Provide an appropriate response.
-Decide if the statement is true or false. If false, explain.
The points $( - 1 , - 1 )$ and $( 1,1 )$ lie on the graph of $f ( x ) = \frac { 1 } { x }$. Therefore, the Mean Value Theorem says that there exists some value $x = c$ on $( - 1,1 )$ for which $f ^ { \prime } ( x ) = \frac { 1 - ( - 1 ) } { 1 - ( - 1 ) } = 1$

Question

Quizplus · Accepted Answer

The Mean Value Theorem states that for a function that is continuous on a closed interval and differentiable on the open interval, there exists a value $c$ in the open interval such that $f'(c)$ is equal to the slope of the secant line connecting the two endpoints of the interval. In this case, the function $f(x)=\frac{1}{x}$ is not continuous on the closed interval $[-1,1]$ because it is not defined at $x=0$. Therefore, the Mean Value Theorem cannot be applied to this interval. Also, even if the function were continuous on the interval $(-1,1)$, the value of $f'(c)$ would not necessarily be equal to the slope of the secant line connecting the two given points, $( - 1 , - 1 )$ and $( 1,1 )$.

Provide an Appropriate Response (−1,−1)( - 1 , - 1 )(−1,−1)

Provide an Appropriate Response $( - 1 , - 1 )$