- Verify that \(f(x) = \sin^{-1} x\) is continuous on \([-\frac{1}{2}, 0]\) and differentiable on \((-\frac{1}{2}, 0)\), confirming the Mean Value Theorem applies.
- Calculate the average rate of change: \(M = \frac{f(0) - f(-1/2)}{0 - (-1/2)} = \frac{\pi}{3}\).
- Solve the equation \(f'(c) = \frac{1}{\sqrt{1-c^2}} = \frac{\pi}{3}\) for \(c\).
- Find the valid point \(c \in (-\frac{1}{2}, 0)\), which is \(c = -\frac{\sqrt{\pi^2 - 9}}{\pi} \approx \boxed{-0.2968}\).
Explanation
- Verification of Mean Value Theorem Applicability
The function provided is \(f(x) = \sin^{-1} x\), and the interval is \(\left[-\frac{1}{2}, 0\right]\). The Mean Value Theorem (MVT) applies to a function \(f\) on an interval \([a, b]\) if the function is continuous on \([a, b]\) and differentiable on \((a, b)\). The arcsine function \(\sin^{-1} x\) is continuous on its domain \([-1, 1]\) and differentiable on its open interval \((-1, 1)\). Since the interval \(\left[-\frac{1}{2}, 0\right]\) is entirely contained within the domain \([-1, 1]\) and the open interval \(\left(-\frac{1}{2}, 0\right)\) is contained within \((-1, 1)\), the Mean Value Theorem applies to \(f(x) = \sin^{-1} x\) on \(\left[-\frac{1}{2}, 0\right]\).
- Calculating the Average Rate of Change
To find the value \(c \in \left(-\frac{1}{2}, 0\right)\) guaranteed by the MVT, we first calculate the average rate of change of the function over the interval. The formula for the average rate of change is \(M = \frac{f(b) - f(a)}{b - a}\), where \(a = -\frac{1}{2}\) and \(b = 0\).
Computing the values:
\(f(a) = \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}\)
\(f(b) = \sin^{-1}(0) = 0\)
Thus, the slope is \(M = \frac{0 - (-\frac{\pi}{6})}{0 - (-\frac{1}{2})} = \frac{\pi/6}{1/2} = \frac{\pi}{3}\).
- Finding the Value of c
According to the MVT, there exists at least one \(c \in \left(-\frac{1}{2}, 0\right)\) such that \(f'(c) = M\). The derivative of \(f(x) = \sin^{-1} x\) is \(f'(x) = \frac{1}{\sqrt{1 - x^2}}\).
Setting the derivative equal to the slope \(M\):
\(\frac{1}{\sqrt{1 - c^2}} = \frac{\pi}{3}\)
Squaring both sides and solving for \(c^2\):
\(\frac{1}{1 - c^2} = \frac{\pi^2}{9}\)
\(1 - c^2 = \frac{9}{\pi^2}\)
\(c^2 = 1 - \frac{9}{\pi^2} = \frac{\pi^2 - 9}{\pi^2}\)
\(c = \pm \frac{\sqrt{\pi^2 - 9}}{\pi}\).
Since we require \(c \in \left(-\frac{1}{2}, 0\right)\), we choose the negative solution:
\(c = -\frac{\sqrt{\pi^2 - 9}}{\pi} \approx -0.2968\).
This value lies within the interval \(\left(-\frac{1}{2}, 0\right)\).
- Final Answer
The Mean Value Theorem applies to the function \(f(x) = \sin^{-1} x\) on \(\left[-\frac{1}{2}, 0\right]\). The point \(c\) that satisfies the condition is \(c = -\frac{\sqrt{\pi^2 - 9}}{\pi} \approx -0.2968\). The final answer is \(\boxed{-\frac{\sqrt{\pi^2 - 9}}{\pi}}\).
Examples
The Mean Value Theorem is essential in motion analysis, such as when driving a car. If you travel 60 miles in one hour, the MVT guarantees that at some instant during that trip, your instantaneous speed was exactly 60 mph. Similarly, in calculus applications, verifying these conditions helps engineers and physicists predict the behavior of complex systems over time, ensuring that the model's average performance is achieved at specific points. This logic helps in tuning control systems or understanding instantaneous rates of change in physical phenomena.