# Change in tandem (function behavior)

> AP Precalculus · Unit 1: Polynomial and Rational Functions
> Source: https://www.owlsprep.com/study/ap-precalculus-u1-change-in-tandem/

How the input and output of a function change together — in tandem: what a function is, its domain, range, and independent/dependent variables; the four ways to represent a function; image and preimage; how to tell where a function is increasing or decreasing by comparing output values; how a verbal description becomes a graph, what makes that graph concave up or concave down, and where its zeros are. This is the foundational vocabulary for all of AP Precalculus.

**Prerequisites:** Plotting and reading points on the coordinate plane; Evaluating an algebraic expression for a given value; Set and interval notation

## Learning objectives

- Define a function and identify its domain, range, and independent and dependent variables.
- Express a function using graphical, numerical, analytical, and verbal representations, and translate between them.
- Find the image of an input value and the preimage of an output value.
- Determine intervals on which a function is increasing or decreasing by comparing output values.

## What a Function Is: Domain, Range, and Variables

Most interesting quantities change **together**. Fill a vase and the water's height rises as the volume poured in rises; the two vary *in tandem*. A **function** is the precise way we pin down that pairing: it takes each input and assigns it one — and only one — output.

**Function** — A relation that maps each input value to **exactly one** output value. If a single input could give two different outputs, the relation is not a function.

*Notation:* f(x)

**Domain and range** — The **domain** is the set of all input values the function accepts. The **range** is the set of all output values it produces.

The input is the **independent variable** (you choose it); the output is the **dependent variable** (its value depends on the input). We write $f(x)$ to mean the output the rule $f$ assigns to the input $x$.

**Equal functions** — Two functions $f$ and $g$ are **equal** when they have the **same domain** and, for every input $a$ in that domain, they give the same output: $f(a) = g(a)$. Same formula but different domains means the functions are *not* equal.

**Worked example:** A stadium sells at most 4 tickets per person at \$25 each. The total cost of $n$ tickets is $C(n) = 25n$. Identify the independent and dependent variables, and state the domain and range.

1. You decide how many tickets to buy, and the cost follows from that choice. So the input $n$ is the independent variable and the output $C$ is the dependent variable.
2. You can buy a whole number of tickets, from 0 up to the limit of 4. The domain is the set of allowed inputs:
3. $$\text{domain} = \{0,\ 1,\ 2,\ 3,\ 4\}$$
4. Apply the rule to each input to get the outputs $C(0)=0,\ C(1)=25,\ C(2)=50,\ C(3)=75,\ C(4)=100$. The range is the set of outputs:
5. $$\text{range} = \{0,\ 25,\ 50,\ 75,\ 100\}$$
6. Because only whole numbers of tickets make sense, both the domain and range are **discrete** sets, not intervals.

> **tip**
>
> Keep inputs and outputs straight: the domain lives on the input side (independent variable), the range on the output side (dependent variable). Swapping them is the single most common early mistake.

## Four Representations, Image, and Preimage

The **function rule** — how each input is paired with its output — can be shown in four equivalent ways. They are four views of the *same* function, and a big part of this course is translating fluently between them.

**Four representations of a function** — **Graphical** (a curve of points $(x, f(x))$), **numerical** (a table of input–output pairs), **analytical** (an equation such as $f(x)=x^2$), and **verbal** (a sentence describing the pairing).

**Image** — The **image** of an input value $a$ is the single output the rule assigns to it, namely $f(a)$. Every input in the domain has exactly one image.

**Preimage** — The **preimage** of an output value $y$ is the **set of all** inputs the rule sends to $y$ — every $x$ with $f(x)=y$. A preimage can contain no inputs, one input, or many.

**Worked example:** A function $f$ is given numerically by the table below. Find (a) the image of $-2$, (b) the preimage of $1$, and (c) the preimage of $3$.

| $x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|---|
| $f(x)$ | $4$ | $1$ | $0$ | $1$ | $4$ |

1. (a) The image of $-2$ is simply the output paired with the input $-2$. Read the column under $x=-2$:
2. $$f(-2) = 4$$
3. (b) The preimage of $1$ is every input whose output is $1$. Scan the bottom row for $1$; it appears under $x=-1$ and $x=1$:
4. $$\text{preimage of } 1 = \{-1,\ 1\}$$
5. Notice the preimage is a **set with two inputs** — very different from an image, which is always one value.
6. (c) The preimage of $3$ is every input with output $3$. No entry in the bottom row equals $3$, so nothing maps to it:
7. $$\text{preimage of } 3 = \{\ \}\ \text{(empty)}$$

> **warning**
>
> Image vs. preimage: an **image** is one output value ($f(a)$); a **preimage** is a *set* of input values. Don't collapse a preimage to a single number — list every input that maps to the target output.

## Increasing and Decreasing: Comparing Outputs

To describe how two quantities change in tandem, we first ask a purely *qualitative* question: as the input increases, does the output go **up** or **down**? We answer it by directly **comparing output values** — no measure of *how fast* is needed yet (that is the next topic).

**Increasing on an interval** — A function is **increasing** on an interval of its domain if larger inputs always give larger outputs: for all $a$ and $b$ in the interval, if $a < b$ then $f(a) < f(b)$.

**Decreasing on an interval** — A function is **decreasing** on an interval of its domain if larger inputs always give smaller outputs: for all $a$ and $b$ in the interval, if $a < b$ then $f(a) > f(b)$.

$$\text{increasing:}\ \ a<b \Rightarrow f(a)<f(b) \qquad \text{decreasing:}\ \ a<b \Rightarrow f(a)>f(b)$$

**Worked example:** A cup of tea cools. The table gives its temperature $T$ (°C) after $m$ minutes. On which interval is $T$ a decreasing function of $m$? Justify by comparing outputs.

| $m$ | $0$ | $2$ | $4$ | $6$ | $8$ |
|---|---|---|---|---|---|
| $T$ | $85$ | $68$ | $57$ | $51$ | $51$ |

1. Read the outputs from left to right and compare each with the next: $85 > 68 > 57 > 51$.
2. From $m=0$ to $m=6$, every larger input gives a strictly smaller output, which is exactly the definition of decreasing:
3. $$0<2<4<6 \ \Rightarrow\ T(0) > T(2) > T(4) > T(6)$$
4. So $T$ is decreasing on the interval $0 \le m \le 6$.
5. From $m=6$ to $m=8$ the output stays at $51$ ($T(6)=T(8)$), so on $6 \le m \le 8$ the function is **constant**, neither increasing nor decreasing.

> **warning**
>
> “Increasing” must hold for **every** pair $a<b$ in the interval, not just the two endpoints. And never merge separate intervals into one claim — a function can increase on one stretch and decrease on another.

## Building the Graph: Concavity and Zeros

A graph is just the set of input–output pairs drawn in the plane, so it shows at a glance how the two values vary together. That means a **verbal description** of how two quantities change is already enough to sketch one — you do not need a formula first.

**Worked example:** Water is poured steadily into each of three vases: (a) one with straight sides, (b) one that gets **wider** toward the top, (c) one that gets **narrower** toward the top. For each, describe the graph of height $H$ against volume poured $V$.

1. (a) Straight sides: every millilitre adds the same height, so the graph rises as a **straight line**.
2. (b) Wider toward the top: as the vase widens, each additional millilitre spreads out and adds **less** height than the one before. The graph still rises, but it flattens out as it goes.
3. (c) Narrower toward the top: each additional millilitre is squeezed into a smaller cross-section and adds **more** height than the one before. The graph rises ever more steeply.
4. All three are increasing — more water always means a higher level. What differs is **how fast** the height is changing, and that is what shapes the curve.

That last distinction has a name. Describing *how fast* the output changes is the subject of Topic 1.2; here we only need the idea of the **rate of change** getting larger or smaller as the input increases.

**Concave up** — The graph of a function is **concave up** on intervals where the **rate of change is increasing** — each step of input raises the output by more than the step before. Vase (c) above is concave up.

**Concave down** — The graph of a function is **concave down** on intervals where the **rate of change is decreasing** — each step of input raises the output by less than the step before. Vase (b) above is concave down.

> **warning**
>
> Concavity is **not** the same as increasing or decreasing. All three vases are increasing; they differ only in concavity. A function can be decreasing and concave up at the same time — falling, but levelling off.

One more feature is read straight off the picture: where the graph meets the horizontal axis.

**Zeros of a function** — The graph intersects the $x$-axis exactly where the **output value is zero**. The corresponding input values are called the **zeros of the function**.

*Notation:* f(a)=0 \iff (a,\,0)\ \text{is on the graph}

**Worked example:** Earlier we found that for $f(x)=x^2-1$ the preimage of the output $0$ is $\{-1,\ 1\}$. What does that tell you about the graph?

1. The preimage of $0$ is the set of inputs whose output is $0$ — so $f(-1)=0$ and $f(1)=0$.
2. An output of $0$ puts the point on the $x$-axis, so the graph passes through $(-1,\ 0)$ and $(1,\ 0)$.
3. **Answer:** the zeros of $f$ are $-1$ and $1$, and they are exactly where the graph crosses the $x$-axis.

> **tip**
>
> “Zero of the function”, “input whose output is $0$” and “$x$-intercept of the graph” all name the same thing seen three ways — analytically, numerically and graphically.

## AP-Style Practice

**Check your understanding**

Test your understanding with this multiple-choice question:

1. A function $h$ is given by the table below. On which interval is $h$ increasing?

| $x$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|
| $h(x)$ | $12$ | $7$ | $4$ | $7$ | $12$ |

   - $1 \le x \le 3$
   - $3 \le x \le 5$
   - $1 \le x \le 5$
   - $2 \le x \le 4$

   *Answer:* $3 \le x \le 5$

   *Why:* From $x=3$ to $x=5$ the outputs rise: $4 < 7 < 12$, so for any $a<b$ there, $h(a)<h(b)$ — increasing. On $1 \le x \le 3$ the outputs fall ($12>7>4$), so that interval is decreasing, and the full interval $1 \le x \le 5$ mixes both.

**Worked example:** Water is poured steadily into an empty vase. Let $H$ be the height of the water (cm) and $V$ the volume poured in (mL). (a) Which is the independent and which the dependent variable? (b) Explain why $H$ is an increasing function of $V$. (c) The vase holds 500 mL and is 20 cm tall when full. State the domain and range of $H(V)$.

1. (a) You control how much water goes in, and the height responds to that. So $V$ is the independent variable and $H$ is the dependent variable.
2. (b) Adding more water can only raise the level. So whenever $V_1 < V_2$, the taller column of water gives $H(V_1) < H(V_2)$. That matches the definition $a<b \Rightarrow f(a)<f(b)$, so $H$ is increasing on its whole domain.
3. (c) The volume runs from empty to full, $0$ to $500$ mL, so the domain is $0 \le V \le 500$.
4. The height correspondingly runs from $0$ to $20$ cm, so the range is $0 \le H \le 20$.

**Worked example:** A function is described **verbally**: “the output is 1 less than the square of the input.” (a) Write it **analytically**. (b) Give the **numerical** representation for inputs $-2,-1,0,1,2$. (c) Find the preimage of the output $0$.

1. (a) “Square of the input” is $x^2$; “1 less than” subtracts 1:
2. $$f(x) = x^2 - 1$$
3. (b) Evaluate the rule at each input to build the table:
4. $$f(-2)=3,\ \ f(-1)=0,\ \ f(0)=-1,\ \ f(1)=0,\ \ f(2)=3$$
5. (c) The preimage of $0$ is every input with output $0$. From the table (or by solving $x^2-1=0$), that is $x=-1$ and $x=1$:
6. $$\text{preimage of } 0 = \{-1,\ 1\}$$

## Common pitfalls

- **Wrong:** Reporting the preimage of an output as a single input value.
  - Why it fails: The preimage is the set of ALL inputs that map to the target output. For $f(x)=x^2$, the preimage of $9$ is $\{-3, 3\}$, not just $3$.
  - Correct: List every input whose output equals the target; the preimage may have zero, one, or several elements.
- **Wrong:** Calling $x = y^2$ a function of $x$.
  - Why it fails: The input $x=4$ would give two outputs, $y=2$ and $y=-2$, but a function must assign exactly one output to each input.
  - Correct: Check that each input yields only one output (graphically, that no vertical line meets the graph twice) before calling a relation a function.
- **Wrong:** Reading the domain off the output side (or the range off the input side).
  - Why it fails: Domain and range get swapped: the domain is the set of inputs (independent variable), the range is the set of outputs (dependent variable).
  - Correct: Anchor on “inputs in, outputs out”: collect inputs for the domain and outputs for the range.
- **Wrong:** Declaring $f(x)=\frac{x^2-4}{x-2}$ and $g(x)=x+2$ equal because the algebra simplifies.
  - Why it fails: $f$ is undefined at $x=2$ while $g$ is defined there, so the two functions have different domains.
  - Correct: Functions are equal only when they share the same domain AND agree on every input; check the domains, not just the formula.
- **Wrong:** Concluding $f$ is increasing on $[0,4]$ just because $f(0) < f(4)$.
  - Why it fails: Increasing requires $f(a)<f(b)$ for EVERY pair $a<b$ in the interval; two matching endpoints can hide a dip in between.
  - Correct: Verify the comparison holds across the whole interval, and never combine separate increasing/decreasing stretches into one claim.

## Cheatsheet

| Concept | Definition / Rule | Key Notes |
| --- | --- | --- |
| Function | Each input maps to exactly one output | Fails if one input gives two outputs (vertical-line test) |
| Domain / Range | Domain = all inputs; range = all outputs | Domain feeds the rule; range comes out of it |
| Independent / Dependent variable | Input = independent; output = dependent | Output value depends on the chosen input |
| Equal functions $f=g$ | Same domain AND $f(a)=g(a)$ for every $a$ | Same formula but different domain ⇒ not equal |
| Four representations | Graphical, numerical, analytical, verbal | Four views of ONE rule; translate between them |
| Image of $a$ | $f(a)$ — the single output for input $a$ | Always exactly one value |
| Preimage of $y$ | All inputs $x$ with $f(x)=y$ | A SET: may be empty, one, or many |
| Increasing on $I$ | For all $a<b$ in $I$, $f(a)<f(b)$ | Compare outputs; must hold for every pair |
| Decreasing on $I$ | For all $a<b$ in $I$, $f(a)>f(b)$ | Compare outputs; must hold for every pair |
| Constant on $I$ | $f(a)=f(b)$ for all $a,b$ in $I$ | Neither increasing nor decreasing |
| Concave up | Rate of change is increasing | Each input step raises the output by more than the last |
| Concave down | Rate of change is decreasing | Each input step raises the output by less than the last |
| Zeros of a function | Inputs $a$ with $f(a)=0$ | Exactly where the graph meets the $x$-axis |

## What's next

You can now describe THAT a function's output rises or falls as its input increases, name a function's domain, range, image, and preimage across all four representations, and read concavity and zeros off a graph. The natural next question is HOW FAST it changes — that is Topic 1.2, Rates of Change, which measures the average rate of change over an interval and makes the idea of concavity precise.

- [Rates of change (average and over equal intervals)](https://www.owlsprep.com/study/ap-precalculus-u1-rates-of-change/)
- [Polynomial functions and rates of change](https://www.owlsprep.com/study/ap-precalculus-u1-polynomial-functions-and-rates-of/)
- [Polynomial functions and complex zeros](https://www.owlsprep.com/study/ap-precalculus-u1-polynomial-functions-and-complex-zeros/)

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