Mathematical-Reasoning · Unit 23 · Video 1 · Interactive Practice

Following Two Arrows: How g(f(x)) Builds a Function

IKey Formulas

Formula	Name	Meaning
$f\colon A \to B$ and $g\colon B \to C$ $\Rightarrow$ $x \mapsto g(f(x))$ is a function $A \to C$	Following two arrows	Each $x \in A$ gets exactly one value $g(f(x))$ in $C$
$x \in A \Rightarrow f(x) \in B \Rightarrow g(f(x)) \in C$	The hand-off	Every output of $f$ lies in $B$ , the set of inputs of $g$
$x \xrightarrow{\times 3} 3x \xrightarrow{+2} 3x + 2 \xrightarrow{\text{cube}} (3x + 2)^3$	A formula as a chain of steps	Three simpler functions applied in turn
$3x^2 + 2 \ge 2 > 0$ for every $x \in \mathbb{R}$	A valid hand-off	The square root in $\sqrt{3x^2 + 2}$ always receives a nonnegative input

Key Insight: A function must give every input exactly one output, but it need not reach every element of its codomain, so $x \mapsto g(f(x))$ may send two inputs to one output and may leave elements of $C$ unreached. Building a function by applying one function to the outputs of another is called composition.

IIVisualization 1 — Following Two Arrows

For each $x \in A$ , $f(x)$ lies in $B$ , so $g(f(x))$ is exactly one element of $C$ .

View

💡 Challenge: redirect arrows so that $x \mapsto g(f(x))$ reaches $u$ . No choice of $f$ and $g$ lets it reach all four elements of $C$ , because every value $g(f(x))$ is $g$ of one of the three elements of $B$ .

IIIVisualization 2 — A Formula as a Chain of Steps

Each step feeds its output to the next, and three different functions can share every step but the last.

Formula Input x 1

IVVisualization 3 — Two Pieces, Then Combined

In $k(x) = \sqrt[3]{\frac{\sin(4x + 3)}{x^2 + 1}}$ , a step may restart from $x$ or use two earlier results.

Step 1 Input x 1

VQuiz Questions

Problem 1 · Follow the Arrows

Given: $A = \{1, 2, 3, 4\}$ , $B = \{p, q, r\}$ , $C = \{s, t, u\}$ , the function $f\colon A \to B$ with $f(1) = q$ , $f(2) = r$ , $f(3) = p$ , $f(4) = q$ , and the function $g\colon B \to C$ with $g(p) = u$ , $g(q) = s$ , $g(r) = s$ . Which list describes the function $x \mapsto g(f(x))$ from $A$ to $C$ ?

✅ Correct! Each input follows one arrow of

f

and then one arrow of

g

. Three inputs share the output

s

and

t

is never reached, both of which a function allows.

❌ Not quite. This list is

f

alone: its outputs are in

B

, not

C

. Follow the second arrow too, for example

g(f(1)) = g(q) = s

❌ Not quite.

f(2) = r

and

g(r) = s

, so

g(f(2)) = s

. No arrow of

g

lands on

t

, and the new function does not have to reach every element of

C

❌ Not quite. Trace each input separately: first find

f(x)

B

, then apply

g

to that element. Here

f(2) = r

g(r) = s

, and

f(3) = p

g(p) = u

Show solution

Follow the arrows two at a time:

g(f(1)) = g(q) = s, \qquad g(f(2)) = g(r) = s,

g(f(3)) = g(p) = u, \qquad g(f(4)) = g(q) = s.

So $x \mapsto g(f(x))$ sends $1$ , $2$ and $4$ to $s$ and sends $3$ to $u$ . Each element of $A$ has exactly one output in $C$ , so this is a function from $A$ to $C$ , even though $t$ is never an output.

Problem 2 · Is Every Hand-Off Valid?

Given: each formula below is computed step by step from an input $x \in \mathbb{R}$ . Which one defines a function from $\mathbb{R}$ to $\mathbb{R}$ , so that every value handed to the next step lies where that step accepts it, for every real $x$ ?

✅ Correct!

x^2 - 2x + 2 = (x - 1)^2 + 1 \ge 1

, so the square root always receives a positive input.

❌ Not quite. At

x = 2

the value handed to the square root is

2 - 4 = -2

, which it does not accept.

❌ Not quite.

x^2 - 4x + 4 = (x - 2)^2

, which is

0

x = 2

, so the division step fails there.

❌ Not quite. At

x = 0

the value handed to the square root is

-1

, which it does not accept.

Show solution

Check the value handed to the last step in each formula.

(a) $2 - x^2 < 0$ when $|x| > \sqrt{2}$ ; for example $x = 2$ gives $-2$ .
(b) $x^2 - 4x + 4 = (x - 2)^2 = 0$ at $x = 2$ , and division by $0$ is undefined.
(c) Completing the square, $x^2 - 2x + 2 = (x - 1)^2 + 1 \ge 1 > 0$ for every real $x$ , so every hand-off is valid.
(d) $x^2 - 1 < 0$ when $-1 < x < 1$ ; for example $x = 0$ gives $-1$ .

Only (c) gives every real input exactly one real output, so only (c) defines a function $\mathbb{R} \to \mathbb{R}$ . The term $-2x$ looks dangerous, but it cannot make the value negative, just as $3x^2 + 2 \ge 2$ protects $\sqrt{3x^2 + 2}$ .

Problem 3 · From Steps to a Formula

Given: a function $h\colon \mathbb{R} \to \mathbb{R}$ is computed by the steps below. Find the formula for $h(x)$ and the value $h(2)$ .

Step	Verbal Description
1	Choose an input, $x$ .
2	Add 1.
3	Square the result.
4	Go back to the original input and multiply it by 5.
5	Subtract the result of step 4 from the result of step 3.
6	Take the cube root of the result.

What is h(x)?

What is h(2)?

✅ Correct!

h(x) = \sqrt[3]{(x + 1)^2 - 5x}

, and

h(2) = \sqrt[3]{9 - 10} = \sqrt[3]{-1} = -1

❌ Not quite. Step 4 goes back to the original input

x

, not to the result of step 3, so it produces

5x

❌ Not quite. Step 3 squares the result of step 2, which is

x + 1

, so it produces

(x + 1)^2

, not

x^2 + 1

❌ Not quite. Step 5 subtracts the result of step 4 from the result of step 3, so it produces

(x + 1)^2 - 5x

❌ Not quite.

\sqrt[3]{-36}

comes from multiplying the result of step 3 by

5

. Step 4 multiplies the original input, giving

5 \cdot 2 = 10

❌ Not quite. A cube root accepts every real number:

(-1)^3 = -1

, so

\sqrt[3]{-1} = -1

❌ Not quite. That is

\sqrt[3]{5 \cdot 2 - 9}

, the subtraction in the wrong order. Step 5 computes

9 - 10 = -1

Show solution

Track the symbolic result of each step:

Step 2: $x + 1$ . Step 3: $(x + 1)^2$ .
Step 4 restarts from the input: $5x$ .
Step 5 uses two earlier results: $(x + 1)^2 - 5x$ .
Step 6: $\sqrt[3]{(x + 1)^2 - 5x}$ .

h(x) = \sqrt[3]{(x + 1)^2 - 5x}.

At $x = 2$ : step 2 gives $3$ , step 3 gives $9$ , step 4 gives $10$ , step 5 gives $9 - 10 = -1$ , and step 6 gives $\sqrt[3]{-1} = -1$ , since $(-1)^3 = -1$ .

h(2) = -1.

Every hand-off is valid for all real $x$ , because a cube root accepts negative inputs, so $h$ really is a function from $\mathbb{R}$ to $\mathbb{R}$ .

Problem 4 · When the Hand-Off Fails

Given: $\mathbb{N} = \{1, 2, 3, \dots\}$ , the function $f\colon \{1, 2, 3\} \to \mathbb{Z}$ with $f(n) = n - 2$ , and the function $g\colon \mathbb{N} \to \mathbb{N}$ with $g(m) = 2m$ . Does $n \mapsto g(f(n))$ define a function from $\{1, 2, 3\}$ to $\mathbb{N}$ ?

✅ Correct! The chain needs every output of

f

to be an input that

g

accepts. Here

f(1) = -1

and

f(2) = 0

lie outside

\mathbb{N}

, the domain of

g

❌ Not quite. The formula

2m

can be evaluated at

-1

and

0

, but

g

is defined only on

\mathbb{N}

, so

g(-1)

and

g(0)

do not exist. The values

-2

and

0

are not in

\mathbb{N}

either.

❌ Not quite. Two functions chain only when the outputs of the first lie in the set of inputs of the second. The outputs

-1

and

0

f

are not in

\mathbb{N}

❌ Not quite. A function does not have to reach every element of its codomain, so missing the odd numbers is allowed. The real failure is the hand-off from

f

g

Show solution

Compute the outputs of $f$ : $f(1) = -1$ , $f(2) = 0$ , $f(3) = 1$ .

The function $g$ accepts only inputs in $\mathbb{N} = \{1, 2, 3, \dots\}$ . Since $-1 \notin \mathbb{N}$ and $0 \notin \mathbb{N}$ , the values $g(f(1))$ and $g(f(2))$ are not defined. Only $g(f(3)) = g(1) = 2$ exists.

A function from $\{1, 2, 3\}$ to $\mathbb{N}$ must give every input an output, so $n \mapsto g(f(n))$ is not such a function. The construction works when $f\colon A \to B$ and $g\colon B \to C$ , that is, when every output of $f$ lies where $g$ accepts it.

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