Using the First Derivative Test to Determine Relative (Local) Extrema - AP Calculus BC Study Guide

The Core Idea: Using the First Derivative Test to Determine Relative (Local) Extrema

The first derivative of a function, $f^{\prime }(x)$, describes the slope of the tangent line to the graph of the original function, $f(x)$. Where $f^{\prime }(x)$ is positive, $f(x)$ is increasing; where $f^{\prime }(x)$ is negative, $f(x)$ is decreasing. Relative (or local) extrema—the "peaks" and "valleys" of a function—occur at points where the function's behavior changes from increasing to decreasing, or vice versa.

The First Derivative Test is a formal method for identifying these extrema. It leverages the fact that such changes can only happen at critical points, which are points where the derivative is either zero (a horizontal tangent) or undefined (a sharp corner, cusp, or vertical tangent). By analyzing the sign (positive or negative) of the first derivative on either side of a critical point, we can determine if that point corresponds to a relative maximum, a relative minimum, or neither.

Key Definitions and The First Derivative Test

The process of finding relative extrema is built upon two core concepts: the definition of a critical point and the First Derivative Test itself.

Definition: Critical Point

A function $f$ has a critical point at $x=c$ if $f^{\prime }(c)=0$ or if $f^{\prime }(c)$ is undefined. Critical points are the only candidates for the location of relative extrema.

The First Derivative Test

Let $c$ be a critical point of a function $f$. The test analyzes the sign of $f^{\prime }(x)$ as $x$ passes through $c$.

1. Relative Maximum: If $f^{\prime }(x)$ changes from positive to negative at $x=c$, then $f$ has a relative maximum at $x=c$.

   • Conceptual link: The function $f$ was increasing ($f^{\prime }>0$) and then switched to decreasing ($f^{\prime }<0$), creating a "peak."

2. Relative Minimum: If $f^{\prime }(x)$ changes from negative to positive at $x=c$, then $f$ has a relative minimum at $x=c$.

   • Conceptual link: The function $f$ was decreasing ($f^{\prime }<0$) and then switched to increasing ($f^{\prime }>0$), creating a "valley."

3. Neither a Maximum nor a Minimum: If $f^{\prime }(x)$does not change sign at $x=c$ (i.e., it is positive on both sides or negative on both sides), then $f$ has neither a relative maximum nor a relative minimum at $x=c$.

   • Conceptual link: The function was increasing, paused, and then continued increasing, or was decreasing, paused, and continued decreasing.

Understanding Critical Points and Sign Changes

The power of the First Derivative Test lies in its systematic approach. The crucial first step is to identify all critical points. A common mistake is to only find where $f^{\prime }(x)=0$ and to ignore points where $f^{\prime }(x)$ is undefined. Functions with cusps (like $f(x)=x^{2/3}$) or corners (like $f(x)=|x|$) have relative extrema at points where their derivatives are undefined.

Once critical points are found, they are used to partition the number line into intervals. The sign of $f^{\prime }(x)$ is constant throughout each of these intervals. By picking a single test value within each interval and evaluating its sign in the derivative, we can determine the behavior of $f(x)$ across its entire domain. The classification of an extremum depends entirely on the change in sign as you move from the interval on the left of a critical point to the interval on the right. No sign change means no relative extremum.

Core Concepts & Rules

• A critical point of a function $f$ exists at $x=c$ if $f^{\prime }(c)=0$ or $f^{\prime }(c)$ is undefined.

• Relative extrema can only occur at critical points.

• The First Derivative Test determines if a critical point is a relative maximum, relative minimum, or neither by analyzing the sign of $f^{\prime }(x)$ around that point.

• Relative Maximum: Occurs at critical point $c$ if the sign of $f^{\prime }(x)$ changes from positive ($+$) to negative ($-$).

• Relative Minimum: Occurs at critical point $c$ if the sign of $f^{\prime }(x)$ changes from negative ($-$) to positive ($+$).

• Neither: Occurs at critical point $c$ if the sign of $f^{\prime }(x)$ does not change (e.g., $+$ to $+$ or $-$ to $-$).

Step-by-Step Example 1: Function with a Smooth Turn

Find and classify all relative extrema for the function $f(x)=x^3-6x^2+5$.

Step 1: Find the derivative.

The derivative gives us information about the slope of $f(x)$.

$$f^{\prime }(x)=3x^2-12x$$

Step 2: Find all critical points.

Critical points occur where $f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined. Since $f^{\prime }(x)$ is a polynomial, it is defined for all real numbers. We only need to set $f^{\prime }(x)=0$.

$$3x^2-12x=0$$

$$3x(x-4)=0$$

The critical points are $x=0$ and $x=4$.

Step 3: Create a sign chart for $f^{\prime }(x)$.

Place the critical points on a number line. These points divide the line into three intervals: $(-\infty ,0)$, $(0,4)$, and $(4,\infty )$.

Step 4: Test the sign of $f^{\prime }(x)$ in each interval.

• Interval $(-\infty ,0)$: Choose a test value, e.g., $x=-1$.

  $f^{\prime }(-1)=3(-1)(-1-4)=(-3)(-5)=15$. The sign is positive (+).

• Interval $(0,4)$: Choose a test value, e.g., $x=1$.

  $f^{\prime }(1)=3(1)(1-4)=(3)(-3)=-9$. The sign is negative (-).

• Interval $(4,\infty )$: Choose a test value, e.g., $x=5$.

  $f^{\prime }(5)=3(5)(5-4)=(15)(1)=15$. The sign is positive (+).

Step 5: Apply the First Derivative Test and state the conclusion.

• At $x=0$, the sign of $f^{\prime }(x)$ changes from positive to negative. Therefore, $f(x)$ has a relative maximum at $x=0$.

• At $x=4$, the sign of $f^{\prime }(x)$ changes from negative to positive. Therefore, $f(x)$ has a relative minimum at $x=4$.

Step-by-Step Example 2: Function with an Undefined Derivative

Find and classify all relative extrema for the function $f(x)=(x+2)x^{2/3}$.

Step 1: Find the derivative.

Use the product rule: $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$.

$$f^{\prime }(x)=(1)(x^{2/3})+(x+2)\left(\frac{2}{3}{x}^{-1/3}\right)$$

$$f^{\prime }(x)=x^{2/3}+\frac{2(x+2)}{3x^{1/3}}$$

To analyze the sign, combine into a single fraction with a common denominator, $3x^{1/3}$.

$$f^{\prime }(x)=\frac{3x^{1/3}\cdot x^{2/3}}{3x^{1/3}}+\frac{2x+4}{3x^{1/3}}=\frac{3x+2x+4}{3x^{1/3}}=\frac{5x+4}{3x^{1/3}}$$

Step 2: Find all critical points.

• Find where $f^{\prime }(x)=0$: This occurs when the numerator is zero.

  $5x+4=0⟹x=-4/5$

• Find where $f^{\prime }(x)$ is undefined: This occurs when the denominator is zero.

  $3x^{1/3}=0⟹x=0$

The critical points are $x=-4/5$ and $x=0$.

Step 3: Create a sign chart for $f^{\prime }(x)$.

Place the critical points $x=-4/5$ and $x=0$ on a number line. This creates three intervals: $(-\infty ,-4/5)$, $(-4/5,0)$, and $(0,\infty )$.

Step 4: Test the sign of $f^{\prime }(x)=\frac{5x+4}{3x^{1/3}}$ in each interval.

• Interval $(-\infty ,-4/5)$: Choose $x=-1$.

  $f^{\prime }(-1)=\frac{5(-1)+4}{3(-1{)}^{1/3}}=\frac{-1}{-3}$. The sign is positive (+).

• Interval $(-4/5,0)$: Choose $x=-1/5$.

  $f^{\prime }(-1/5)=\frac{5(-1/5)+4}{3(-1/5{)}^{1/3}}=\frac{3}{\text{negative}}$. The sign is negative (-).

• Interval $(0,\infty )$: Choose $x=1$.

  $f^{\prime }(1)=\frac{5(1)+4}{3(1{)}^{1/3}}=\frac{9}{3}$. The sign is positive (+).

Step 5: Apply the First Derivative Test and state the conclusion.

• At $x=-4/5$, the sign of $f^{\prime }(x)$ changes from positive to negative. Therefore, $f(x)$ has a relative maximum at $x=-4/5$.

• At $x=0$, the sign of $f^{\prime }(x)$ changes from negative to positive. Therefore, $f(x)$ has a relative minimum at $x=0$.

Using Your Calculator

The First Derivative Test is an analytical method. Your justification on the AP Exam must be based on the sign change of the derivative, not on a picture from your calculator. However, a graphing calculator can be a powerful tool for verification and exploration, especially on calculator-active questions.

To analyze $f(x)$ using its derivative $f^{\prime }(x)$:

1. Graph the Derivative: Input the expression for $f^{\prime }(x)$ into your calculator's graphing utility (e.g., into Y1).

2. Find Critical Points:

   • Use the $zero$ or $root$ function on the graph of $f^{\prime }(x)$ to find the x-intercepts. These are the critical points where $f^{\prime }(x)=0$.

   • Visually inspect the graph of $f^{\prime }(x)$ for discontinuities or vertical asymptotes. These may correspond to critical points where $f^{\prime }(x)$ is undefined.

3. Analyze the Sign of $f^{\prime }(x)$:

   • Where the graph of $f^{\prime }(x)$ is above the x-axis, $f^{\prime }(x)$ is positive.

   • Where the graph of $f^{\prime }(x)$ is below the x-axis, $f^{\prime }(x)$ is negative.

4. Apply the First Derivative Test:

   • A relative maximum for $f(x)$ occurs where the graph of $f^{\prime }(x)$ crosses the x-axis from above to below (positive to negative).

   • A relative minimum for $f(x)$ occurs where the graph of $f^{\prime }(x)$ crosses the x-axis from below to above (negative to positive).

AP Exam Quick Hit

Common Question Types

• Analytical Function Analysis: Given a function $f(x)$ (often involving products, quotients, or chain rule), you will be asked to find the x-coordinates of any relative extrema and provide a justification.

  • Example: "For the function $f(x)=x^2{e}^{-x}$, find all values of $x$ at which $f$ has a relative minimum. Justify your answer."

• Analysis of the Graph of $f^{\prime }(x)$: You will be given the graph of the derivative, $f^{\prime }(x)$, and asked to determine properties of the original function, $f(x)$.

  • Example: "The graph of $f^{\prime }$, the derivative of a function $f$, is shown above. At which of the following x-values does $f$ have a relative maximum? (A) a (B) b (C) c (D) d" (The correct answer would be where the graph of $f^{\prime }$ crosses the x-axis from positive to negative).

Common Mistakes

• Incorrect Justification: Stating that $f(x)$ has a maximum because $f^{\prime }(x)=0$. This is insufficient; you must state that $f^{\prime }(x)$ changes sign from positive to negative.

• Confusing $f$ and $f^{\prime }$: Writing justifications like "$f(x)$ changes from positive to negative," when you mean the derivative $f^{\prime }(x)$ changes sign. The sign of $f(x)$ itself tells you if the graph is above or below the x-axis, which is irrelevant to this test.

• Ignoring Undefined Derivatives: Forgetting to find critical points where the derivative is undefined. This is a common feature in functions involving fractional exponents or absolute values.

• Stopping After Finding Critical Points: Finding the correct critical points is only the first step. The analysis of the sign change of $f^{\prime }(x)$ around those points is required to classify them.

• Using the Calculator as Justification: Writing "the function has a minimum at $x=2$ because the calculator's graph shows a valley there" will not earn credit. The justification must be based on calculus principles, specifically the sign change of the derivative.