Using the Mean Value Theorem - AP Calculus BC Study Guide

The Core Idea: Using the Mean Value Theorem

The Mean Value Theorem (MVT) provides a powerful connection between the average rate of change of a function over an interval and the instantaneous rate of change at a specific point within that interval. In essence, the theorem guarantees that for any "well-behaved" function (one that is continuous and smooth), there must be at least one moment where the instantaneous rate of change is precisely equal to the overall average rate of change.

Imagine driving a car on a trip. Your average speed is the total distance divided by the total time. The Mean Value Theorem guarantees that at some specific instant during your trip, your speedometer must have read exactly that average speed. Geometrically, this means that for any secant line connecting two points on a smooth curve, there exists at least one point on the curve between them where the tangent line is parallel to that secant line. Rolle's Theorem is a special case of this idea, where if the function's values at the endpoints are equal (a zero average rate of change), there must be a point between them with a horizontal tangent line (a zero instantaneous rate of change).

Key Theorems

The Mean Value Theorem

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 value $c$ in the open interval $(a,b)$ such that:

$$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$$

Rolle's Theorem

Rolle's Theorem is a special case of the Mean Value Theorem. It states that if a function $f$ is continuous on the closed interval $[a,b]$, differentiable on the open interval $(a,b)$, and $f(a)=f(b)$, then there exists at least one value $c$ in the open interval $(a,b)$ such that:

$$f^{\prime }(c)=0$$

This follows directly from the Mean Value Theorem, because if $f(a)=f(b)$, the average rate of change $\frac{f(b)-f(a)}{b-a}$ is zero.

Understanding the Conditions

The two conditions for applying the Mean Value Theorem—continuity and differentiability—are not optional technicalities; they are essential for the theorem's conclusion to hold true.

1. Continuity on the closed interval $[a,b]$: This ensures there are no jumps, holes, or vertical asymptotes within the interval, including at the endpoints. If a function has a discontinuity, it's possible to draw a secant line connecting the endpoints that has no parallel tangent line anywhere on the curve. For example, a function with a jump discontinuity could "skip over" the required slope.

2. Differentiability on the open interval $(a,b)$: This ensures the function is "smooth" and has no sharp corners or cusps between the endpoints. At a sharp corner, the derivative is undefined, meaning there is no tangent line. A function could have a corner whose slopes on either side are both greater (or both less) than the slope of the secant line, thus never achieving the required instantaneous rate of change.

It is critical to always verify that both conditions are met before applying the Mean Value Theorem or Rolle's Theorem. If either condition fails, the theorem cannot be applied, and its conclusion is not guaranteed.

Core Concepts & Rules

• Connecting Average and Instantaneous Rates: The MVT establishes that for a function satisfying the conditions, the average rate of change over $[a,b]$ must be achieved by the instantaneous rate of change at some point $c$ in $(a,b)$.

• Geometric Interpretation: The value $\frac{f(b)-f(a)}{b-a}$ represents the slope of the secant line connecting the points $(a,f(a))$ and $(b,f(b))$. The value $f^{\prime }(c)$ represents the slope of the tangent line to the curve at $x=c$. The MVT guarantees the existence of at least one point $c$ where the tangent line is parallel to the secant line.

• Conditions are Mandatory: The function $f$ must be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. If these conditions are not met, the theorem does not apply.

• Existence, Not Location: The MVT is an "existence theorem." It guarantees that at least one such value of $c$ exists, but it does not provide a method for finding it. Finding $c$ typically requires solving the equation $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$ algebraically.

• Rolle's Theorem as a Special Case: If the function values at the endpoints are equal ($f(a)=f(b)$), the slope of the secant line is zero. Rolle's Theorem then guarantees a point $c$ where the tangent line is horizontal ($f^{\prime }(c)=0$).

Step-by-Step Example 1: Basic Application

Problem: For the function $f(x)=x^3-4x$, find the value(s) of $c$ on the interval $[0,3]$ that satisfy the conclusion of the Mean Value Theorem.

Step 1: Verify the Conditions

• The function $f(x)=x^3-4x$ is a polynomial.

• Continuity: Polynomials are continuous everywhere, so $f(x)$ is continuous on the closed interval $[0,3]$.

• Differentiability: Polynomials are differentiable everywhere, so $f(x)$ is differentiable on the open interval $(0,3)$.

• Since both conditions are met, the Mean Value Theorem can be applied.

Step 2: Calculate the Average Rate of Change

First, find the function values at the endpoints $a=0$ and $b=3$.

• $f(0)=(0{)}^3-4(0)=0$

• $f(3)=(3{)}^3-4(3)=27-12=15$

Now, calculate the average rate of change using the formula $\frac{f(b)-f(a)}{b-a}$.

$$\text{Average Rate of Change}=\frac{f(3)-f(0)}{3-0}=\frac{15-0}{3}=5$$

Step 3: Find the Derivative of the Function

Find the instantaneous rate of change, $f^{\prime }(x)$.

$$f^{\prime }(x)=\frac{d}{dx}(x^3-4x)=3x^2-4$$

Step 4: Set $f^{\prime }(c)$ Equal to the Average Rate of Change and Solve

The MVT guarantees a value $c$ such that $f^{\prime }(c)=5$.

$$3c^2-4=5$$

$$3c^2=9$$

$$c^2=3$$

$$c=\pm \sqrt{3}$$

Step 5: Verify that $c$ is in the Open Interval $(a,b)$

The interval is $(0,3)$. We must check which of our solutions, $c=\sqrt{3}$ and $c=-\sqrt{3}$, lies within this interval.

• $c=\sqrt{3}\approx 1.732$. Since $0<1.732<3$, this value is in the interval.

• $c=-\sqrt{3}\approx -1.732$. This value is not in the interval $(0,3)$.

Therefore, the only value of $c$ that satisfies the conclusion of the Mean Value Theorem is $c=\sqrt{3}$.

Step-by-Step Example 2: Exam-Style Application

Problem: Consider the function $f(x)=\sqrt{25-x^2}$.

(a) Explain why the Mean Value Theorem can be applied on the interval $[0,3]$.

(b) Find the value(s) of $c$ guaranteed by the theorem on $[0,3]$.

(c) Explain why Rolle's Theorem can be applied on the interval $[-4,4]$ and find the value of $c$ guaranteed by it.

(a) Verifying Conditions for MVT on $[0,3]$

• Continuity: The function $f(x)=\sqrt{25-x^2}$ is continuous on its domain, which is $-5\le x\le 5$. Since the interval $[0,3]$ is within this domain, $f(x)$ is continuous on $[0,3]$.

• Differentiability: First, find the derivative.

  $$f^{\prime }(x)=\frac{d}{dx}(25-x^2{)}^{1/2}=\frac{1}{2}(25-x^2{)}^{-1/2}(-2x)=\frac{-x}{\sqrt{25-x^2}}$$

  The derivative $f^{\prime }(x)$ is defined for all $x$ such that $25-x^2>0$, which means $-5<x<5$. Since the open interval $(0,3)$ is within $(-5,5)$, the function is differentiable on $(0,3)$.

• Because $f(x)$ is continuous on $[0,3]$ and differentiable on $(0,3)$, the MVT applies.

(b) Finding $c$ on $[0,3]$

• Average Rate of Change:

  • $f(0)=\sqrt{25-0^2}=5$

  • $f(3)=\sqrt{25-3^2}=\sqrt{16}=4$

  • $\frac{f(3)-f(0)}{3-0}=\frac{4-5}{3}=-\frac{1}{3}$

• Set $f^{\prime }(c)$ Equal to Average Rate of Change:

  $$\frac{-c}{\sqrt{25-c^2}}=-\frac{1}{3}$$

  $$3c=\sqrt{25-c^2}$$

  Square both sides:

  $$9c^2=25-c^2$$

  $$10c^2=25$$

  $$c^2=\frac{25}{10}=\frac{5}{2}$$

  $$c=\pm \sqrt{\frac{5}{2}}=\pm \frac{\sqrt{10}}{2}$$

• Check the Interval: The interval is $(0,3)$.

  • $c=\frac{\sqrt{10}}{2}\approx \frac{3.16}{2}=1.58$. This is in $(0,3)$.

  • $c=-\frac{\sqrt{10}}{2}\approx -1.58$. This is not in $(0,3)$.

  • The value is $c=\frac{\sqrt{10}}{2}$.

(c) Applying Rolle's Theorem on [-4, 4]

• Verify Conditions:

  • Continuity: The interval $[-4,4]$ is within the domain $[-5,5]$, so $f(x)$ is continuous on $[-4,4]$.

  • Differentiability: The interval $(-4,4)$ is within $(-5,5)$, so $f(x)$ is differentiable on $(-4,4)$.

  • Check Endpoints:$f(a)=f(b)$?

    • $f(-4)=\sqrt{25-(-4{)}^2}=\sqrt{25-16}=\sqrt{9}=3$

    • $f(4)=\sqrt{25-(4{)}^2}=\sqrt{25-16}=\sqrt{9}=3$

    • Since $f(-4)=f(4)=3$, this condition is met.

• Find $c$: Rolle's Theorem guarantees a $c$ in $(-4,4)$ such that $f^{\prime }(c)=0$.

  $$f^{\prime }(c)=\frac{-c}{\sqrt{25-c^2}}=0$$

  A fraction is zero only when its numerator is zero.

  $$-c=0⟹c=0$$

• Check the Interval: The value $c=0$ is in the open interval $(-4,4)$. Thus, $c=0$ is the value guaranteed by Rolle's Theorem.

Using Your Calculator

The Mean Value Theorem is primarily an analytical tool, meaning problems are typically solved by hand using algebra and calculus. A calculator is not used to find the value of $c$ directly but can be used to support the analytical process.

1. Verifying Conditions:

• You can graph the function $f(x)$ on the interval $[a,b]$ to visually check for discontinuities (jumps, holes) or non-differentiable points (sharp corners, cusps).

2. Calculating the Average Rate of Change:

• A calculator is useful for quickly and accurately finding $f(a)$ and $f(b)$, especially for complex functions, to compute $\frac{f(b)-f(a)}{b-a}$.

3. Solving for $c$:

• If the equation $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$ is difficult to solve algebraically, you can use a calculator's numeric solver or graphical intersection feature.

• Example using Graphical Intersection (TI-84 Style):

  • Let's say you need to solve $3c^2-4=5$ from Example 1.

  • In your calculator, set Y1 = 3X^2 - 4 (the derivative).

  • Set Y2 = 5 (the average rate of change).

  • Graph both functions and use the $calc$ menu (2nd -> TRACE) and select $5:intersect$.

  • The calculator will find the x-coordinates of the intersection points, which correspond to the values of $c$. You must still manually check which of these values fall within the specified open interval.

AP Exam Quick Hit

Common Question Types

• Algebraic Application: Given a function $f(x)$ and an interval $[a,b]$, find the value of $c$ that satisfies the conclusion of the Mean Value Theorem. This is the most direct type of question.

  • Example: "For $f(x)=\sin (x)$ on $[0,\pi /2]$, find the value of $c$ guaranteed by the MVT."

• Justification and Condition Checking: Given a function (often piecewise or with potential discontinuities/cusps), determine if the MVT can be applied and justify your reasoning based on the conditions.

  • Example: "Can the Mean Value Theorem be applied to $f(x)=|x-2|$ on the interval $[1,4]$? Justify your answer." (Answer: No, because $f(x)$ is not differentiable at $x=2$, which is in the interval $(1,4)$.)

• Conceptual Application with a Table or Graph: Given a table of values for a differentiable function $f$ or a graph of $f$, use the MVT to draw a conclusion about its derivative, $f^{\prime }$.

  • Example: "The function $g(x)$ is continuous and differentiable. Selected values of $g(x)$ are given in the table below. What is the minimum number of times $g^{\prime }(x)$ must equal 2 on the interval $(0,10)$?" You would calculate the average rate of change on subintervals to apply the MVT.

(On $[0,4]$, the average rate of change is $(13-5)/(4-0)=2$. So the MVT guarantees $g^{\prime }(c)=2$ for some $c$ in $(0,4)$.)

Common Mistakes

• Forgetting to State and Check Conditions: The most common error is immediately setting $f^{\prime }(c)$ equal to the average rate of change without first explicitly stating and verifying that the function is continuous on the closed interval and differentiable on the open interval. Justification is key on the AP Exam.

• Using the Wrong Interval for Conditions: Stating that the function is differentiable on the closed interval $[a,b]$. The theorem only requires differentiability on the open interval $(a,b)$.

• Incorrectly Calculating the Average Rate of Change: Simple arithmetic errors or mixing up the numerator and denominator in the slope formula $\frac{f(b)-f(a)}{b-a}$.

• Solving for $c$ but Not Checking if it is in the Interval: After finding one or more potential values for $c$, students forget the final, critical step of ensuring the value lies within the open interval$(a,b)$. Any solution outside this interval must be discarded.

• Confusing MVT with Rolle's Theorem: Applying Rolle's Theorem ($f^{\prime }(c)=0$) when the condition $f(a)=f(b)$ is not met, or not recognizing that Rolle's Theorem can be used when it is applicable.