Rolle's Theorem and the Mean Value Theorem
Rolle's Theorem:
 Concept: Rolle's Theorem states that if a function $f(x)$ is continuous on a closed interval $[a,b]$, differentiable on the open interval $(a,b)$, and $f(a)=f(b)$, then there exists at least one $c$ in $(a,b)$ where ${f}^{\mathrm{\prime}}(c)=0$.
 This means that between any two points where the function has the same value, there exists at least one point where the derivative of the function is zero.
The Mean Value Theorem (MVT):
 Concept: The Mean Value Theorem states that if a function $f(x)$ is continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists at least one $c$ in $(a,b)$ where: ${f}^{\mathrm{\prime}}(c)=\frac{f(b)f(a)}{ba}$ In other words, the average rate of change of $f(x)$ over the interval $[a,b]$ is equal to the instantaneous rate of change at some point $c$ in the interval.
Calculation Using Derivative Rules:
Example 1: Applying Rolle's Theorem
Given $f(x)={x}^{3}3{x}^{2}9x+5$ on the interval $[2,3]$:
 Show that the conditions of Rolle's Theorem are satisfied on the given interval.
 Find the value of $c$ that satisfies Rolle's Theorem.
Solution:

Conditions of Rolle's Theorem:
 $f(x)$ is continuous on $[2,3]$ and differentiable on $(2,3)$.
 $f(2)=13$ and $f(3)=13$.

Applying Rolle's Theorem:
 Since $f(2)=f(3)=13$, Rolle's Theorem applies.
 To find $c$, where ${f}^{\mathrm{\prime}}(c)=0$:
 Find ${f}^{\mathrm{\prime}}(x)$: ${f}^{\mathrm{\prime}}(x)=3{x}^{2}6x9$.
 Solve ${f}^{\mathrm{\prime}}(c)=0$: $3{c}^{2}6c9=0$.
 Factor and solve: ${c}^{2}2c3=0$ $\text{\hspace{0.25em}\hspace{0.05em}}\u27f9\text{\hspace{0.25em}\hspace{0.05em}}(c3)(c+1)=0$.
 $c=3$ or $c=1$, but $c=3$ is within $[2,3]$, satisfying the conditions.
Example: Mean Value Theorem Application
Consider the function $f(x)={x}^{2}4x$ on the interval $[1,5]$.
1. Verification of MVT Conditions:
 Continuity: The function $f(x)={x}^{2}4x$ is a polynomial and is continuous on the closed interval $[1,5]$.
 Differentiability: The function is also differentiable on the open interval $(1,5)$ as it's a polynomial function.
 Endpoints: $f(1)={1}^{2}4(1)=3$and $f(5)={5}^{2}4(5)=520=15$.
2. Application of the Mean Value Theorem:
 Compute the derivative ${f}^{\mathrm{\prime}}(x)$: ${f}^{\mathrm{\prime}}(x)=\frac{d}{dx}({x}^{2}4x)=2x4$
 Find the average rate of change of $f(x)$ over the interval $[1,5]$: $\text{Averagerateofchange}=\frac{f(5)f(1)}{51}=\frac{15(3)}{51}=\frac{12}{4}=3$
Conclusion:
By the Mean Value Theorem, there exists at least one $c$ in the interval $(1,5)$ where the derivative ${f}^{\mathrm{\prime}}(c)$ equals the average rate of change, which is $3$.
 Solve for ${f}^{\mathrm{\prime}}(c)=3$ to find $c$: $2c4=3$ $2c=1$ $c=\frac{1}{2}$
Interpretation:
According to the Mean Value Theorem, there exists at least one $c$ in the interval $[1,5]$ such that ${f}^{\mathrm{\prime}}(c)=3$. In this case, $c=\frac{1}{2}$, indicating that at $x=\frac{1}{2}$, the derivative of the function $f(x)$ is equal to the average rate of change of $f(x)$ over the interval $[1,5]$, which is $3$.
Strategies:
 Function Analysis: Ensure the function meets the specified conditions (continuity, differentiability, equality of function values) for Rolle's Theorem or MVT.
 Derivative Calculations: Compute the derivative ${f}^{\mathrm{\prime}}(x)$ to find points where ${f}^{\mathrm{\prime}}(c)=0$ or to evaluate the average rate of change in MVT.