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 $$f(x)=\left\\{\begin{array}{ll}{\cos x,} & {0 \leq x<\pi / 2} \\ {\sin x,} & {\pi / 2 \leq x \leq \pi}\end{array}\right. \quad \text { on }[0, \pi]$$

Short answer

The given function does not satisfy the Mean Value Theorem on the interval [0, π] due to the lack of differentiability at x = π/2. Although the Mean Value Theorem can be applied to the ranges [0, π/2) and (π/2, π], no 'c' values that satisfy the Mean Value Theorem can be found in these intervals.

Step-by-step solution

Identify if the function satisfies the Mean Value Theorem

To apply the Mean Value Theorem, the function must be both continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Here, the function f(x) = cos(x) for 0 ≤ x < π/2 and f(x) = sin(x) for π/2 ≤ x ≤ π is continuous on the interval [0, π]. The function is differentiable for all x, except at x = π/2 where there is a sharp corner. Therefore, the function does not satisfy the Mean Value theorem over the interval [0, π].

Find the value of 'c' where the Mean Value Theorem is satisfied.

Since the function does not satisfy the Mean Value Theorem over the interval [0, π], we cannot find a 'c' value. However, over the intervals [0, π/2) and (π/2, π], the function will satisfy the Mean Value Theorem.

Apply Mean Value Theorem to intervals [0, π/2) and (π/2, π].

For [0, π/2), the function is f(x) = cos(x). The Mean Value Theorem says that f'(c) = (f(b) - f(a))/(b - a). Substituting the values gives: -sin(c) = (cos(π/2) - cos(0)) / (π/2 - 0) = 0, which implies sin(c) = 0. This equation is satisfied for c = 0 on the interval [0, π/2). Similarly, for (π/2, π], the function is f(x) = sin(x). Substituting the values for this function and interval gives: cos(c) = (sin(π) - sin(π/2)) / (π - π/2) = 0, which implies cos(c) = 0. This equation is satisfied for c = π/2 on the interval (π/2, π].

Tune the results.

The values of c found don't exactly fall in the original intervals of the Mean Value Theorem application. Since 0 is not in [0, π/2) and π/2 is not in (π/2, π], there are no values of c that satisfy the Mean Value Theorem in the given intervals.