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.
-$f ( x ) = x ^ { 2 } + 3 x + 1 , [ - 1,2 ]$
A) $0 , \frac { 1 } { 2 }$
B) $- 1,2$
C) $\frac { 1 } { 2 }$
D) $- \frac { 1 } { 2 } , \frac { 1 } { 2 }$

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 $f(x) = x^2 + 3x + 1$, first calculate $f'(x) = 2x + 3$. Then, find $f(b) - f(a)$ where $a = -1$ and $b = 2$: $f(2) = 2^2 + 3(2) + 1 = 11$ and $f(-1) = (-1)^2 + 3(-1) + 1 = -1$, so $\frac{f(b) - f(a)}{b - a} = \frac{11 - (-1)}{2 - (-1)} = \frac{12}{3} = 4$. Setting $f'(c) = 2c + 3 = 4$ and solving for $c$ gives $2c = 1$, so $c = \frac{1}{2}$.

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