Change in tandem (function behavior) — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Covers average and instantaneous rates of change, increasing/decreasing behavior, local extrema, concavity, and inflection points for polynomial and rational functions, relating changes in input to corresponding changes in output.

You should already know: Limits of polynomial and rational functions, basic derivative rules for power functions, how to factor and find roots of polynomials.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Change in tandem (function behavior)?

Change in tandem (also referred to as co-variation of function behavior) is the core study of how the output value of a polynomial or rational function changes in lockstep with changes to its input value, connecting algebraic, graphical, and numerical descriptions of function behavior. Per the AP Precalculus Course and Exam Description, this topic is part of Unit 1: Polynomial and Rational Functions, which accounts for 14-16% of the total AP exam score, with change in tandem itself making up approximately 4-6% of the total exam. This topic is tested in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam, often as part of multi-step questions that ask you to analyze function behavior or interpret change in real-world contexts. Unlike basic graphing, change in tandem focuses on quantifying and describing how change in one variable drives change in the other, building the intuitive foundation for key calculus concepts you will encounter later in the course.

2. Average and Instantaneous Rates of Change

The foundation of analyzing change in tandem is quantifying how much output changes for a given change in input. The average rate of change (AROC) of a function $f(x)$ over the interval $[a,b]$ is the total change in output divided by the total change in input: $$\text{AROC}=\frac{f(b)-f(a)}{b-a}$$ AROC equals the slope of the secant line connecting the points $(a,f(a))$ and $(b,f(b))$, and it describes the average behavior of the function over the entire interval.

The instantaneous rate of change (IROC) at a point $x=a$ is the rate of change at that exact input value, defined as the limit of AROC as the interval width approaches zero: $$\text{IROC}=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}=f^{\prime }(a)$$ For all polynomials and all points on rational functions that are not at vertical asymptotes, $f(x)$ is differentiable, so IROC exists and can be calculated directly with basic derivative rules.

Worked Example

For $f(x)=2x^3-6x+1$, find (a) the average rate of change over $[1,3]$, and (b) the instantaneous rate of change at $x=2$.

1. First calculate the function values at the endpoints of the interval: $f(1)=2(1{)}^3-6(1)+1=-3$, and $f(3)=2(3{)}^3-6(3)+1=37$.
2. Apply the AROC formula: $\text{AROC}=\frac{f(3)-f(1)}{3-1}=\frac{37-(-3)}{2}=\frac{40}{2}=20$.
3. For IROC, compute the first derivative using the power rule: $f^{\prime }(x)=6x^2-6$.
4. Evaluate the derivative at $x=2$: $f^{\prime }(2)=6(2{)}^2-6=24-6=18$.

Final results: AROC = 20, IROC = 18.

Exam tip: On AP Precalculus MCQ, if asked to estimate IROC from a table of values, always use the AROC of the smallest interval centered at the point of interest, which is almost always the correct answer.

3. Increasing/Decreasing Behavior and Local Extrema

We can use the sign of the instantaneous rate of change (first derivative) to describe whether a function is growing or shrinking over an interval. A function $f(x)$ is increasing on an interval $I$ if for any $x_1<x_2$ in $I$, $f(x_1)<f(x_2)$, which corresponds to $f^{\prime }(x)>0$ for all $x$ in $I$. A function is decreasing on $I$ if for any $x_1<x_2$ in $I$, $f(x_1)>f(x_2)$, which corresponds to $f^{\prime }(x)<0$ for all $x$ in $I$.

Local extrema are points where the function changes from increasing to decreasing (local maximum) or decreasing to increasing (local minimum). These occur at critical points, where $f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined, but the function must be defined at the point to count as an extremum. For a degree $n$ polynomial, the maximum number of local extrema (turning points) is $n-1$.

Worked Example

Find all intervals where $f(x)=x^3-3x^2-9x+10$ is increasing or decreasing, and identify all local extrema.

1. Compute the first derivative and factor it: $f^{\prime }(x)=3x^2-6x-9=3(x-3)(x+1)$.
2. Critical points occur where $f^{\prime }(x)=0$, so $x=-1$ and $x=3$. These split the number line into three intervals: $(-\infty ,-1)$, $(-1,3)$, $(3,\infty )$.
3. Test the sign of $f^{\prime }(x)$ in each interval:
   • $(-\infty ,-1)$: $f^{\prime }(-2)=3(-5)(-1)=15>0\to$ increasing
   • $(-1,3)$: $f^{\prime }(0)=3(-3)(1)=-9<0\to$ decreasing
   • $(3,\infty )$: $f^{\prime }(4)=3(1)(5)=15>0\to$ increasing
4. Apply the first derivative test: $f$ changes from increasing to decreasing at $x=-1$, so $f(-1)=15$ is a local maximum. $f$ changes from decreasing to increasing at $x=3$, so $f(3)=-17$ is a local minimum.

Exam tip: When identifying intervals of increase/decrease for rational functions, always exclude points at vertical asymptotes, as the function is not defined there and cannot have behavior on those points.

4. Concavity and Inflection Points

Concavity describes how the rate of change itself changes as input changes, so it is change in the rate of change, a core concept for change in tandem. A function is concave up on an interval $I$ if $f^{\prime \prime }(x)>0$, meaning the slope of the tangent line increases as $x$ increases (the graph curves upward). A function is concave down on $I$ if $f^{\prime \prime }(x)<0$, meaning the slope of the tangent line decreases as $x$ increases (the graph curves downward).

An inflection point is a point on the graph where concavity changes from up to down or down to up. For a point to be an inflection point, $f(x)$ must be defined at the point, and the sign of $f^{\prime \prime }(x)$ must change around the point. A zero of $f^{\prime \prime }(x)$ is not automatically an inflection point. For a degree $n$ polynomial, the maximum number of inflection points is $n-2$.

Worked Example

For $f(x)=x^4-4x^3+10$, find all intervals of concavity and all inflection points.

1. Compute the first and second derivatives: $f^{\prime }(x)=4x^3-12x^2$, $f^{\prime \prime }(x)=12x^2-24x=12x(x-2)$.
2. Find candidate inflection points where $f^{\prime \prime }(x)=0$: $x=0$ and $x=2$. $f(x)$ is defined at both points.
3. Test the sign of $f^{\prime \prime }(x)$ on intervals split by the candidates:
   • $(-\infty ,0)$: $f^{\prime \prime }(-1)=12(-1)(-3)=36>0\to$ concave up
   • $(0,2)$: $f^{\prime \prime }(1)=12(1)(-1)=-12<0\to$ concave down
   • $(2,\infty )$: $f^{\prime \prime }(3)=12(3)(1)=36>0\to$ concave up
4. Confirm sign change at both candidates, so both are inflection points. Calculate their $y$-coordinates: $f(0)=10$, $f(2)=-6$.

Final results: Concave up on $(-\infty ,0)\cup (2,\infty )$, concave down on $(0,2)$, inflection points at $(0,10)$ and $(2,-6)$.

Exam tip: On FRQ, you will not get full credit for identifying an inflection point just by showing $f^{\prime \prime }(x)=0$; you must explicitly confirm that the sign of $f^{\prime \prime }$ changes around the candidate point.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming $x=0$ is an inflection point for $f(x)=x^4$ because $f^{\prime \prime }(0)=0$. Why: Students assume any zero of the second derivative is automatically an inflection point, without checking for a sign change. For $f(x)=x^4$, $f^{\prime \prime }(x)=12x^2$ which is positive on both sides of $x=0$, so concavity does not change. Correct move: Always test the sign of the second derivative on both sides of a candidate inflection point to confirm a concavity change.
• Wrong move: Claiming $f(x)=\frac{1}{x-2}$ is increasing on the entire domain $(-\infty ,2)\cup (2,\infty )$, so you write "increasing on $(-\infty ,2)\cup (2,\infty )$ as a global claim. Why: Students confuse interval-by-interval increasing behavior with global behavior, ignoring that $f(1)=-1<f(3)=1$, which violates the definition of an increasing function. Correct move: Always state increasing/decreasing behavior for rational functions separately over each interval separated by asymptotes, and never combine disjoint intervals into a single global claim of monotonic behavior.
• Wrong move: Calculating average rate of change over $[2,5]$ as $\frac{f(2)-f(5)}{5-2}$. Why: Students mix up the order of subtraction in the numerator, leading to a sign error. Correct move: Always write AROC as $\frac{\Delta f}{\Delta x}=\frac{f(\text{end of interval})-f(\text{start of interval})}{\text{end input}-\text{start input}}$ to preserve the order of change.
• Wrong move: Claiming $f(x)=\frac{x^2-4}{x-2}$ has a local extremum at $x=2$ because the first derivative changes sign there. Why: Students forget that $x=2$ is a hole, so the function is not defined at that point and cannot be an extremum. Correct move: Always exclude points where the function is undefined (holes, asymptotes) from consideration for extrema or inflection points.
• Wrong move: Stating that a function is increasing because $f^{\prime \prime }(x)>0$. Why: Students confuse concavity (change in the rate of change) with increasing/decreasing behavior (the sign of the rate of change itself). Correct move: Memorize that first derivative sign determines increase/decrease, second derivative sign determines concavity.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following correctly describes the behavior of $f(x)=x^3-3x$ on the interval $(-1,0)$? A) Increasing and concave up B) Decreasing and concave up C) Increasing and concave down D) Decreasing and concave down

Worked Solution: First, we calculate the first derivative to check increasing/decreasing behavior: $f^{\prime }(x)=3x^2-3=3(x^2-1)$. For all $x$ in $(-1,1)$, $x^2<1$, so $f^{\prime }(x)<0$, meaning $f$ is decreasing on $(-1,0)$. Next, we calculate the second derivative to check concavity: $f^{\prime \prime }(x)=6x$. For all $x$ in $(-1,0)$, $x$ is negative, so $f^{\prime \prime }(x)<0$, meaning $f$ is concave down on the interval. This matches option D. Correct answer: $\boxed{D}$

Question 2 (Free Response)

Given $f(x)=-x^4+8x^2$, (a) Find all critical points of $f(x)$. (b) Classify each critical point as a local maximum, local minimum, or neither, and state the intervals where $f(x)$ is increasing and decreasing. (c) Find all inflection points of $f(x)$ and state the intervals of concavity.

Worked Solution: (a) Calculate the first derivative: $$f^{\prime }(x)=-4x^3+16x=-4x(x-2)(x+2)$$ Critical points occur where $f^{\prime }(x)=0$ (polynomials are defined everywhere), so critical points are $x=-2$, $x=0$, and $x=2$.

(b) Test the sign of $f^{\prime }(x)$ across intervals:

• $(-\infty ,-2)$: $f^{\prime }(x)>0\to$ increasing
• $(-2,0)$: $f^{\prime }(x)<0\to$ decreasing
• $(0,2)$: $f^{\prime }(x)>0\to$ increasing
• $(2,\infty )$: $f^{\prime }(x)<0\to$ decreasing By the first derivative test: $x=-2$ is a local maximum, $x=0$ is a local minimum, $x=2$ is a local maximum. Increasing on $(-\infty ,-2)\cup (0,2)$; decreasing on $(-2,0)\cup (2,\infty )$.

(c) Calculate the second derivative: $$f^{\prime \prime }(x)=-12x^2+16=-4(3x^2-4)$$ Set $f^{\prime \prime }(x)=0$: $x=\pm \frac{2\sqrt{3}}{3}$. Sign test:

• $(-\infty ,-\frac{2\sqrt{3}}{3})$: $f^{\prime \prime }(x)<0\to$ concave down
• $(-\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3})$: $f^{\prime \prime }(x)>0\to$ concave up
• $(\frac{2\sqrt{3}}{3},\infty )$: $f^{\prime \prime }(x)<0\to$ concave down Sign changes occur at both candidates, so inflection points are $\left(-\frac{2\sqrt{3}}{3},\frac{80}{9}\right)$ and $\left(\frac{2\sqrt{3}}{3},\frac{80}{9}\right)$.

Question 3 (Application / Real-World Style)

A small bakery estimates that the daily profit $P$ (in hundreds of dollars) from selling $x$ hundred loaves of sourdough bread is given by $P(x)=-x^3+9x^2-15x+10$ for $x\ge 0$. For what values of $x$ is profit increasing at an increasing rate? Interpret your result in context.

Worked Solution: Profit increasing means $P^{\prime }(x)>0$, and profit increasing at an increasing rate means the rate of change of profit is itself increasing, so $P^{\prime \prime }(x)>0$. First, factor $P^{\prime }(x)$: $$P^{\prime }(x)=-3x^2+18x-15=-3(x-1)(x-5)$$ $P^{\prime }(x)>0$ when $1<x<5$. Next, calculate $P^{\prime \prime }(x)$: $$P^{\prime \prime }(x)=-6x+18=-6(x-3)$$ $P^{\prime \prime }(x)>0$ when $x<3$. The intersection of $1<x<5$ and $x<3$ is $1<x<3$. In context, when the bakery sells between 100 and 300 loaves of bread per day, each additional hundred loaves sold adds more profit than the previous hundred loaves, so profit is growing faster as more loaves are sold.

7. Quick Reference Cheatsheet

8. What's Next

This topic builds the core understanding of co-variation that you will apply immediately to modeling with polynomial functions and analyzing rational function end behavior in the rest of Unit 1. Understanding how change in input relates to change in output is the foundation for all calculus-related topics that come later in AP Precalculus, including rates of change of trigonometric functions and modeling with differential equations. Without mastering how to identify extrema, concavity, and intervals of increase/decrease, you will struggle to graph rational functions and solve optimization problems that appear on both MCQ and FRQ sections of the exam.

Follow-on topics: Modeling with polynomial functions Rational function end behavior Rates of change in context