# Rolle's Theorem and the Mean Value Theorem

### Rolle's Theorem:

• Concept: Rolle's Theorem states that if a function $f\left(x\right)$ is continuous on a closed interval $\left[a,b\right]$, differentiable on the open interval $\left(a,b\right)$, and $f\left(a\right)=f\left(b\right)$, then there exists at least one $c$ in $\left(a,b\right)$ where ${f}^{\mathrm{\prime }}\left(c\right)=0$.
• 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 $f\left(x\right)$ is continuous on a closed interval $\left[a,b\right]$ and differentiable on the open interval $\left(a,b\right)$, then there exists at least one $c$ in $\left(a,b\right)$ where: ${f}^{\mathrm{\prime }}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ In other words, the average rate of change of $f\left(x\right)$ over the interval $\left[a,b\right]$ 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 $f\left(x\right)={x}^{3}-3{x}^{2}-9x+5$ on the interval $\left[-2,3\right]$:

1. Show that the conditions of Rolle's Theorem are satisfied on the given interval.
2. Find the value of $c$ that satisfies Rolle's Theorem.

Solution:

1. Conditions of Rolle's Theorem:

• $f\left(x\right)$ is continuous on $\left[-2,3\right]$ and differentiable on $\left(-2,3\right)$.
• $f\left(-2\right)=13$ and $f\left(3\right)=13$.
2. Applying Rolle's Theorem:

• Since $f\left(-2\right)=f\left(3\right)=13$, Rolle's Theorem applies.
• To find $c$, where ${f}^{\mathrm{\prime }}\left(c\right)=0$:
• Find ${f}^{\mathrm{\prime }}\left(x\right)$: ${f}^{\mathrm{\prime }}\left(x\right)=3{x}^{2}-6x-9$.
• Solve ${f}^{\mathrm{\prime }}\left(c\right)=0$: $3{c}^{2}-6c-9=0$.
• Factor and solve: ${c}^{2}-2c-3=0$ $\text{ }⟹\text{ }\left(c-3\right)\left(c+1\right)=0$.
• $c=3$ or $c=-1$, but $c=3$ is within $\left[-2,3\right]$, satisfying the conditions.

### Example: Mean Value Theorem Application

Consider the function $f\left(x\right)={x}^{2}-4x$ on the interval $\left[1,5\right]$.

#### 1. Verification of MVT Conditions:

• Continuity: The function $f\left(x\right)={x}^{2}-4x$ is a polynomial and is continuous on the closed interval $\left[1,5\right]$.
• Differentiability: The function is also differentiable on the open interval $\left(1,5\right)$ as it's a polynomial function.
• Endpoints: $f\left(1\right)={1}^{2}-4\left(1\right)=-3$ and $f\left(5\right)={5}^{2}-4\left(5\right)=5-20=-15$.

#### 2. Application of the Mean Value Theorem:

• Compute the derivative ${f}^{\mathrm{\prime }}\left(x\right)$: ${f}^{\mathrm{\prime }}\left(x\right)=\frac{d}{dx}\left({x}^{2}-4x\right)=2x-4$
• Find the average rate of change of $f\left(x\right)$ over the interval $\left[1,5\right]$:

#### Conclusion:

By the Mean Value Theorem, there exists at least one $c$ in the interval $\left(1,5\right)$ where the derivative ${f}^{\mathrm{\prime }}\left(c\right)$ equals the average rate of change, which is $-3$.

• Solve for ${f}^{\mathrm{\prime }}\left(c\right)=-3$ to find $c$: $2c-4=-3$ $2c=1$ $c=\frac{1}{2}$

### Interpretation:

According to the Mean Value Theorem, there exists at least one $c$ in the interval $\left[1,5\right]$ such that ${f}^{\mathrm{\prime }}\left(c\right)=-3$. In this case, $c=\frac{1}{2}$, indicating that at $x=\frac{1}{2}$, the derivative of the function $f\left(x\right)$ is equal to the average rate of change of $f\left(x\right)$ over the interval $\left[1,5\right]$, 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 ${f}^{\mathrm{\prime }}\left(x\right)$ to find points where ${f}^{\mathrm{\prime }}\left(c\right)=0$ or to evaluate the average rate of change in MVT.