Question

In Exercises \(1-8,(a)\) state whether or not the function satisfies the hypotheses of the Mean Value Theorem on the given interval, and (b) if it does, find each value of \(c\) in the interval \((a, b)\) that satisfies the equation $$f(x)=\sin ^{-1} x \quad \text { on }[-1,1]$$

Short answer

Yes, the function \( f(x) = \sin^{-1}x \) satisfies the hypotheses of the Mean Value Theorem on the interval [-1, 1]. The points in the interval that satisfy the Mean Value Theorem are \( c = \sqrt{1 - (2/\pi)^2} \) and \( c = -\sqrt{1 - (2/\pi)^2} \).

Step-by-step solution

Function Continuity

The function \( f(x) = \sin^{-1}x \) is continuous on the interval [-1,1] because the inverse sine function is defined and continuous for all values of x in the interval [-1,1].

Function Differentiability

Differentiability of the function \( f(x) = \sin^{-1}x \) on the interval (-1, 1) is achievable as the derivative of \( f(x) = \sin^{-1}x \) is \( f'(x) = \frac{1}{\sqrt{1 - x^2}} \), which is defined on (-1, 1). Therefore, function is differentiable on the interval (-1, 1).

Apply Mean Value Theorem

The Mean Value Theorem states that if a function \( f(x) \) is continuous on [a, b] and differentiable on (a, b), then there exists a number \( c \) in the interval (a, b) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). Here, \( a = -1, b = 1, f(a) = f(-1) = -\pi/2, f(b) = f(1) = \pi/2 \). Thus, \( f'(c) = \frac{f(1) - f(-1)}{1 - (-1)} = \frac{\pi/2 - (-\pi/2)}{2} = \frac{\pi}{2} \). Solve the equation \( \frac{1}{\sqrt{1 - c^2}} = \frac{\pi}{2} \) to find the values of \( c \).

Solve for \( c \)

Solving \( \frac{1}{\sqrt{1 - c^2}} = \frac{\pi}{2} \) for \( c \) results in \( c = \sqrt{1 - (2/\pi)^2} \) and \( c = -\sqrt{1 - (2/\pi)^2} \). Both these values lie within the interval (-1, 1). Hence, these are valid solutions.