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) $0.320$
B) $0,0.320$
C) $- 0.320,0,0.320$
D) $\pm 0.320$

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 $f(x) = \tan^{-1}x$ on $\left[-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right]$, we first find $f'(x) = \frac{1}{1+x^2}$. Then, we calculate the average rate of change from $a = -\frac{\sqrt{3}}{3}$ to $b = \frac{\sqrt{3}}{3}$:$$\frac{f(b) - f(a)}{b - a} = \frac{\tan^{-1}\left(\frac{\sqrt{3}}{3}\right) - \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)}{\frac{\sqrt{3}}{3} - \left(-\frac{\sqrt{3}}{3}\right)} = \frac{\frac{\pi}{6} - \left(-\frac{\pi}{6}\right)}{\frac{2\sqrt{3}}{3}} = \frac{\pi}{3} \cdot \frac{3}{2\sqrt{3}} = \frac{\pi}{2\sqrt{3}}.$$Setting this equal to $f'(c) = \frac{1}{1+c^2}$ and solving for $c$, we get:$$\frac{\pi}{2\sqrt{3}} = \frac{1}{1+c^2}.$$Solving this equation for $c$ gives $c = \pm \sqrt{\frac{2\sqrt{3}}{\pi} - 1}$. Plugging in the values, we find that $c \approx \pm 0.320$, which corresponds to choice D.

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 $\frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c )$