Using the Mean Value Theorem - AP Calculus AB 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 simple terms, if a function has a smooth, unbroken path between two points, there must be at least one point on that path where the tangent line is parallel to the secant line connecting the endpoints. This means that the instantaneous rate of change ($f^{\prime }(c)$) at some point $c$ must be equal to the average rate of change over the entire interval. The theorem guarantees the existence of such a point, which is a fundamental concept in calculus for proving other theorems and justifying conclusions about a function's behavior.

The Mean Value Theorem and Rolle's Theorem

The Mean Value Theorem and its special case, Rolle's Theorem, are formally defined as follows.

The Mean Value Theorem (MVT)

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. If a function $f$ meets the conditions of the MVT (continuous on $[a,b]$ and differentiable on $(a,b)$) AND if $f(a)=f(b)$, then there exists at least one value $c$ in the open interval $(a,b)$ such that:

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

This is because if $f(a)=f(b)$, the average rate of change $\frac{f(b)-f(a)}{b-a}=\frac{0}{b-a}=0$.

Understanding the Conditions

The Mean Value Theorem is only guaranteed to apply if two critical conditions are met. It is essential to verify these conditions before applying the theorem.

1. Continuity on the closed interval $[a,b]$: The function must be continuous for all values from $a$ to $b$, including the endpoints. This means there are no holes, jumps, or vertical asymptotes within the interval. If a function is not continuous, it's possible for it to "jump" over the value of the average rate of change, so the conclusion of the theorem may not hold.

2. Differentiability on the open interval $(a,b)$: The function must be differentiable at every point between$a$ and $b$. This means the graph is smooth and has no sharp corners, cusps, or vertical tangents. If a function is not differentiable at a point, the instantaneous rate of change is undefined there, and the theorem cannot guarantee that the slope will equal the average rate of change.

If either of these conditions is not met, the Mean Value Theorem cannot be applied, and its conclusion is not guaranteed.

Core Concepts & Rules

• The Mean Value Theorem establishes that for a function satisfying the necessary conditions, the instantaneous rate of change at some interior point $c$ is equal to the average rate of change over the entire interval.

• The expression $\frac{f(b)-f(a)}{b-a}$ represents the average rate of change of the function $f$ on the interval $[a,b]$. Geometrically, this is the slope of the secant line connecting the points $(a,f(a))$ and $(b,f(b))$.

• The expression $f^{\prime }(c)$ represents the instantaneous rate of change of the function $f$ at the point $x=c$. Geometrically, this is the slope of the tangent line to the curve at $x=c$.

• Required Conditions: Before applying the MVT to a function $f$ on an interval $[a,b]$, you must confirm that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$.

• Rolle's Theorem is a specific application of the MVT where the function values at the endpoints are equal ($f(a)=f(b)$). This guarantees the existence of at least one point $c$ in $(a,b)$ where the tangent line is horizontal ($f^{\prime }(c)=0$).

Step-by-Step Example 1: Applying the Mean Value Theorem

Problem: For the function $f(x)=x^3-2x^2+1$ on the interval $[0,3]$, find the value(s) of $c$ guaranteed by the Mean Value Theorem.

Step 1: Check the conditions.

The function $f(x)$ is a polynomial. Polynomial functions are continuous and differentiable for all real numbers. Therefore, $f(x)$ is continuous on the closed interval $[0,3]$ and differentiable on the open interval $(0,3)$. The Mean Value Theorem applies.

Step 2: Calculate the average rate of change.

First, find the function values at the endpoints.

$f(0)=(0{)}^3-2(0{)}^2+1=1$

$f(3)=(3{)}^3-2(3{)}^2+1=27-18+1=10$

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

$$\text{Average Rate of Change}=\frac{f(3)-f(0)}{3-0}=\frac{10-1}{3}=\frac{9}{3}=3$$

Step 3: Find the derivative of the function.

Using the power rule, we find the derivative $f^{\prime }(x)$.

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

Step 4: Set $f^{\prime }(c)$ equal to the average rate of change and solve for $c$

We need to find the value $c$ such that $f^{\prime }(c)=3$.

$3c^2-4c=3$

$3c^2-4c-3=0$

This is a quadratic equation. We can use the quadratic formula to solve for $c$: $c=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

$c=\frac{-(-4)\pm \sqrt{(-4{)}^2-4(3)(-3)}}{2(3)}$

$c=\frac{4\pm \sqrt{16+36}}{6}$

$c=\frac{4\pm \sqrt{52}}{6}=\frac{4\pm 2\sqrt{13}}{6}=\frac{2\pm \sqrt{13}}{3}$

This gives two possible values for $c$:

$c_1=\frac{2+\sqrt{13}}{3}\approx \frac{2+3.606}{3}\approx 1.869$

$c_2=\frac{2-\sqrt{13}}{3}\approx \frac{2-3.606}{3}\approx -0.535$

Step 5: Verify that $c$ is in the open interval $(0,3)$.

The value $c_1\approx 1.869$ is between 0 and 3.

The value $c_2\approx -0.535$ is not in the interval $(0,3)$.

Therefore, the value of $c$ guaranteed by the Mean Value Theorem is $c=\frac{2+\sqrt{13}}{3}$.

Step-by-Step Example 2: Justifying with the Mean Value Theorem

Problem: The velocity of a particle moving along a straight line is given by a differentiable function $v(t)$, where $t$ is measured in seconds and $v(t)$ is in meters/second. Selected values of $v(t)$ are shown in the table below.

Must there be a time $t=c$ in the interval $(0,8)$ for which the particle's acceleration is equal to $\frac{15}{8}$ m/s^2? Justify your answer.

Step 1: Identify the relevant theorem and check conditions.

The question asks about acceleration, which is the derivative of velocity ($a(t)=v^{\prime }(t)$). We are asked if the instantaneous rate of change of velocity ($v^{\prime }(c)$) must equal a specific value. This is an application of the Mean Value Theorem.

The problem states that $v(t)$ is a differentiable function. If a function is differentiable, it must also be continuous. Therefore, $v(t)$ is continuous on $[0,8]$ and differentiable on $(0,8)$. The MVT applies.

Step 2: Calculate the average rate of change over the interval.

The interval in question is $[0,8]$. We will calculate the average acceleration (average rate of change of velocity) on this interval.

$a=\frac{v(b)-v(a)}{b-a}=\frac{v(8)-v(0)}{8-0}$

Using the values from the table:

$a=\frac{20-5}{8}=\frac{15}{8}$

Step 3: Apply the Mean Value Theorem to form a conclusion.

According to the Mean Value Theorem, since $v(t)$ is continuous on $[0,8]$ and differentiable on $(0,8)$, there must exist at least one time $c$ in the open interval $(0,8)$ such that the instantaneous acceleration $v^{\prime }(c)$ is equal to the average acceleration over the interval.

$$v^{\prime }(c)=\frac{v(8)-v(0)}{8-0}=\frac{15}{8}$$

Justification:

Yes, there must be such a time $c$. Since $v(t)$ is differentiable, it is also continuous. By the Mean Value Theorem, there is a time $c$ in $(0,8)$ such that $v^{\prime }(c)=\frac{v(8)-v(0)}{8-0}=\frac{20-5}{8}=\frac{15}{8}$ m/s^2.

Using Your Calculator

The Mean Value Theorem is an existence theorem, and problems involving it are primarily solved analytically (by hand). A calculator is not typically used to find the value of $c$ directly but can be a powerful tool for verification.

To verify a result for $f(x)$ on $[a,b]$:

1. Graphically check conditions: Graph $Y_1=f(x)$ in a window that includes the interval $[a,b]$. Visually inspect the graph for any discontinuities (jumps, holes) or non-differentiable points (sharp corners, cusps).

2. Verify the value of $c$:

   • Calculate the average rate of change, $m=\frac{f(b)-f(a)}{b-a}$.

   • In your calculator, set $Y_1=f^{\prime }(x)$. You can use the numerical derivative feature, e.g., `nDeriv(f(x), X, X)or $d/dx(f(x))|x=X.

   • Set $Y_2=m$ (the average rate of change you calculated).

   • Graph both $Y_1$ and $Y_2$ on the interval $[a,b]$.

   • Use the calculator's "intersect" feature (2nd -> CALC -> $5:intersect$) to find the x-coordinate of the point where the graphs cross. This x-coordinate is the value of $c$. This can confirm the value you found algebraically.

AP Exam Quick Hit

Common Question Types

• Find the value of $c$: Given an explicit function (e.g., $f(x)=\sqrt{x}$) and an interval (e.g., $[1,9]$), you will be asked to find the specific value of $c$ that satisfies the conclusion of the MVT.

• Justify the existence of a rate: Given a table of values or a graph for a differentiable function $f$, you will be asked to explain why there must be a point $c$ where $f^{\prime }(c)$ equals a certain value. This requires you to identify the correct interval, calculate the average rate of change, and invoke the MVT.

• Determine if the MVT applies: You may be given a function, such as a piecewise function or one with an absolute value, and asked to explain why the MVT does or does not apply on a given interval by checking the continuity and differentiability conditions.

Common Mistakes

• Forgetting to state and check the conditions: A full justification requires you to explicitly state that the function is continuous on the closed interval and differentiable on the open interval. Simply finding a value for $c$ is not sufficient.

• Using the wrong interval for $c$: The value $c$ guaranteed by the MVT must lie in the open interval $(a,b)$. If you solve for $c$ and find a value that is an endpoint ($a$ or $b$) or outside the interval, it is not a valid solution under the theorem.

• Algebraic errors in solving for $c$: After setting $f^{\prime }(c)$ equal to the average rate of change, students often make simple algebraic mistakes, especially when solving quadratic or radical equations.

• Confusing MVT and IVT: The Mean Value Theorem guarantees a specific slope (derivative), while the Intermediate Value Theorem guarantees a specific function value (y-value). Be sure to apply the correct theorem based on what the question is asking for.

• Incorrectly applying Rolle's Theorem: Forgetting to verify the third condition for Rolle's Theorem, $f(a)=f(b)$, before concluding that there must be a point $c$ where $f^{\prime }(c)=0$.