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

Question

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

Quizplus · Accepted Answer

The function $f(\theta)$ is not continuous on the closed interval $[- \pi, 0]$ because it is not defined at $\theta = 0$ in a way that makes it continuous. The function has a different value for $\theta = 0$ than the limit of $\frac{\cos\theta}{\theta}$ as $\theta$ approaches 0 from the left, which does not exist. Therefore, it does not satisfy the hypotheses of the Mean Value Theorem for the interval $[- \pi, 0]$.

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