Analytical Applications of Differentiation

Mean Value Theorem

L’Hopital’s Rule

Intermediate Value Theorem

Fundamental Theorem of Calculus

There exists a value $c$ in $[0, 3]$ such that $h'(c) = 9$.

There exists a value $c$ in $(0, 3)$ such that $h(c) = 9$.

There exists a value $c$ in $(0, 3)$ such that $h'(c) = 9$.

There exists a value $c$ in $[0, 3]$ such that $h'(c) = 3$.

There exists a value $c$ in $(-\frac{\pi}{2}, \frac{\pi}{2})$ such that $h'(c) = -1$.

There exists a value $c$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $h'(c) = 0$.

There exists a value $c$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $h'(c) = 1$.

There exists a value $c$ in $(-\frac{\pi}{2}, \frac{\pi}{2})$ such that $h'(c) = 0$.

There exists a value $c$ in $(0, 2)$ such that $g'(c) = \frac{e^2 - 1}{2}$.

There exists a value $c$ in $[0, 2]$ such that $g'(c) = 0$.

There exists a value $c$ in $[0, 2]$ such that $g(c) = \frac{e^2 - 1}{2}$.

There exists a value $c$ in $(0, 2)$ such that $g'(c) = e$.

$\frac{f(b) + f(a)}{2}$

$\frac{f(b) - f(a)}{b - a}$

$f(b) - f(a)$

$\frac{b - a}{f(b) - f(a)}$

Infinitely many values of c exist.

Two distinct values of c exist.

Exactly one value of c exists.

No values of c satisfy this equation.

There exists at least one number $c$ in $(a, b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.

There exists at least one number $c$ in $(a, b)$ such that $\frac{f(b)-f(a)}{b-a} = 0$.

There exists at least one number $c$ in $(a, b)$ such that $f''(c) = 0$.

There exists at least one number $c$ in $(a, b)$ such that $f(c) = f(a)$.

No C exists in (-3,7) as P''(x) has a constant slope of 0

Every x in (-3,7), the P''(x) = $\frac{P(7) - P(-3)}{14}$

There are two distinct numbers C in (-3,7) such that P''(C) = $\frac{P(7) - P(-3)}{14}$

There is at least one C in (-3,7) such that P'(C) = $\frac{P(7) - P(-3)}{14}$

There exists a value $c$ in $(0, \\pi]$ such that $h(c) = 1$.

There exists a value $c$ in $(0, \\pi)$ such that $h'(c) = 1$.

There exists a value $c$ in $[0, \\pi]$ such that $h'(c) = -1$.

There exists a value $c$ in $(0, \\pi)$ such that $h'(c) = 0$.

There must be some number c in (4,10) with g'(c)=12.

There must be some number c either less than 4 or greater than 10 with g'(c)=2.

For all numbers x in (4,10), g'(x)=2.

There must be some number c in (4,10) such that g'(c)=2.