Mathematical-Reasoning · Unit 21 · Video 1 · Interactive Practice

Polygon Diagonals and Derivatives: A Formula Is Not a Function

IKey Formulas

Formula	Name	Meaning
$D = \mathbb{N} - \{1, 2\}$ , $\;d\colon D \to \mathbb{N} \cup \{0\}$	Diagonal count	$d(n)$ is the number of diagonals of a convex polygon with $n$ sides
$d(n) = \dfrac{n(n-3)}{2}$	Counting the diagonals	Each of the $n$ vertices starts $n - 3$ diagonals, and each diagonal is counted once at each of its two ends
$f\colon \mathbb{R} \to \mathbb{R}$ , $\;f(x) = \dfrac{x(x-3)}{2}$	Same expression, new function	$d(n) = f(n)$ for every $n \in D$ , yet the domains and codomains differ, so $d \neq f$
$f \mapsto f'$	Differentiation as a function	Inputs: differentiable real functions; outputs: real functions. $\lvert x \rvert$ is not an input

Key Insight: A function is its domain, its codomain and its rule together. A formula that matches a function's outputs does not settle the domain or the codomain, so it does not by itself make two functions the same.

IIVisualization 1 — Counting Diagonals Vertex by Vertex

Each vertex starts $n - 3$ diagonals and every diagonal has two ends, so $d(n) = \frac{n(n-3)}{2}$ .

Sides n 6 Vertices visited 3

IIIVisualization 2 — The Formula Beyond the Polygons

The formula $f(x) = \frac{x(x-3)}{2}$ accepts every real $x$ ; the count $d$ accepts only $n = 3, 4, 5, \dots$

Input x 1

IVVisualization 3 — Differentiation as a Function

Differentiation sends each differentiable input $f$ to exactly one output $f'$ ; different inputs may share an output.

Input

VQuiz Questions

Problem 1 · Handshakes Versus a Formula

Given: for $n \in \mathbb{N}$ , let $h(n)$ be the number of handshakes when each pair of $n$ people shakes hands exactly once, so $h\colon \mathbb{N} \to \mathbb{N} \cup \{0\}$ . Each person shakes $n - 1$ hands and each handshake involves two people, so $h(n) = \frac{n(n-1)}{2}$ . Let $g\colon \mathbb{R} \to \mathbb{R}$ be $g(x) = \frac{x(x-1)}{2}$ . Which statement is true?

✅ Correct! The rules give the same number at every

n \in \mathbb{N}

, but

h

accepts only natural numbers while

g

accepts every real number, and the codomains

\mathbb{N} \cup \{0\}

and

\mathbb{R}

differ too.

❌ Not quite. Agreeing on

\mathbb{N}

is not enough:

g

also accepts inputs such as

\tfrac{1}{2}

, with

g\left(\tfrac{1}{2}\right) = -\tfrac{1}{8}

, and

h

has no output there at all.

❌ Not quite. The two rules never disagree on

\mathbb{N}

: the handshake count really is

\frac{n(n-1)}{2}

, for instance

h(1) = 0 = g(1)

and

h(4) = 6 = g(4)

. The difference lies elsewhere.

❌ Not quite. Changing the codomain leaves the domains unequal:

h

would still be defined only on

\mathbb{N}

, while

g

is defined on

\mathbb{R}

Show solution

Compare the three parts of each function.

Rule: $h(n)$ counts handshakes. Each of the $n$ people shakes $n - 1$ hands, which gives $n(n-1)$ handshake ends, and each handshake has two ends, so $h(n) = \frac{n(n-1)}{2} = g(n)$ for every $n \in \mathbb{N}$ .
Domain: $\mathbb{N}$ for $h$ , but $\mathbb{R}$ for $g$ . For example $g(-2) = \frac{(-2)(-3)}{2} = 3$ , while $h(-2)$ is not defined.
Codomain: $\mathbb{N} \cup \{0\}$ for $h$ , but $\mathbb{R}$ for $g$ ; $g\left(\tfrac{1}{2}\right) = -\tfrac{1}{8}$ is not a count of anything.

Matching outputs on $\mathbb{N}$ do not make the functions equal, because the domains and codomains differ. So $h \neq g$ , exactly as $d \neq f$ in the video.

Problem 2 · Triangles from Points on a Circle

Given: for $n \in \mathbb{N}$ , let $t(n)$ be the number of triangles whose three vertices are chosen from $n$ points marked on a circle, so $t\colon \mathbb{N} \to \mathbb{N} \cup \{0\}$ . For $n \ge 3$ , counting gives $t(n) = \frac{n(n-1)(n-2)}{6}$ . Let $p\colon \mathbb{R} \to \mathbb{R}$ be $p(x) = \frac{x(x-1)(x-2)}{6}$ .

What is $p(-1)$ ?

For which inputs $x$ is $t(x) = p(x)$ a true statement?

✅ Correct!

p(-1) = -1

is a value of

p

at an input

t

does not accept, and

-1 \notin \mathbb{N} \cup \{0\}

anyway. The two functions agree on all of

\mathbb{N}

, including

n = 1

and

n = 2

, where both give

0

❌ Check the sign. The product

(-1)(-2)(-3)

has three negative factors, so it is

-6

, not

6

❌ Not quite. "Fewer than three points, so no triangles" is a thought about

t

, and

t

does not even accept

-1

. The question asks about

p

, a formula on

\mathbb{R}

: substitute

x = -1

into

\frac{x(x-1)(x-2)}{6}

❌ Almost. The product

(-1)(-2)(-3) = -6

is right, but the formula still divides it by

6

❌ Not quite. Check

n = 1

and

n = 2

: with fewer than three points there are no triangles, so

t(1) = t(2) = 0

, and

p(1) = p(2) = 0

as well.

❌ Not quite.

t

accepts only natural numbers, so

t\left(\tfrac{3}{2}\right)

is not defined even though

p\left(\tfrac{3}{2}\right) = -\tfrac{1}{16}

is.

❌ Not quite. At

x = -1

the formula gives

p(-1) = -1

, but

t(-1)

does not exist: no one marks

-1

points on a circle.

❌ Not quite. Substitute

x = -1

into each factor and multiply.

❌ Not quite. The statement needs both

t(x)

and

p(x)

to exist and be equal.

Show solution

The value:

p(-1) = \frac{(-1)(-1-1)(-1-2)}{6} = \frac{(-1)(-2)(-3)}{6} = \frac{-6}{6} = -1

The input $-1$ is not in the domain $\mathbb{N}$ of $t$ , and the output $-1$ is not in its codomain $\mathbb{N} \cup \{0\}$ .

Where they agree: $t(x)$ exists only for $x \in \mathbb{N}$ . For $n \ge 3$ the stem gives $t(n) = p(n)$ . For the two remaining natural numbers, one or two points form no triangle, so $t(1) = t(2) = 0$ , and

p(1) = \frac{1 \cdot 0 \cdot (-1)}{6} = 0, \qquad p(2) = \frac{2 \cdot 1 \cdot 0}{6} = 0.

So $t(n) = p(n)$ for every natural number $n$ . Even so, $t \neq p$ : the domains $\mathbb{N}$ and $\mathbb{R}$ differ.

Problem 3 · Which Function Is an Input?

Given: let $\mathcal{F}_D$ be the set of functions from $\mathbb{R}$ to $\mathbb{R}$ that have a derivative at every real number; it is the domain of differentiation in this problem. Which function is an element of $\mathcal{F}_D$ ?

✅ Correct!

s(x) = x^2

for

x \ge 0

and

s(x) = -x^2

for

x \lt 0

; both pieces have slope

0

0

, so

s

has a derivative everywhere, with

s'(x) = 2\lvert x \rvert

❌ Not quite.

u

is defined on all of

\mathbb{R}

, but its graph has a vertical tangent at

0

\frac{u(h) - u(0)}{h} = h^{-2/3}

grows without bound as

h \to 0

, so

u'(0)

does not exist.

❌ Not quite.

w

is the absolute value graph moved to the right by

2

: it has a corner at

x = 2

, with slope

-1

from the left and

1

from the right.

❌ Not quite. The pieces meet at

(1, 1)

, so there is no gap, but the slope from the left is

2 \cdot 1 = 2

and the slope from the right is

1

. No single slope fits both sides at

x = 1

Show solution

An element of $\mathcal{F}_D$ must be defined on all of $\mathbb{R}$ and have a derivative at every real number. Each wrong option fails at one point.

$u(x) = \sqrt[3]{x}$ : for $x \neq 0$ , $u'(x) = \frac{1}{3}x^{-2/3}$ , but at $0$ the difference quotient $\frac{\sqrt[3]{h}}{h} = h^{-2/3}$ has no finite limit. Not an input.
$w(x) = \lvert x - 2 \rvert$ : slope $-1$ for $x \lt 2$ and $1$ for $x \gt 2$ ; no derivative at $2$ . Not an input.
$q$ : slope from the left at $1$ is $\frac{d}{dx}x^2 = 2x = 2$ , slope from the right is $1$ ; no derivative at $1$ . Not an input.

For $s(x) = x\lvert x \rvert$ : if $x \gt 0$ then $s(x) = x^2$ and $s'(x) = 2x$ ; if $x \lt 0$ then $s(x) = -x^2$ and $s'(x) = -2x$ . At $0$ ,

\frac{s(h) - s(0)}{h} = \frac{h\lvert h \rvert}{h} = \lvert h \rvert \to 0,

so $s'(0) = 0$ . Thus $s \in \mathcal{F}_D$ , and differentiation sends it to the single output $s'(x) = 2\lvert x \rvert$ . That output is itself not in $\mathcal{F}_D$ , but it is a real function, so it lies in the codomain.

Problem 4 · Inputs and Outputs of Differentiation

Given: $\mathcal{F}_D$ as in Problem 3, $\mathcal{F}$ the set of all functions from $\mathbb{R}$ to $\mathbb{R}$ , and $T\colon \mathcal{F}_D \to \mathcal{F}$ defined by $T(f) = f'$ .

For $m(x) = x^2 e^{3x}$ , what is $T(m)(x)$ ?

For $f(x) = x^3 + 4x$ and $g(x) = x^3 + 4x + 10$ , which statement is true?

✅ Correct!

T(m)(x) = 2xe^{3x} + 3x^2e^{3x}

, and

f

and

g

are two different inputs sent to the same output

3x^2 + 4

, which a function allows.

❌ Not quite. That differentiates only the factor

x^2

. The product rule also needs

x^2

times the derivative of

e^{3x}

❌ Not quite. The derivative of a product is not the product of the derivatives

2x \cdot 3e^{3x}

; use the product rule.

❌ Close. The chain rule gives

\frac{d}{dx}e^{3x} = 3e^{3x}

, not

e^{3x}

❌ Not quite. The constant

10

has derivative

0

, so

T(g)(x) = 3x^2 + 4 = T(f)(x)

❌ Not quite. A function must give each input exactly one output. Nothing forbids two different inputs from sharing an output; each of

f

and

g

still has exactly one derivative.

❌ Not quite.

f(0) = 0

but

g(0) = 10

, so

f \neq g

. Equal outputs tell us nothing about whether the inputs were equal.

❌ Not quite. Use the product rule on

x^2 \cdot e^{3x}

❌ Not quite. Differentiate both

f

and

g

before deciding.

Show solution

Part 1. The product rule, with the chain rule for $e^{3x}$ :

T(m)(x) = \frac{d}{dx}\left(x^2\right) e^{3x} + x^2 \frac{d}{dx}\left(e^{3x}\right) = 2x e^{3x} + 3x^2 e^{3x}.

Part 2. By the power rule term by term, and since a constant has derivative $0$ ,

T(f)(x) = 3x^2 + 4, \qquad T(g)(x) = 3x^2 + 4 + 0 = 3x^2 + 4.

So $T(f) = T(g)$ even though $f \neq g$ (for instance $f(0) = 0$ and $g(0) = 10$ ). This does not conflict with $T$ being a function: the requirement is that each input $f$ in $\mathcal{F}_D$ has exactly one output, and each differentiable function has exactly one derivative. Two inputs sharing one output is allowed.

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