AP Calculus AB 5.1 The Mean Value Theorem on a Polynomial

### AP Calculus AB 5.1: The Mean Value Theorem on a Polynomial

**Overview**: The Mean Value Theorem (MVT) is a fundamental concept in calculus that applies to continuous and differentiable functions. For polynomials, which are smooth and differentiable on their entire domain, the MVT provides a powerful tool to connect average and instantaneous rates of change.

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### **Key Concepts**:

#### **1. Statement of the Mean Value Theorem**:
- **Theorem**: If a function \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \( (a, b) \), then there exists at least one point \( c \) in \( (a, b) \) such that:
\[
f'(c) = \frac{f(b) - f(a)}{b - a}
\]
- **Interpretation**: The MVT guarantees that there is at least one point \( c \) where the instantaneous rate of change (slope of the tangent line) equals the average rate of change (slope of the secant line) over the interval.

#### **2. Applying the MVT to a Polynomial**:
- Polynomials are both continuous and differentiable everywhere, so they automatically satisfy the conditions of the MVT on any closed interval \([a, b]\).

- **Steps to Apply**:
1. **Find the Average Rate of Change**:
- Compute the slope of the secant line connecting the endpoints of the interval \([a, b]\):
\[
\frac{f(b) - f(a)}{b - a}
\]

2. **Differentiate the Polynomial**:
- Take the derivative of the polynomial \( f(x) \) to find \( f'(x) \), which represents the instantaneous rate of change.

3. **Set the Derivative Equal to the Average Rate of Change**:
- Solve \( f'(c) = \frac{f(b) - f(a)}{b - a} \) for the value(s) of \( c \) within the interval \( (a, b) \).

#### **3. Example**:
- Let’s consider a polynomial \( f(x) = x^3 - 3x^2 + 4 \) on the interval \([1, 4]\).

1. **Find the Average Rate of Change**:
\[
f(4) = 4^3 - 3(4^2) + 4 = 64 - 48 + 4 = 20
\]
\[
f(1) = 1^3 - 3(1^2) + 4 = 1 - 3 + 4 = 2
\]
\[
\text{Average rate of change} = \frac{f(4) - f(1)}{4 - 1} = \frac{20 - 2}{3} = \frac{18}{3} = 6
\]

2. **Find the Derivative**:
\[
f'(x) = 3x^2 - 6x
\]

3. **Set the Derivative Equal to the Average Rate**:
\[
3x^2 - 6x = 6
\]
Simplifying:
\[
x^2 - 2x - 2 = 0
\]
Using the quadratic formula, we find:
\[
x = 1 \pm \sqrt{3}
\]
Therefore, \( c \approx 2.732 \) lies in the interval \( (1, 4) \).

#### **4. Conclusion**:
- The MVT confirms that, on a polynomial's interval, there exists a point where the slope of the tangent matches the slope of the secant. This theorem provides both a theoretical foundation and a practical tool for understanding how rates of change behave over an interval.

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### **Applications**:
- The Mean Value Theorem is used in various real-world scenarios, including analyzing rates of change in physics, economics, and engineering, where understanding how an instantaneous rate compares to an average rate can yield important insights into a system's behavior.

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Nick Perich
Norristown Area High School
Norristown Area School District
Norristown, Pa

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