Determine whether the function satisfies the hypotheses of the Mean Value Theorem for the given interval.
-$f ( \theta ) = \left\{ \begin{array} { c l }
\frac { \cos \theta } { \theta } , & - \pi \leq \theta

Question

Determine whether the function satisfies the hypotheses of the Mean Value Theorem for the given interval.
-$f ( \theta ) = \left\{ \begin{array} { c l } 
\frac { \cos \theta } { \theta } , & - \pi \leq \theta < 0 \
0 , & \theta = 0
\end{array} \right.$

Quizplus · Accepted Answer

The function $f(\theta)$ is not continuous at $\theta = 0$ because it is defined piecewise with a discontinuity at $\theta = 0$. For the Mean Value Theorem to apply, the function must be continuous on the closed interval $[- \pi, 0]$ and differentiable on the open interval $(- \pi, 0)$. The discontinuity at $\theta = 0$ violates the continuity requirement.

Determine Whether the Function Satisfies the Hypotheses of the Mean f(θ)={cos⁡θθ,−π≤θ<00,θ=0f ( \theta ) = \left\{ \begin{array} { c l } \frac { \cos \theta } { \theta } , & - \pi \leq \theta < 0 \\0 , & \theta = 0\end{array} \right.f(θ)={θcosθ​,0,​−π≤θ<0θ=0​

Determine Whether the Function Satisfies the Hypotheses of the Mean $f ( \theta ) = \left\{ \begin{array} { c l } \frac { \cos \theta } { \theta } , & - \pi \leq \theta < 0 \\0 , & \theta = 0\end{array} \right.$