MATH 430 - Lesson 22: Rolle's Theorem and Mean Value Theorem

Topics Covered:

Warm-up: Common Features of Graphs

Consider two graphs that share the following characteristics:

  1. Continuous
  2. Differentiable
  3. Same endpoint values

Observation: Both graphs shown have a point in the middle with a horizontal tangent line. This observation leads us to Rolle's Theorem.

Rolle's Theorem

Rolle's Theorem: Let \(f(x)\) be a function that satisfies the following conditions:

  1. \(f(x)\) is continuous on the closed interval \([a, b]\)
  2. \(f(x)\) is differentiable on the open interval \((a, b)\) (the derivative exists)
  3. \(f(a) = f(b)\) (same endpoint values)

Then there exists at least one number \(c\) such that:

Graph Description: The graph shows a smooth curve starting at point \((a, f(a))\) and ending at point \((b, f(b))\) where both endpoints have the same y-value. Between these points, there is at least one point \(c\) where the tangent line is horizontal, meaning the derivative equals zero.

When Rolle's Theorem Fails

Rolle's Theorem will not apply if any of the three conditions are violated:

Case 1: Not Continuous - A graph with a jump discontinuity or break may not have a horizontal tangent even if the endpoints have the same value.

Case 2: Not Differentiable - A graph with a sharp corner or cusp may not have a point where \(f'(c) = 0\), even though it might have a horizontal tangent visually.

Case 3: \(f(a) \neq f(b)\) - If the endpoint values are different, Rolle's Theorem does not apply. This situation is handled by the Mean Value Theorem instead.

Examples Using Rolle's Theorem

Example 1: Finding c for \(f(x) = x^3 - 4x + 6\) on \([0, 2]\)

Step 1: Verify the conditions

  1. Continuous: Yes, it's a polynomial
  2. Differentiable: Yes, polynomials are differentiable everywhere
  3. Check endpoints: \(f(0) = 0^3 - 4(0) + 6 = 6\) and \(f(2) = 2^3 - 4(2) + 6 = 8 - 8 + 6 = 6\)

Since \(f(0) = f(2) = 6\), all three conditions are satisfied.

Step 2: Find c by solving \(f'(c) = 0\)

First, find the derivative:

\[f'(x) = 3x^2 - 4\]

Set the derivative equal to zero:

\[f'(c) = 3c^2 - 4 = 0\] \[3c^2 = 4\] \[c^2 = \frac{4}{3}\] \[c = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}\]

Since we need \(0 < c < 2\), we check which value(s) are in this interval:

\[c = \frac{2\sqrt{3}}{3} \approx 1.15\]

This value is between 0 and 2, so \(c = \frac{2\sqrt{3}}{3}\) is the answer. Note that \(c = -\frac{2\sqrt{3}}{3}\) is not in the interval \([0, 2]\).

Example 2: Finding c for \(f(x) = e^{x^2}\) on \([-3, 3]\)

Step 1: Verify the conditions

  1. Continuous: Yes, composition of exponential and polynomial functions
  2. Differentiable: Yes, both exponential and polynomial functions are differentiable
  3. Check endpoints: \(f(-3) = e^{(-3)^2} = e^9\) and \(f(3) = e^{3^2} = e^9\)

All three conditions are satisfied.

Step 2: Find c by solving \(f'(c) = 0\)

Using the chain rule:

\[f'(x) = \frac{d}{dx}e^{x^2} = e^{x^2} \cdot 2x\]

Set the derivative equal to zero:

\[f'(c) = e^{c^2} \cdot 2c = 0\]

Since \(e^{c^2} > 0\) for all values of \(c\), we must have:

\[2c = 0\] \[c = 0\]

Since \(-3 < 0 < 3\), the value \(c = 0\) satisfies Rolle's Theorem.

Mean Value Theorem (MVT)

Mean Value Theorem: Let \(f(x)\) be a function that satisfies:

  1. \(f(x)\) is continuous on the closed interval \([a, b]\)
  2. \(f(x)\) is differentiable on the open interval \((a, b)\)

Then there exists at least one number \(c\) such that:

This means there is a point in the middle where the tangent line is parallel to the secant line connecting the endpoints.

Connection to Rolle's Theorem: Rolle's Theorem is a special case of the Mean Value Theorem where \(f(a) = f(b)\). In this case, the slope of the secant line is \(\frac{f(b) - f(a)}{b - a} = \frac{0}{b - a} = 0\), which gives us \(f'(c) = 0\).

Graph Description: The graph shows a curve from point \((a, f(a))\) to point \((b, f(b))\). A secant line connects these two endpoints with slope \(\frac{f(b) - f(a)}{b - a}\). At some point \(c\) between \(a\) and \(b\), there is a tangent line that is parallel to this secant line, meaning \(f'(c)\) equals the slope of the secant line.

Example Using Mean Value Theorem

Example: Finding c for \(f(x) = x^2\) on \([0, 3]\)

We can find a value \(c\) such that \(0 < c < 3\) and the tangent line at \(c\) is parallel to the secant line.

Step 1: Find the slope of the secant line

\[\text{Slope} = \frac{f(3) - f(0)}{3 - 0} = \frac{9 - 0}{3} = 3\]

Step 2: Find the derivative

\[f'(x) = 2x\]

Step 3: Set \(f'(c)\) equal to the slope of the secant line

\[f'(c) = 3\] \[2c = 3\] \[c = \frac{3}{2}\]

Since \(0 < \frac{3}{2} < 3\), this value satisfies the Mean Value Theorem.

What the Derivative Tells Us

The derivative provides crucial information about the behavior of a function:

Increasing and Decreasing:

Graph Interpretation: Consider a function with critical points at \(x = 2\) and \(x = 5\) where \(f'(2) = 0\) and \(f'(5) = 0\):

Local Extrema

Local/Relative Maximum: At \(x = 2\), if the function is increasing on the left (\(f'(x) > 0\)) and decreasing on the right (\(f'(x) < 0\)), then \(x = 2\) is a local maximum.

Local/Relative Minimum: At \(x = 5\), if the function is decreasing on the left (\(f'(x) < 0\)) and increasing on the right (\(f'(x) > 0\)), then \(x = 5\) is a local minimum.

First Derivative Test

First Derivative Test for Local Maximum:

A function \(f(x)\) has a relative/local maximum at \(x = c\) if:

  1. \(f'(c) = 0\)
  2. \(f(x)\) is increasing on the left of \(c\) (i.e., \(f'(x) > 0\) for \(x < c\))
  3. \(f(x)\) is decreasing on the right of \(c\) (i.e., \(f'(x) < 0\) for \(x > c\))

First Derivative Test for Local Minimum:

A function \(f(x)\) has a relative/local minimum at \(x = c\) if:

  1. \(f'(c) = 0\)
  2. \(f(x)\) is decreasing on the left of \(c\) (i.e., \(f'(x) < 0\) for \(x < c\))
  3. \(f(x)\) is increasing on the right of \(c\) (i.e., \(f'(x) > 0\) for \(x > c\))

Example Using First Derivative Test

Example: Find relative maximum and minimum for \(f(x) = 3x^5 - 5x^3\)

Step 1: Find the derivative

\[f'(x) = 15x^4 - 15x^2 = 15x^2(x^2 - 1) = 15x^2(x - 1)(x + 1)\]

Step 2: Find critical points by setting \(f'(x) = 0\)

\[15x^2(x - 1)(x + 1) = 0\] \[x = 0, \quad x = 1, \quad x = -1\]

Step 3: Create a sign chart for \(f'(x)\)

Test the sign of \(f'(x)\) in each interval:

Interval Test Point Sign of \(f'(x)\) Behavior
\((-\infty, -1)\) \(x = -2\) \(+\) Increasing ↑
\((-1, 0)\) \(x = -0.5\) \(-\) Decreasing ↓
\((0, 1)\) \(x = 0.5\) \(-\) Decreasing ↓
\((1, \infty)\) \(x = 2\) \(+\) Increasing ↑

Step 4: Apply the First Derivative Test

Key Takeaway: Not every critical point (where \(f'(x) = 0\)) is a local maximum or minimum. We must check the sign changes of the derivative to determine the nature of each critical point.