Determining Intervals on Which a Function Is Increasing or Decreasing - AP Calculus AB Study Guide

The Core Idea: Determining Intervals on Which a Function Is Increasing or Decreasing

The central concept of this topic is the fundamental relationship between a function and its first derivative. The first derivative, $f^{\prime }(x)$, represents the slope of the tangent line to the function $f(x)$ at any given point $x$. By analyzing the sign (positive or negative) of the first derivative over an interval, we can determine the behavior of the original function on that same interval.

If the slope of the tangent line is positive ($f^{\prime }(x)>0$), the function is rising, which means it is increasing. Conversely, if the slope of the tangent line is negative ($f^{\prime }(x)<0$), the function is falling, meaning it is decreasing. The points that separate these intervals of increasing and decreasing behavior are called critical points, which are essential for this analysis.

Key Definitions and Theorems

This topic is governed by a direct application of the first derivative to describe the behavior of a function.

The First Derivative Test for Increasing and Decreasing Functions

Let $f$ be a function that is continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b)$.

• If $f^{\prime }(x)>0$ for all $x$ in $(a,b)$, then $f$ is increasing on $[a,b]$.

• If $f^{\prime }(x)<0$ for all $x$ in $(a,b)$, then $f$ is decreasing on $[a,b]$.

Definition of a Critical Point

A critical point of a function $f$ is a point $c$ in the domain of $f$ for which either $f^{\prime }(c)=0$ or $f^{\prime }(c)$ is undefined.

Understanding the Role of Critical Points

The intervals on which a function is strictly increasing or strictly decreasing are separated by its critical points. These are the only points where the first derivative can potentially change its sign from positive to negative or vice versa.

To determine the intervals of increase and decrease for a function $f(x)$, we follow a systematic process:

1. Find the domain of $f(x)$.

2. Calculate the first derivative, $f^{\prime }(x)$.

3. Identify all critical points by finding where $f^{\prime }(x)=0$ and where $f^{\prime }(x)$ is undefined. These points must be within the domain of the original function $f(x)$.

4. Use the critical points to partition the domain of the function into distinct open intervals.

5. Choose a test value within each interval and evaluate the sign of $f^{\prime }(x)$ at that value. The sign of $f^{\prime }(x)$ for that single test value will be the sign for the entire interval.

6. Conclude which intervals correspond to increasing behavior ($f^{\prime }(x)>0$) and which correspond to decreasing behavior ($f^{\prime }(x)<0$).

This process, often organized using a sign chart, provides a complete analysis of the function's increasing and decreasing behavior.

Core Concepts & Rules

• The sign of the first derivative, $f^{\prime }(x)$, determines whether the original function, $f(x)$, is increasing or decreasing.

• A positive derivative ($f^{\prime }(x)>0$) on an interval indicates that $f(x)$ is increasing on that interval.

• A negative derivative ($f^{\prime }(x)<0$) on an interval indicates that $f(x)$ is decreasing on that interval.

• Critical points are the specific $x$-values in the function's domain where the derivative is either zero or undefined.

• Critical points are the boundaries that partition the number line into intervals, over which the sign of $f^{\prime }(x)$ is constant.

Step-by-Step Example 1: Basic Application

Problem: Find the open intervals on which the function $f(x)=2x^3+3x^2-36x+7$ is increasing or decreasing.

Step 1: Find the first derivative.

The function is a polynomial, so it is differentiable everywhere.

$$f^{\prime }(x)=\frac{d}{dx}(2x^3+3x^2-36x+7)$$

$$f^{\prime }(x)=6x^2+6x-36$$

Step 2: Find the critical points.

Set $f^{\prime }(x)=0$ to find where the tangent line is horizontal. Since $f^{\prime }(x)$ is a polynomial, it is never undefined.

$$6x^2+6x-36=0$$

$$6(x^2+x-6)=0$$

$$6(x+3)(x-2)=0$$

The critical points are $x=-3$ and $x=2$.

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

The critical points $x=-3$ and $x=2$ divide the number line into three intervals: $(-\infty ,-3)$, $(-3,2)$, and $(2,\infty )$.

Step 4: Test a value from each interval in $f^{\prime }(x)$.

• Interval $(-\infty ,-3)$: Let's test $x=-4$.

  $f^{\prime }(-4)=6(-4+3)(-4-2)=6(-1)(-6)=36$. The sign is positive ($+$).

• Interval $(-3,2)$: Let's test $x=0$.

  $f^{\prime }(0)=6(0+3)(0-2)=6(3)(-2)=-36$. The sign is negative ($-$).

• Interval $(2,\infty )$: Let's test $x=3$.

  $f^{\prime }(3)=6(3+3)(3-2)=6(6)(1)=36$. The sign is positive ($+$).

Step 5: Write the conclusion with justification.

Based on the sign analysis of $f^{\prime }(x)$:

• $f(x)$ is increasing on the intervals $(-\infty ,-3)$ and $(2,\infty )$ because $f^{\prime }(x)>0$ on these intervals.

• $f(x)$ is decreasing on the interval $(-3,2)$ because $f^{\prime }(x)<0$ on this interval.

Step-by-Step Example 2: Exam-Style Application

Problem: Find the open intervals on which the function $g(x)=\frac{x^2}{x-2}$ is increasing or decreasing.

Step 1: Find the first derivative.

The function is a rational function. Its domain is all real numbers except $x=2$. We use the quotient rule.

$$g^{\prime }(x)=\frac{(2x)(x-2)-(x^2)(1)}{(x-2{)}^2}$$

$$g^{\prime }(x)=\frac{2x^2-4x-x^2}{(x-2{)}^2}$$

$$g^{\prime }(x)=\frac{x^2-4x}{(x-2{)}^2}=\frac{x(x-4)}{(x-2{)}^2}$$

Step 2: Find the critical points.

First, find where $g^{\prime }(x)=0$ by setting the numerator to zero.

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

This gives $x=0$ and $x=4$. Both are in the domain of $g(x)$.

Next, find where $g^{\prime }(x)$ is undefined by setting the denominator to zero.

$$(x-2{)}^2=0$$

This gives $x=2$. However, $x=2$ is not in the domain of the original function $g(x)$, so it is not a critical point. It is a vertical asymptote and must still be used to partition the number line for our analysis.

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

The critical points $x=0,x=4$ and the discontinuity $x=2$ divide the number line into four intervals: $(-\infty ,0)$, $(0,2)$, $(2,4)$, and $(4,\infty )$.

Step 4: Test a value from each interval in $g^{\prime }(x)$.

Note that the denominator $(x-2{)}^2$ is always positive for $x\ne 2$. Therefore, the sign of $g^{\prime }(x)$ is determined entirely by the sign of the numerator, $x(x-4)$.

• Interval $(-\infty ,0)$: Test $x=-1$. Numerator: $(-1)(-1-4)=5$. Sign is positive ($+$).

• Interval $(0,2)$: Test $x=1$. Numerator: $(1)(1-4)=-3$. Sign is negative ($-$).

• Interval $(2,4)$: Test $x=3$. Numerator: $(3)(3-4)=-3$. Sign is negative ($-$).

• Interval $(4,\infty )$: Test $x=5$. Numerator: $(5)(5-4)=5$. Sign is positive ($+$).

Step 5: Write the conclusion with justification.

• $g(x)$ is increasing on $(-\infty ,0)$ and $(4,\infty )$ because $g^{\prime }(x)>0$ on these intervals.

• $g(x)$ is decreasing on $(0,2)$ and $(2,4)$ because $g^{\prime }(x)<0$ on these intervals.

Using Your Calculator

For problems where you are permitted to use a graphing calculator, you can determine intervals of increase and decrease by analyzing the graph of the first derivative. The calculator is primarily a tool for verification or for problems with functions whose derivatives are very complex to compute by hand.

Procedure:

1. Given a function $f(x)$, you can graph its derivative, f'(x)`. 2. In the `Y=` editor, enter the derivative into `Y1`. You can either calculate the derivative by hand and enter its formula, or use the calculator's numerical derivative function. For $f(x) = 2x^3 + 3x^2 - 36x + 7, you would enter:

   Y1 = 6X^2 + 6X - 36

   or

   Y1 = nDeriv(2X^3 + 3X^2 - 36X + 7, X, X)

2. Press GRAPH to view the graph of $f^{\prime }(x)$. Adjust the window as needed.

3. Identify the intervals where the graph of Y1 is above the x-axis. This is where $f^{\prime }(x)>0$, so $f(x)$ is increasing.

4. Identify the intervals where the graph of Y1 is below the x-axis. This is where $f^{\prime }(x)<0$, so $f(x)$ is decreasing.

5. Use the CALC menu (2nd + TRACE) and select 2:zero to find the x-intercepts of the graph of $f^{\prime }(x)$. These are the critical points where $f^{\prime }(x)=0$.

This graphical method provides a visual confirmation of the analytical work done with a sign chart.

AP Exam Quick Hit

Common Question Types

• Analytical Function: You will be given an equation for a function $f(x)$ and asked to find and justify the intervals on which it is increasing or decreasing. This is the most common type and mirrors the examples above.

• Graph of the Derivative: You will be given the graph of $f^{\prime }(x)$ and asked to determine the intervals where the original function $f(x)$ is increasing.

  • Example: "The graph of $f^{\prime }(x)$ is shown above. On which intervals is $f(x)$ decreasing?" You must identify the intervals where the provided graph is below the x-axis.

• Table of Derivative Values: You may be given a table of selected values for $f^{\prime }(x)$ and asked about the behavior of $f(x)$.

  • Example: "A table of values for $f^{\prime }(x)$ is given. If $f^{\prime }(x)>0$ for $0<x<4$, what can be concluded about $f(x)$ on this interval?" The conclusion is that $f(x)$ is increasing on that interval.

Common Mistakes

• Confusing $f(x)$ and $f^{\prime }(x)$: When given the graph of $f^{\prime }(x)$, students mistakenly look at where the graph of $f^{\prime }(x)$ is increasing or decreasing. You must look at the sign (y-values) of $f^{\prime }(x)$—whether it is above or below the x-axis—to determine the behavior of $f(x)$.

• Forgetting Undefined Derivatives: Students often find critical points by setting $f^{\prime }(x)=0$ but forget to check for x-values where $f^{\prime }(x)$ is undefined (e.g., denominators of zero or negative numbers inside even roots).

• Incomplete Justification: On free-response questions, simply stating the intervals is not enough. You must provide a justification that connects the sign of the first derivative to the behavior of the function. For example: "$f(x)$ is increasing on $(a,b)$ because $f^{\prime }(x)>0$ on this interval."

• Algebraic Errors: Simple mistakes in differentiation (e.g., product/quotient rule) or in solving $f^{\prime }(x)=0$ will lead to incorrect critical points and an entirely incorrect analysis.

• Ignoring the Domain: A critical point must be in the domain of the original function $f(x)$. Points of discontinuity, like vertical asymptotes, must be included on the sign chart but are not critical points.