(b) - f(a)) / (b - a) represents the average rate of change of the function over the interval [a, b], also known as the slope of the secant line connecting the points (a, f(a)) and (b, f(b)).
Geometrically, the Mean Value Theorem guarantees the existence of a point c where the tangent line to the graph of f(x) is parallel to the secant line connecting the endpoints of the interval. This visual representation provides a strong intuitive understanding of the theorem.
To apply the Mean Value Theorem effectively, it's crucial to verify that the function meets the two key conditions: continuity and differentiability. Continuity ensures that there are no breaks or jumps in the function's graph within the interval, while differentiability ensures the existence of a well-defined tangent line at every point within the open interval. These conditions are essential for the theorem's conclusion to hold true. If either of these conditions is not met, the Mean Value Theorem cannot be applied, and we cannot guarantee the existence of a point c satisfying the theorem's equation.
H2: Verifying the Conditions for f(x) = arcsin(x) on [0, 1/2]
Now, let's turn our attention to the function f(x) = arcsin(x) on the interval [0, 1/2]. To determine if the Mean Value Theorem applies, we need to verify that f(x) satisfies the two aforementioned conditions: continuity on the closed interval [0, 1/2] and differentiability on the open interval (0, 1/2).
H3: Continuity of arcsin(x)
The function f(x) = arcsin(x) is the inverse sine function, also known as the arcsine function. It represents the angle whose sine is x. The domain of arcsin(x) is [-1, 1], and its range is [-π/2, π/2]. A crucial property of inverse trigonometric functions, including arcsin(x), is that they are continuous over their entire domain. This means that there are no breaks, jumps, or holes in the graph of arcsin(x) within its domain.
Since our interval of interest, [0, 1/2], is entirely contained within the domain of arcsin(x), we can confidently conclude that f(x) = arcsin(x) is continuous on the closed interval [0, 1/2]. This satisfies the first condition of the Mean Value Theorem. The continuity of arcsin(x) is a fundamental property that stems from the continuity of the sine function itself and the nature of inverse functions. The graph of arcsin(x) smoothly connects all points within its domain, without any abrupt changes or discontinuities.
H3: Differentiability of arcsin(x)
To check the differentiability of f(x) = arcsin(x), we need to examine its derivative. The derivative of arcsin(x) is given by:
f'(x) = 1 / √(1 - x²)
The derivative f'(x) exists for all x in the open interval (-1, 1). This is because the expression under the square root, (1 - x²), must be positive to avoid taking the square root of a negative number, and the denominator cannot be zero. Therefore, the derivative is defined as long as -1 < x < 1.
Our interval of interest, (0, 1/2), is entirely contained within the interval (-1, 1) where the derivative exists. Therefore, we can conclude that f(x) = arcsin(x) is differentiable on the open interval (0, 1/2). This satisfies the second condition of the Mean Value Theorem. The differentiability of arcsin(x) is a consequence of its smooth, continuous nature and the properties of its derivative. The existence of a well-defined derivative at every point within the open interval ensures that the function has a tangent line at each of those points.
H2: Applying the Mean Value Theorem
Having established that f(x) = arcsin(x) is both continuous on [0, 1/2] and differentiable on (0, 1/2), we can now confidently apply the Mean Value Theorem. The theorem guarantees the existence of at least one point c in the interval (0, 1/2) such that:
f'(c) = (f(1/2) - f(0)) / (1/2 - 0)
Let's break down this equation and solve for c.
First, we need to calculate f(1/2) and f(0):
- f(1/2) = arcsin(1/2) = π/6 (since sin(π/6) = 1/2)
- f(0) = arcsin(0) = 0 (since sin(0) = 0)
Next, we need to find the derivative of f(x), which we already determined to be:
- f'(x) = 1 / √(1 - x²)
Now we can substitute these values into the Mean Value Theorem equation:
1 / √(1 - c²) = (π/6 - 0) / (1/2 - 0)
Simplifying the right side of the equation:
1 / √(1 - c²) = (π/6) / (1/2) = π/3
Now, we need to solve for c. Let's take the reciprocal of both sides:
√(1 - c²) = 3/π
Square both sides:
1 - c² = 9/π²
Isolate c²:
c² = 1 - 9/π²
Take the square root of both sides:
c = ±√(1 - 9/π²)
Since we are looking for a value of c in the interval (0, 1/2), we only consider the positive square root:
c = √(1 - 9/π²)
Approximating this value, we get:
c ≈ √(1 - 9/9.8696) ≈ √(1 - 0.912) ≈ √0.088 ≈ 0.2966
Therefore, the point c guaranteed by the Mean Value Theorem in the interval (0, 1/2) is approximately 0.2966.
H2: Conclusion
In conclusion, we have successfully demonstrated that the Mean Value Theorem applies to the function f(x) = arcsin(x) on the interval [0, 1/2]. We verified that arcsin(x) is continuous on the closed interval and differentiable on the open interval. By applying the theorem, we found that there exists a point c ≈ 0.2966 within the interval (0, 1/2) where the instantaneous rate of change of arcsin(x) is equal to its average rate of change over the interval. This exercise highlights the power and applicability of the Mean Value Theorem in connecting the local and global behavior of functions. Understanding and applying such theorems are fundamental skills in calculus and mathematical analysis, paving the way for more advanced concepts and problem-solving techniques.
