Find the value or values of $c$ that satisfy the equation $\frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c )$ in the conclusion of the Mean Value Theorem for the function and interval.
-$f ( x ) = \tan ^ { - 1 } x , \left[ - \frac { \sqrt { 3 } } { 3 } , \frac { \sqrt { 3 } } { 3 } \right] \quad$ Round to the nearest thousandth.
A) $\pm 0.320$
B) $- 0.320,0,0.320$
C) $0.320$
D) $0,0.320$

Question

Quizplus · Accepted Answer

The Mean Value Theorem states that if $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one $c$ in $(a, b)$ such that $\frac{f(b) - f(a)}{b - a} = f'(c)$. For $f(x) = \tan^{-1}x$, the derivative $f'(x) = \frac{1}{1+x^2}$.Given interval: $\left[-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right]$1. Calculate the average rate of change from $a = -\frac{\sqrt{3}}{3}$ to $b = \frac{\sqrt{3}}{3}$:$$\frac{f(b) - f(a)}{b - a} = \frac{\tan^{-1}\left(\frac{\sqrt{3}}{3}\right) - \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)}{\frac{\sqrt{3}}{3} - \left(-\frac{\sqrt{3}}{3}\right)} = \frac{\frac{\pi}{6} - \left(-\frac{\pi}{6}\right)}{\frac{2\sqrt{3}}{3}} = \frac{\pi/3}{2\sqrt{3}/3} = \frac{\pi}{2\sqrt{3}} = \frac{\sqrt{3}\pi}{6}$$2. Set this equal to the derivative and solve for $x$:$$\frac{1}{1+x^2} = \frac{\sqrt{3}\pi}{6}$$Solving for $x$ gives:$$1+x^2 = \frac{6}{\sqrt{3}\pi} \Rightarrow x^2 = \frac{6}{\sqrt{3}\pi} - 1$$Plugging in the value of $\pi$ and solving for $x$ gives $x \approx \pm 0.320$, which means $c$ can be either $0.320$ or $-0.320$, matching choice A.

Find the Value or Values Of ccc That Satisfy the Equation f(b)−f(a)b−a=f′(c)\frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c )b−af(b)−f(a)​=f′(c)

Find the Value or Values Of $c$ That Satisfy the Equation $\frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c )$