Consider the function $f ( x ) = x ^ { 2 }$ on the interval $\left[ 0 , \frac { 1 } { 2 } \right]$ . According to the Mean Value Theorem, there must be a number c in $\left( 0 , \frac { 1 } { 2 } \right)$ such that $f ^ { \prime } ( c )$ is equal to a particular value d. What is d?
A) $\frac { 3 } { 2 }$

B)1
C) $\frac { 1 } { 2 }$
D)2
E) $\frac { 2 } { 3 }$
F) $\frac { 3 } { 4 }$

G) $\frac { 1 } { 4 }$

H)3

Question

Consider the function $f ( x ) = x ^ { 2 }$ on the interval $\left[ 0 , \frac { 1 } { 2 } \right]$ . According to the Mean Value Theorem, there must be a number c in $\left( 0 , \frac { 1 } { 2 } \right)$ such that $f ^ { \prime } ( c )$ is equal to a particular value d. What is d?
A) $\frac { 3 } { 2 }$

B)1
C) $\frac { 1 } { 2 }$ 
D)2
E) $\frac { 2 } { 3 }$
F) $\frac { 3 } { 4 }$

G) $\frac { 1 } { 4 }$

H)3

Quizplus · Accepted Answer

The Mean Value Theorem states that if a function $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one number $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$. For $f(x) = x^2$ on $\left[0, \frac{1}{2}\right]$, $f'(x) = 2x$. The average rate of change from $0$ to $\frac{1}{2}$ is $\frac{f(\frac{1}{2}) - f(0)}{\frac{1}{2} - 0} = \frac{\left(\frac{1}{2}\right)^2 - 0^2}{\frac{1}{2}} = \frac{1}{4} \times 2 = \frac{1}{2}$.

Consider the Function f(x)=x2f ( x ) = x ^ { 2 }f(x)=x2

Consider the Function $f ( x ) = x ^ { 2 }$