Connecting a Function, Its First Derivative, and Its Second Derivative - AP Calculus BC Study Guide

The Core Idea: Connecting a Function, Its First Derivative, and Its Second Derivative

The fundamental goal of this topic is to understand the deep relationship between a function, $f(x)$, and its first and second derivatives, $f^{\prime }(x)$ and $f^{\prime \prime }(x)$. By analyzing the signs and behavior of the derivatives, we can construct a detailed picture of the original function's graph without necessarily plotting every point. The first derivative, $f^{\prime }(x)$, reveals where the function is increasing or decreasing and allows us to locate and classify local (or relative) extrema—the peaks and valleys of the graph.

The second derivative, $f^{\prime \prime }(x)$, describes the function's concavity—whether the graph is shaped like a cup opening upwards (concave up) or a cap opening downwards (concave down). This is directly related to how the slope of the function is changing. The second derivative also helps us identify points of inflection, which are specific points on the graph where the concavity changes. By synthesizing information from both derivatives, we can accurately describe the shape and key features of the original function's graph.

Key Theorems and Tests

The First Derivative Test

This test uses the sign of the first derivative, $f^{\prime }(x)$, to classify critical points of a function $f(x)$ as local maxima, local minima, or neither. A critical point $c$ is a point in the domain of $f$ where $f^{\prime }(c)=0$ or $f^{\prime }(c)$ is undefined.

Let $c$ be a critical point of a continuous function $f$.

• If $f^{\prime }(x)$ changes from positive to negative at $x=c$, then $f$ has a local maximum at $x=c$.

• If $f^{\prime }(x)$ changes from negative to positive at $x=c$, then $f$ has a local minimum at $x=c$.

• If $f^{\prime }(x)$does not change sign at $x=c$, then $f$ has neither a local maximum nor a local minimum at $x=c$.

The Second Derivative Test

This test uses the sign of the second derivative at a critical point to classify it as a local maximum or minimum. This test is often faster than the First Derivative Test but is less universally applicable.

Let $f$ be a twice-differentiable function such that $f^{\prime }(c)=0$.

• If $f^{\prime \prime }(c)>0$, then $f$ has a local minimum at $x=c$.

• If $f^{\prime \prime }(c)<0$, then $f$ has a local maximum at $x=c$.

• If $f^{\prime \prime }(c)=0$, the test is inconclusive. You must use the First Derivative Test to classify the critical point.

Concavity and Points of Inflection

The second derivative provides direct information about the concavity of the graph of $f(x)$. A point of inflection is a point on the graph where the concavity changes.

• If $f^{\prime \prime }(x)>0$ on an interval, then the graph of $f$ is concave up on that interval.

• If $f^{\prime \prime }(x)<0$ on an interval, then the graph of $f$ is concave down on that interval.

• If $f(x)$ has a point of inflection at $x=c$, then $f^{\prime \prime }(c)=0$ or $f^{\prime \prime }(c)$ is undefined. Note that this condition identifies candidates for points of inflection; you must confirm that the sign of $f^{\prime \prime }(x)$ actually changes at $x=c$.

Understanding the Connections

The rules for derivatives are not arbitrary; they describe geometric relationships between the graphs of $f$, $f^{\prime }$, and $f^{\prime \prime }$.

The most crucial connection is that the derivative of a function describes the rate of change (or slope) of that function. Therefore, $f^{\prime \prime }(x)$ is the derivative of $f^{\prime }(x)$. This single fact is the key to understanding concavity.

• Why Concavity Relates to $f^{\prime \prime }$:

  • The graph of $f$ is concave up when its slope is increasing. The slope of $f$ is given by $f^{\prime }(x)$. For $f^{\prime }(x)$ to be an increasing function, its derivative must be positive. The derivative of $f^{\prime }(x)$ is $f^{\prime \prime }(x)$. Therefore, $f(x)$ is concave up where $f^{\prime \prime }(x)>0$.

  • The graph of $f$ is concave down when its slope is decreasing. For $f^{\prime }(x)$ to be a decreasing function, its derivative must be negative. Therefore, $f(x)$ is concave down where $f^{\prime \prime }(x)<0$.

• Why the Second Derivative Test Works:

  • Imagine a point $c$ where the tangent line is horizontal ($f^{\prime }(c)=0$).

  • If the function is also concave up at that point ($f^{\prime \prime }(c)>0$), the graph is shaped like the bottom of a cup. A horizontal tangent at the bottom of a cup must be a local minimum.

  • If the function is concave down at that point ($f^{\prime \prime }(c)<0$), the graph is shaped like the top of a cap. A horizontal tangent at the top of a cap must be a local maximum.

  • If $f^{\prime \prime }(c)=0$, the graph is neither definitively cup-shaped nor cap-shaped at that point (it might be flattening out), so the test is inconclusive.

Core Concepts & Rules

• Local Extrema: Local (relative) maxima and minima can only occur at critical points, where $f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined.

• First Derivative Test: The key to this test is a sign change in $f^{\prime }(x)$ at a critical point. No sign change means no local extremum.

• Second Derivative Test: This test uses the sign of $f^{\prime \prime }(x)$ at a critical point where $f^{\prime }(x)=0$. It is a test for concavity at a single point with a horizontal tangent.

• Inconclusive Second Derivative Test: If $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)=0$, the Second Derivative Test fails. You must revert to the First Derivative Test to determine if $c$ is a local extremum.

• Concavity: The sign of $f^{\prime \prime }(x)$ over an interval determines the concavity of $f(x)$ on that interval.

  • $f^{\prime \prime }(x)>0⟹f^{\prime }(x)$ is increasing $⟹f(x)$ is concave up.

  • $f^{\prime \prime }(x)<0⟹f^{\prime }(x)$ is decreasing $⟹f(x)$ is concave down.

• Points of Inflection: A point of inflection is a point on the graph of $f(x)$ where the concavity changes. This requires a sign change in $f^{\prime \prime }(x)$. Candidates for points of inflection occur where $f^{\prime \prime }(x)=0$ or $f^{\prime \prime }(x)$ is undefined.

Step-by-Step Example 1: Analyzing a Polynomial Function

Problem: For the function $f(x)=x^4-4x^3$, find all local extrema and points of inflection. Justify your answers.

Step 1: Find the first and second derivatives.

$$f^{\prime }(x)=4x^3-12x^2$$

$$f^{\prime \prime }(x)=12x^2-24x$$

Step 2: Find critical points.

Set $f^{\prime }(x)=0$ and solve for $x$.

$$4x^3-12x^2=0$$

$$4x^2(x-3)=0$$

The critical points are $x=0$ and $x=3$.

Step 3: Classify critical points using the First Derivative Test.

Create a sign chart for $f^{\prime }(x)$ using the critical points $x=0$ and $x=3$.

• Test $x=-1$: $f^{\prime }(-1)=4(-1{)}^2(-1-3)=4(1)(-4)=-16$ (Negative)

• Test $x=1$: $f^{\prime }(1)=4(1{)}^2(1-3)=4(1)(-2)=-8$ (Negative)

• Test $x=4$: $f^{\prime }(4)=4(4{)}^2(4-3)=4(16)(1)=64$ (Positive)

• At $x=0$, $f^{\prime }(x)$ does not change sign. Therefore, there is neither a local maximum nor a local minimum at $x=0$.

• At $x=3$, $f^{\prime }(x)$ changes from negative to positive. Therefore, $f(x)$ has a local minimum at $x=3$.

Step 4: Find potential points of inflection.

Set $f^{\prime \prime }(x)=0$ and solve for $x$.

$$12x^2-24x=0$$

$$12x(x-2)=0$$

The potential points of inflection are at $x=0$ and $x=2$.

Step 5: Determine intervals of concavity and confirm points of inflection.

Create a sign chart for $f^{\prime \prime }(x)$ using the potential inflection points $x=0$ and $x=2$.

• Test $x=-1$: $f^{\prime \prime }(-1)=12(-1)(-1-2)=(-12)(-3)=36$ (Positive)

• Test $x=1$: $f^{\prime \prime }(1)=12(1)(1-2)=(12)(-1)=-12$ (Negative)

• Test $x=3$: $f^{\prime \prime }(3)=12(3)(3-2)=(36)(1)=36$ (Positive)

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

• At $x=2$, $f^{\prime \prime }(x)$ changes sign (from negative to positive). Therefore, $f(x)$ has a point of inflection at $x=2$.

Summary:

• Local Minimum: at $x=3$ because $f^{\prime }(x)$ changes from negative to positive there.

• Points of Inflection: at $x=0$ and $x=2$ because $f^{\prime \prime }(x)$ changes sign at these points.

Step-by-Step Example 2: Analysis from a Graph of the Derivative

Problem: The figure below shows the graph of $g^{\prime }(x)$, the derivative of a twice-differentiable function $g(x)$, on the interval $[-3,4]$.

(Imagine a graph of $g^{\prime }(x)$ that starts at $(-3,-2)$, increases to a local max at $(-1,2)$, decreases, crossing the x-axis at $x=1$, to a local min at $(2,-1)$, and then increases, ending at $(4,1)$.)

(a) At what values of $x$ does $g(x)$ have a local minimum? Justify your answer.

(b) On what open intervals is the graph of $g(x)$ concave down? Justify your answer.

(c) At what values of $x$ does the graph of $g(x)$ have a point of inflection? Justify your answer.

Solution:

(a) Local Minimum for $g(x)$

• Analysis: A local minimum for $g(x)$ occurs at a critical point where $g^{\prime }(x)$ changes from negative to positive.

• Applying to the graph: We need to find where the graph of $g^{\prime }(x)$ crosses the x-axis from below to above. The graph shows that $g^{\prime }(x)<0$ for $x<1$ (in the vicinity of 1) and $g^{\prime }(x)>0$ for $x>1$. The sign change from negative to positive occurs at $x=1$.

• Answer: The function $g(x)$ has a local minimum at $x=1$ because $g^{\prime }(x)$ changes from negative to positive at that point.

(b) Concavity of $g(x)$

• Analysis: The graph of $g(x)$ is concave down on intervals where $g^{\prime \prime }(x)<0$. Since $g^{\prime \prime }(x)$ is the derivative of $g^{\prime }(x)$, this is equivalent to finding where $g^{\prime }(x)$ is a decreasing function.

• Applying to the graph: We look for the intervals where the graph of $g^{\prime }(x)$ has a negative slope. The graph is decreasing from its local maximum at $x=-1$ to its local minimum at $x=2$.

• Answer: The graph of $g(x)$ is concave down on the interval $(-1,2)$ because $g^{\prime }(x)$ is decreasing on that interval.

(c) Points of Inflection for $g(x)$

• Analysis: A point of inflection for $g(x)$ occurs where the concavity changes. This happens when $g^{\prime \prime }(x)$ changes sign, which is equivalent to $g^{\prime }(x)$ changing from increasing to decreasing or vice versa.

• Applying to the graph: We look for the points where the slope of $g^{\prime }(x)$ changes sign. These are the locations of the local extrema on the graph of $g^{\prime }(x)$. The graph of $g^{\prime }(x)$ changes from increasing to decreasing at $x=-1$, and it changes from decreasing to increasing at $x=2$.

• Answer: The graph of $g(x)$ has points of inflection at $x=-1$ and $x=2$ because $g^{\prime }(x)$ changes from increasing to decreasing at $x=-1$ and from decreasing to increasing at $x=2$.

Using Your Calculator

While the justifications required on the AP Exam are analytical, a graphing calculator is a powerful tool for exploring function behavior and confirming your results.

Problem: Analyze $f(x)=x^3{e}^{-x}$ for local extrema and points of inflection.

Steps on a TI-84 Style Calculator:

1. Enter the functions:

   • In Y1, enter the original function: Y1 = X^3 * e^(-X)

   • In Y2, enter the first derivative: Y2 = nDeriv(Y1, X, X)

   • In Y3, enter the second derivative: Y3 = nDeriv(Y2, X, X) or nDeriv(Y1, X, X, 2)

   Pro-Tip: Deselect Y1 by moving the cursor over the = sign and pressing ENTER. This will let you see the derivatives more clearly.

2. Find Critical Points:

   • Graph Y2 (the first derivative).

   • Use the 2nd -> TRACE (CALC) menu and select 2:zero.

   • Find the x-intercepts of Y2. You will find a zero at $x=0$ and $x=3$. These are your critical points.

   • Observe the graph of Y2: it is positive, then touches the x-axis at $x=0$ (but stays positive), and then crosses from positive to negative at $x=3$. This confirms a local maximum at $x=3$ and no extremum at $x=0$.

3. Find Potential Points of Inflection and Concavity:

   • Now, graph Y3 (the second derivative). Deselect Y2.

   • Use the $CALC$ -> 2:zero function to find the x-intercepts of Y3. You will find zeros at`x=0$, $x=3-\sqrt{3}$ (approx. 1.268), and $x=3+\sqrt{3}$ (approx. 4.732).

   • Observe the graph of Y3:

     • It is negative for $x<0$. ($f$ is concave down)

     • It changes to positive at $x=0$. ($f$ is concave up)

     • It changes to negative at $x=3-\sqrt{3}$. ($f$ is concave down)

     • It changes to positive at $x=3+\sqrt{3}$. ($f$ is concave up)

   • This confirms that points of inflection exist at all three x-values where $f^{\prime \prime }(x)=0$ because the sign of $f^{\prime \prime }(x)$ changes at each one.

AP Exam Quick Hit

Common Question Types

• Full Analysis of a Given Function: Given a function $f(x)=\frac{\ln x}{x}$, find the coordinates of any local extrema and points of inflection. You must show the derivative calculations and use a test (First or Second Derivative) to justify your claims.

• Analysis from the Graph of $f^{\prime }(x)$: Given the graph of the derivative $f^{\prime }(x)$, identify the x-values for local minima of $f(x)$ (where $f^{\prime }$ crosses the x-axis from negative to positive) and the x-values for points of inflection of $f(x)$$ (where $f^{\prime }$ has local extrema).

• Analysis from a Table of Values: Given a table of selected values for a twice-differentiable function $f(x)$ and its derivatives, determine if $f(x)$ must have a local extremum or point of inflection on a given interval. For example, if $f^{\prime }(2)=0$ and $f^{\prime \prime }(2)=-5$, you can conclude $f$ has a local maximum at $x=2$.

Common Mistakes

• Confusing the Function and its Derivatives: A very common error is to mix up the properties. For example, stating "$f(x)$ is concave up because $f^{\prime }(x)$ is positive." The correct statement is "$f(x)$ is concave up because $f^{\prime \prime }(x)$ is positive" or "$f(x)$ is concave up because $f^{\prime }(x)$ is increasing."

• Insufficient Justification for Extrema: Stating "$f(x)$ has a local minimum at $x=c$ because $f^{\prime }(c)=0$." This is not enough. The justification must be based on a sign change in $f^{\prime }(x)$ (First Derivative Test) or the sign of $f^{\prime \prime }(c)$ (Second Derivative Test).

• Misinterpreting $f^{\prime \prime }(c)=0$: Incorrectly concluding that if $f^{\prime \prime }(c)=0$, there must be a point of inflection at $x=c$. This condition only identifies a candidate. You must show that the sign of $f^{\prime \prime }(x)$ changes at $x=c$ to confirm it is a point of inflection.

• Misapplying the Second Derivative Test: Forgetting that the Second Derivative Test is inconclusive if $f^{\prime \prime }(c)=0$. In this case, you cannot conclude anything about extrema from the second derivative and must use the First Derivative Test.

• Confusing Slopes with y-values on a Derivative Graph: When looking at the graph of $f^{\prime }(x)$, the concavity of $f(x)$ is determined by the slope of the $f^{\prime }(x)$ graph, not by whether the $f^{\prime }(x)$ graph is above or below the x-axis.