AP Calculus AB Unit 5: Analytical Applications of Differentiation Study Guide

5.1 The Mean Value Theorem (MVT) & Rolle’s Theorem

Theoretical Framework

The Mean Value Theorem is one of the most critical existence theorems in Calculus. It establishes a connection between the average rate of change over an interval and the instantaneous rate of change at a specific point within that interval.

The Mean Value Theorem (MVT)

Hypotheses (Conditions):
For a function $f(x)$ defined on an interval $[a, b]$:

1. $f(x)$ must be continuous on the closed interval $[a, b]$.
2. $f(x)$ must be differentiable on the open interval $(a, b)$.

Conclusion:
There exists at least one number $c$ such that $a < c < b$ where:
f'(c) = \frac{f(b) - f(a)}{b - a}

Translation: There is at least one point where the slope of the tangent line equals the slope of the secant line connected endpoints $a$ and $b$.

Rolle's Theorem

Rolle’s Theorem is a special case of the MVT where the starting and ending $y$-values are the same ($f(a) = f(b)$).

Conclusion:
If $f(a) = f(b)$, then there exists at least one $c$ in $(a, b)$ such that:
f'(c) = 0
Translation: Somewhere between two points with the same height, the graph must turn around, meaning there is a horizontal tangent.

5.2 The Extreme Value Theorem (EVT)

Absolute (Global) Extrema

The Extreme Value Theorem guarantees the existence of maximums and minimums under specific conditions.

The Theorem:
If a function $f(x)$ is continuous on a closed interval $[a, b]$, then $f(x)$ has both an absolute maximum and an absolute minimum on that interval.

Local (Relative) vs. Absolute (Global)

• Absolute Extrema: The highest or lowest $y$-value on the entire domain or interval.
• Relative Extrema: The highest or lowest $y$-value relative to the points immediately surrounding it (hills and valleys).

5.3 First Derivative Analysis: Increasing & Decreasing Intervals

The sign of the first derivative tells us the direction of the function.

Definitions

• If $f'(x) > 0$ on an interval, $f(x)$ is increasing.
• If $f'(x) < 0$ on an interval, $f(x)$ is decreasing.

Finding Critical Points

A Critical Point is a point in the domain of $f$ where:

1. $f'(x) = 0$ (Horizontal tangent)
2. $f'(x)$ is undefined (Corner, Cusp, or Vertical Tangent)

Procedure: The Number Line Test

1. Find the domain of $f(x)$.
2. Find $f'(x)$ and determine all Critical Points.
3. Place Critical Points on a number line.
4. Test a value in each sub-interval in the derivative $f'(x)$.

Example: $f(x) = x^3 - 6x^2 + 9x + 2$

1. Differentiate: $f'(x) = 3x^2 - 12x + 9$
2. Find Critical Points:
   3(x^2 - 4x + 3) = 0 \rightarrow 3(x-3)(x-1)=0
   Critical points at $x=1$ and $x=3$.
3. Test Intervals:
   • $x < 1$ (Test 0): $f'(0) = 9$ (Pos) $\rightarrow$ Increasing
   • $1 < x < 3$ (Test 2): $f'(2) = 12 - 24 + 9 = -3$ (Neg) $\rightarrow$ Decreasing
   • $x > 3$ (Test 4): $f'(4) = 48 - 48 + 9 = 9$ (Pos) $\rightarrow$ Increasing

5.4 Using the First Derivative Test for Local Extrema

Once sign analysis is complete, we classify the critical points.

The First Derivative Test:
At a critical point $c$:

• If $f'$ changes from Positive to Negative $\rightarrow$ Relative Maximum (The graph goes up, then down).
• If $f'$ changes from Negative to Positive $\rightarrow$ Relative Minimum (The graph goes down, then up).
• If $f'$ does not change signs $\rightarrow$ Not an extrema (Plateau).

Applying to Example Above:

• At $x=1$, signs shift $+ \to -$, so $x=1$ is a Relative Max.
• At $x=3$, signs shift $- \to +$, so $x=3$ is a Relative Min.

5.5 Finding Absolute Extrema: The Candidates Test

When looking for the Absolute Max/Min on a closed interval, you cannot rely solely on the First Derivative Test. You must use the Candidates Test.

Steps for Candidates Test:

1. Confirm $f(x)$ is continuous on the closed interval $[a, b]$.
2. Find all critical points inside $(a, b)$.
3. Evaluate the original function $f(x)$ at:
   • All Critical Points
   • The Endpoints ($a$ and $b$)
4. Compare the $y$-values. The highest is the Absolute Max; the lowest is the Absolute Min.

Common Mistake: Students often forget to check the endpoints. The absolute max often occurs at the start or end of the interval, not just at the peaks/valleys.

5.6 Concavity and Points of Inflection

Concavity refers to the curvature of the graph—whether it opens up or down. This is determined by the second derivative, $f''(x)$.

Definitions

• Concave Up: $f(x)$ acts like a cup holding water. Slopes ($f'$) are increasing.
  • Condition: $f''(x) > 0$
• Concave Down: $f(x)$ acts like a frown. Slopes ($f'$) are decreasing.
  • Condition: $f''(x) < 0$

Point of Inflection (POI)

A Point of Inflection is a point where the graph changes concavity.

Conditions for a POI at $x=c$:

1. The tangent line must exist at $c$ ($f$ is differentiable or has a vertical tangent).
2. $f''(x)$ must change signs at $x=c$.

Note: $f''(c)=0$ is NOT sufficient to prove a POI. You must verify the sign change.

Example from previous function:
$f'(x) = 3x^2 - 12x + 9$

1. Second Derivative: $f''(x) = 6x - 12$
2. Find PIPs (Possible Inflection Points): Set $f''(x) = 0 \rightarrow 6x=12 \rightarrow x=2$.
3. Sign Check:
   • $x < 2$: $f''$ is Negative (Concave Down)
   • $x > 2$: $f''$ is Positive (Concave Up)
   • Since sign changes, $x=2$ is a Point of Inflection.

5.7 The Second Derivative Test (For Extrema)

This is an alternative method to determining if a critical point is a max or min without creating a full sign chart for $f'$.

The Rule:
Let $c$ be a critical point where $f'(c) = 0$.

1. If $f''(c) > 0$ (Concave Up) $\rightarrow$ $c$ is a Relative Minimum.
2. If $f''(c) < 0$ (Concave Down) $\rightarrow$ $c$ is a Relative Maximum.
3. If $f''(c) = 0$, the test is inconclusive. (You must go back and use the First Derivative Test).

Mnemonic:

• $f''$ is Positive $(+)$ $\rightarrow$ Happy Face $\cup$ $\rightarrow$ Minimum at the bottom.
• $f''$ is Negative $(-)$ $\rightarrow$ Sad Face $\cap$ $\rightarrow$ Maximum at the top.

5.8 Connecting f, f', and f''

Understanding the relationship between the graph of a function and its derivatives is crucial for multiple-choice questions.

5.9 Optimization

Optimization involves finding the "best" value (maximum profit, minimum cost, minimum distance, max area) in real-world scenarios.

General Strategy

1. Draw a Picture: Label constant values and variables.
2. Primary Equation: Write the formula for the quantity you want to maximize/minimize.
3. Constraint: Find an equation relating the variables (e.g., restricted amount of fencing).
4. Substitution: Use the constraint to rewrite the Primary Equation in terms of a single variable.
5. Domain: Identify the feasible interval $[a, b]$ for the variable.
6. Calculus: Find $f'$, set to 0, and find critical points.
7. Verification: Use the First Derivative Test or Candidates Test (if closed interval) to confirm max/min.
8. Answer the Question: Check if it asks for the $x$-value or the resulting max/min dimension.

Summary of Common Mistakes & Pitfalls

1. Forgetting Conditions: You cannot apply MVT, Rolle's, or EVT without stating that the function is continuous (and differentiable for MVT) on the specific interval.
2. Relative vs. Absolute: Finding a local max does not mean you found the absolute max. Always check endpoints on closed intervals.
3. Second Derivative Logic: A common error is thinking $f''(c)=0$ automatically means an inflection point. It is only a candidate; the sign must change.
4. Misinterpreting Graphs: Do not confuse the graph of $f'$ with the graph of $f$. If looking at the graph of $f'$, a maximum represents a point of inflection for $f$, not a max for $f$.
5. Incorrect Inconclusive Test: If the Second Derivative Test yields 0, do not say "there is no extrema." Say "the test fails" and use the First Derivative test instead.