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.
-$f ( x ) = x + \frac { 12 } { x } , [ 3,4 ]$
A) 3,4
B) $- 2 \sqrt { 3 } , 2 \sqrt { 3 }$
C) $0,2 \sqrt { 3 }$
D) $2 \sqrt { 3 }$

Question

Quizplus · Accepted Answer

To find the value of $c$ that satisfies the equation given by the Mean Value Theorem, we first need to calculate $f'(x)$, the derivative of $f(x) = x + \frac{12}{x}$. Differentiating $f(x)$ with respect to $x$ gives $f'(x) = 1 - \frac{12}{x^2}$. The Mean Value Theorem states that there exists at least one $c$ in the open interval $(a, b)$ such that $\frac{f(b) - f(a)}{b - a} = f'(c)$. For the given function and interval $[3,4]$, we calculate:1. $f(4) = 4 + \frac{12}{4} = 4 + 3 = 7$2. $f(3) = 3 + \frac{12}{3} = 3 + 4 = 7$So, $\frac{f(4) - f(3)}{4 - 3} = \frac{7 - 7}{1} = 0$.We set $f'(c) = 0$ to find $c$:$$1 - \frac{12}{c^2} = 0$$$$\frac{12}{c^2} = 1$$$$c^2 = 12$$$$c = \pm \sqrt{12} = \pm 2\sqrt{3}$$However, since $c$ must be in the interval $(3,4)$, only $c = 2\sqrt{3}$ is valid, because $-2\sqrt{3}$ is not in the interval $(3,4)$.

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