---
title: "Reference Functions"
book: "High School Mathematics"
subject: math
language: en
chapter: 4
exercises: 10
source: https://one-course.com/books/math/2/en/chapter/4-reference-functions
---

# Chapter 4 — Reference Functions

A handful of simple [functions](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) — affine, square, inverse, square root, cube — appear everywhere, alone or combined. Knowing their [graphs](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) and variations by heart turns many problems into a quick sketch. This chapter studies each of them in turn and proves their variations with the comparison technique of [Chapter 3](https://one-course.com/books/math/2/en/chapter/3-functions#ch-g10-functions).

## 4.1 Affine functions

**Definition 4.1 (Affine function).**

An *affine function* is a [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) of the form

$$
f(x) = mx + p,
$$

where $m$ and $p$ are fixed [real numbers](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets). Its [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) is a straight line: $m$ is the *slope* and $p$ the *$y$-intercept* (the line crosses the vertical axis at $(0, p)$). When $p = 0$ the [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) is *linear*: $f(x) = mx$ expresses proportionality.

**Proposition 4.2 (Slope and variations).**

Let $f(x) = mx + p$.

1. For any two distinct inputs $u \neq v$ : $m = \dfrac{f(v) - f(u)}{v - u}$ (the [slope](#def-g10-reffunc-affine) is the change of the output per unit change of the input).
2. If $m > 0$ , $f$ is strictly [increasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\R$ ; if $m < 0$ , strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) ; if $m = 0$ , constant.

**Proof.** 1. Compute $f(v) - f(u) = (mv + p) - (mu + p) = m(v - u)$, then divide by $v - u \neq 0$.

2. If $u < v$, then $f(v) - f(u) = m(v-u)$ has the sign of $m$, since $v - u > 0$: positive $m$ gives $f(u) < f(v)$ ([increasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations)), negative $m$ gives $f(u) > f(v)$ ([decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations)). ∎

![Two affine functions: increasing for m = 1/2 > 0, decreasing for m = -1 < 0. Moving one unit to the right changes y by m.](https://one-course.com/images/onecourse/chapters/math-2/g10-reffunc/fig-9e20879f061b.svg)

*Two [affine functions](#def-g10-reffunc-affine): [increasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) for $m = \frac12 > 0$, [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) for $m = -1 < 0$. Moving one unit to the right changes $y$ by $m$.*

**Example 4.3.**

Find the [affine function](#def-g10-reffunc-affine) whose [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) passes through $A(1, 3)$ and $B(4, 9)$. The [slope](#def-g10-reffunc-affine) is

$$
m = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2 .
$$

Then $f(x) = 2x + p$, and $f(1) = 3$ gives $2 + p = 3$, so $p = 1$: $f(x) = 2x + 1$. Check with $B$: $f(4) = 9$.

## 4.2 The square function

**Proposition 4.4 (Square function).**

The [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) $f(x) = x^2$, defined on $\R$:

1. is strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoc{-\infty}{0}$ and strictly [increasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intco{0}{+\infty}$ , with [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $0$ at $x = 0$ ;
2. satisfies $f(-x) = f(x)$ for all $x$ : its [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) , a *parabola* , is symmetric about the vertical axis.

**Proof.** 1. Let $0 \leq u < v$. Then

$$
f(v) - f(u) = v^2 - u^2 = (v - u)(v + u) > 0,
$$

since $v - u > 0$ and $v + u > 0$: $f$ is strictly [increasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intco{0}{+\infty}$. If $u < v \leq 0$, then $v - u > 0$ but $v + u < 0$, so $f(v) - f(u) < 0$: strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations). Finally $x^2 \geq 0 = f(0)$ for all $x$.

2. $(-x)^2 = x^2$; so the points $(x, x^2)$ and $(-x, x^2)$, mirror [images](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) across the vertical axis, are both on the [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph). ∎

**Remark 4.5 (Squares and order).**

Because the square [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) decreases on the negative side, taking squares *reverses* the order of negative numbers: $-3 < -2$ but $9 > 4$. Never square both sides of an inequality without checking signs.

## 4.3 The inverse function

**Proposition 4.6 (Inverse function).**

The [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) $f(x) = \dfrac1x$, defined for $x \neq 0$:

1. is strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoo{-\infty}{0}$ and strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoo{0}{+\infty}$ ;
2. satisfies $f(-x) = -f(x)$ : its [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) , a *hyperbola* , is symmetric about the origin.

**Proof.** 1. Let $0 < u < v$. Then

$$
f(v) - f(u) = \frac1v - \frac1u = \frac{u - v}{uv} < 0,
$$

since $u - v < 0$ and $uv > 0$: strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoo{0}{+\infty}$. On $\intoo{-\infty}{0}$ the same quotient has $u - v < 0$ and $uv > 0$ again (product of two negatives), so $f$ decreases there too.

2. $\frac{1}{-x} = -\frac1x$. ∎

**Remark 4.7.**

Careful: $\frac1x$ is *not* [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on its whole [domain](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function). From $u = -1$ to $v = 1$ the value jumps from $-1$ up to $1$. The two branches must be studied separately.

![The parabola y = x2 (symmetric about the vertical axis) and the hyperbola y = 1x (two branches, symmetric about the origin).](https://one-course.com/images/onecourse/chapters/math-2/g10-reffunc/fig-c9fe377b98a0.svg)

![The parabola y = x2 (symmetric about the vertical axis) and the hyperbola y = 1x (two branches, symmetric about the origin).](https://one-course.com/images/onecourse/chapters/math-2/g10-reffunc/fig-7be6853a2a0e.svg)

*The parabola $y = x^2$ (symmetric about the vertical axis) and the hyperbola $y = \frac1x$ (two branches, symmetric about the origin).*

## 4.4 Square root and cube

**Proposition 4.8 (Square root function).**

The [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) $f(x) = \sqrt{x}$, defined on $\intco{0}{+\infty}$, is strictly [increasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations).

**Proof.** Let $0 \leq u < v$. Multiply and divide by the *conjugate*:

$$
\sqrt v - \sqrt u
= \frac{(\sqrt v - \sqrt u)(\sqrt v + \sqrt u)}{\sqrt v + \sqrt u}
= \frac{v - u}{\sqrt v + \sqrt u} > 0,
$$

since $v - u > 0$ and $\sqrt v + \sqrt u > 0$ (note $v > 0$, so $\sqrt v > 0$). ∎

**Proposition 4.9 (Cube function).**

The [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) $f(x) = x^3$, defined on $\R$, is strictly [increasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations), and its [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) is symmetric about the origin.

**Proof.** *Admitted at this level.* ∎

![The square root (defined for x ≥ 0 only) and the cube, both increasing.](https://one-course.com/images/onecourse/chapters/math-2/g10-reffunc/fig-ec6a3c7fb7bc.svg)

![The square root (defined for x ≥ 0 only) and the cube, both increasing.](https://one-course.com/images/onecourse/chapters/math-2/g10-reffunc/fig-5d553e336fa1.svg)

*The square root (defined for $x \geq 0$ only) and the cube, both [increasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations).*

**Proposition 4.10 (Comparing xxx, x2x^2x2 and x\sqrt xx​).**

For $0 < x < 1$: $x^2 < x < \sqrt x$. For $x > 1$: $\sqrt x < x < x^2$. At $x = 0$ and $x = 1$ the three values are equal.

**Proof.** Suppose $0 < x < 1$. Multiplying $x < 1$ by $x > 0$ gives $x^2 < x$. Next, $x < \sqrt x$: since both sides are positive and squaring is [increasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on positive numbers, this inequality is equivalent to $x^2 < x$, which we just proved. For $x > 1$, multiplying $x > 1$ by $x$ gives $x^2 > x$, and $\sqrt x < x$ follows by the same squaring argument. ∎

![The three curves cross at (0,0) and (1,1) and swap order there: between 0 and 1, x2 is the smallest and √ x the largest; beyond 1 the order is reversed.](https://one-course.com/images/onecourse/chapters/math-2/g10-reffunc/fig-2f2ec319808a.svg)

*The three curves cross at $(0,0)$ and $(1,1)$ and swap order there: between $0$ and $1$, $x^2$ is the smallest and $\sqrt x$ the largest; beyond $1$ the order is reversed.*

**Method 4.11 (Comparing images).**

To compare $f(a)$ and $f(b)$ for a reference [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) $f$:

1. place $a$ and $b$ on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) where the variations of $f$ are known;
2. if $f$ increases there, [images](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) are in the same order as the inputs; if $f$ decreases, the order is reversed;
3. if $a$ and $b$ are not in the same [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) of monotonicity, deal with signs first (e.g. for the square: a negative and a positive input).

**Example 4.12.**

Compare $\dfrac{1}{2.3}$ and $\dfrac{1}{2.4}$: both inputs are in $\intoo{0}{+\infty}$, where the inverse [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) decreases, and $2.3 < 2.4$, so $\dfrac{1}{2.3} > \dfrac{1}{2.4}$. Compare $(-1.2)^2$ and $(-1.5)^2$: on $\intoc{-\infty}{0}$ the square decreases, and $-1.5 < -1.2$, so $(-1.5)^2 > (-1.2)^2$ — indeed $2.25 > 1.44$.

## 4.5 Exercises

**Exercise 4.1 ★.**

For each [affine function](#def-g10-reffunc-affine), give its [slope](#def-g10-reffunc-affine), its $y$-intercept, and say whether it is [increasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) or [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations):

$$
f(x) = 3x - 2, \qquad
g(x) = -\tfrac12 x + 4, \qquad
h(x) = 7, \qquad
k(x) = -x .
$$

**Solution of Exercise 4.1.**

$f$: [slope](#def-g10-reffunc-affine) $3$, intercept $-2$, [increasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations). $g$: [slope](#def-g10-reffunc-affine) $-\frac12$, intercept $4$, [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations). $h$: [slope](#def-g10-reffunc-affine) $0$, intercept $7$, constant. $k$: [slope](#def-g10-reffunc-affine) $-1$, intercept $0$, [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) (and linear).

**Exercise 4.2 ★.**

Find the [affine function](#def-g10-reffunc-affine) whose [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) passes through $(2, 1)$ and $(5, 10)$; then the one through $(-1, 4)$ and $(3, -4)$.

**Solution of Exercise 4.2.**

Through $(2,1)$ and $(5,10)$: [slope](#def-g10-reffunc-affine) $m = \frac{10 - 1}{5 - 2} = 3$; then $1 = 3 \times 2 + p$ gives $p = -5$: $f(x) = 3x - 5$.

Through $(-1,4)$ and $(3,-4)$: [slope](#def-g10-reffunc-affine) $m = \frac{-4 - 4}{3 - (-1)} =
\frac{-8}{4} = -2$; then $4 = -2 \times (-1) + p$ gives $p = 2$: $f(x) = -2x + 2$.

**Exercise 4.3 ★.**

Without a calculator, compare:

$$
(3.1)^2 \text{ and } (3.2)^2; \qquad
(-2.7)^2 \text{ and } (-2.8)^2; \qquad
\frac{1}{5.1} \text{ and } \frac{1}{5.2}; \qquad
\sqrt{17} \text{ and } \sqrt{15}.
$$

**Solution of Exercise 4.3.**

$(3.1)^2 < (3.2)^2$: the square increases on positives.

$(-2.7)^2 < (-2.8)^2$: the square decreases on negatives and $-2.8 < -2.7$.

$\frac{1}{5.1} > \frac{1}{5.2}$: the inverse decreases on positives.

$\sqrt{17} > \sqrt{15}$: the square root increases.

**Exercise 4.4 ★.**

Using the [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) of the square [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function), solve $x^2 = 16$, then $x^2 \leq 16$, then $x^2 > 9$.

**Solution of Exercise 4.4.**

$x^2 = 16$: two solutions, $x = 4$ and $x = -4$ (the horizontal line $y = 16$ cuts the parabola twice).

$x^2 \leq 16$: the parabola is below the line between the two [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) points: $x \in \intcc{-4}{4}$.

$x^2 > 9$: the parabola is strictly above $y = 9$ outside $\intcc{-3}{3}$: $x \in \intoo{-\infty}{-3} \cup \intoo{3}{+\infty}$.

**Exercise 4.5 ★.**

Let $x \in \intoo{0}{1}$. Order the numbers $x$, $x^2$, $x^3$ and $\sqrt x$ from smallest to largest, and check your answer with $x = 0.25$.

**Solution of Exercise 4.5.**

For $0 < x < 1$: multiplying $x < 1$ repeatedly by $x$ gives $x^3 < x^2 < x$, and [Proposition 4.10](#prop-g10-reffunc-compare) gives $x < \sqrt x$, so

$$
x^3 < x^2 < x < \sqrt x .
$$

Check with $x = 0.25$: $x^3 = 0.015625$, $x^2 = 0.0625$, $x = 0.25$, $\sqrt x = 0.5$.

**Exercise 4.6 ★★.**

Solve graphically, then algebraically: $\dfrac1x = 2$, and $\dfrac1x < 2$ for $x > 0$. What changes if we also allow $x < 0$?

**Solution of Exercise 4.6.**

$\frac1x = 2$ has the unique solution $x = \frac12$ (the hyperbola meets the horizontal line $y = 2$ once, on the positive branch).

For $x > 0$: $\frac1x < 2$. Multiplying by $x > 0$: $1 < 2x$, so $x > \frac12$: solution set $\intoo{\frac12}{+\infty}$.

If $x < 0$ is allowed: every negative $x$ satisfies $\frac1x < 0 < 2$, so the full solution set is $\intoo{-\infty}{0} \cup \intoo{\frac12}{+\infty}$.

**Exercise 4.7 ★★.**

Knowing that the square [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) decreases on $\intoc{-\infty}{0}$ and increases on $\intco{0}{+\infty}$, frame $x^2$ when:

$$
\text{(a) } x \in \intcc{2}{5};
\qquad
\text{(b) } x \in \intcc{-3}{-1};
\qquad
\text{(c) } x \in \intcc{-2}{3}.
$$

(In case (c), be careful: the [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) of $x^2$ is not at an endpoint.)

**Solution of Exercise 4.7.**

(a) On $\intcc{2}{5}$, all inputs are positive and squaring increases: $4 \leq x^2 \leq 25$.

(b) On $\intcc{-3}{-1}$, squaring decreases: the largest square comes from $-3$: $1 \leq x^2 \leq 9$.

(c) On $\intcc{-2}{3}$, the [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) contains $0$, where the square reaches its [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $0$; the [maximum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) is the larger of $(-2)^2 = 4$ and $3^2 = 9$. So $0 \leq x^2 \leq 9$.

**Exercise 4.8 ★★.**

A taxi company charges a fixed fee of $4$ plus $1.5$ per kilometer; another charges no fee and $2$ per kilometer. Model each price by an [affine function](#def-g10-reffunc-affine) of the distance $x$, plot both lines, and find from which distance the first company is cheaper.

**Solution of Exercise 4.8.**

First company: $f(x) = 1.5x + 4$; second: $g(x) = 2x$. The first is cheaper when

$$
1.5x + 4 < 2x
\ \Longleftrightarrow\
4 < 0.5x
\ \Longleftrightarrow\
x > 8 .
$$

Beyond $8$ km, the first company is cheaper; at exactly $8$ km both charge $16$.

**Exercise 4.9 ★★.**

Show that for all $a, b \geq 0$: $\sqrt{a + b} \leq \sqrt a + \sqrt b$. (Hint: both sides are nonnegative, so compare their squares.) When is there equality?

**Solution of Exercise 4.9.**

Both sides are nonnegative, so the inequality is equivalent to the one between their squares:

$$
\left(\sqrt{a+b}\right)^2 = a + b
\qquad\text{and}\qquad
\left(\sqrt a + \sqrt b\right)^2 = a + 2\sqrt a \sqrt b + b .
$$

Since $2\sqrt a\sqrt b \geq 0$, the second square is at least the first, which proves $\sqrt{a+b} \leq \sqrt a + \sqrt b$. Equality holds exactly when $\sqrt a \sqrt b = 0$, i.e. when $a = 0$ or $b = 0$.

**Exercise 4.10 ★★★.**

Let $f(x) = \dfrac{2x + 1}{x - 1}$, defined for $x \neq 1$.

1. Show that $f(x) = 2 + \dfrac{3}{x - 1}$ for all $x \neq 1$ .
2. Deduce the variations of $f$ on $\intoo{1}{+\infty}$ from those of the inverse [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) .

**Solution of Exercise 4.10.**

*1.* Put the right-hand side over a common denominator:

$$
2 + \frac{3}{x-1} = \frac{2(x-1) + 3}{x - 1} = \frac{2x + 1}{x - 1}
= f(x).
$$

*2.* On $\intoo{1}{+\infty}$, as $x$ increases, $x - 1$ increases and stays positive, so $\frac{3}{x-1}$ decreases (inverse [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function), times the positive constant $3$), so $f(x) = 2 + \frac{3}{x-1}$ decreases: $f$ is strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoo{1}{+\infty}$.

## 4.6 Problem: The laws of nature speak in reference functions

**Problem 4.1.**

Weekend problem — braking distances grow as $v^2$, levers obey $1/d$, planets follow $a^{3/2}$: reading physics with the reference [functions](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function)

Open a physics book and the same few [graphs](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) appear on every page: the parabola, the hyperbola, the root curve. Nature writes its laws with this chapter’s reference [functions](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) — and knowing their shapes ([Proposition 4.4](#prop-g10-reffunc-square), [Proposition 4.6](#prop-g10-reffunc-inverse), [Proposition 4.8](#prop-g10-reffunc-sqrt), [Proposition 4.9](#prop-g10-reffunc-cube)) is enough to brake a car, balance a lever, and time the planets. The grand finale is Kepler’s third law, read straight off the solar system’s data.

**Part I — The square law of braking.** A rule of thumb for a car’s braking distance on dry road: $d = \left(\frac{v}{10}\right)^{2}$ meters, at speed $v$ km/h.

1. Compute the braking distances at $30$ , $50$ , $90$ and $130$ km/h.
2. Doubling the speed multiplies the braking distance by how much? Explain from the square [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) ’s scaling, and check on $50 \to 100$ km/h.
3. Forensics: skid marks measure $50$ m. What speed does the formula convict the driver of (to the km/h)? Which reference [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) answered — and on which [domain](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) is the reading unambiguous ( [Proposition 4.8](#prop-g10-reffunc-sqrt) )?
4. Compare the extra distance caused by *one* km/h more, at low and at high speed: compute $d(31) - d(30)$ and $d(91) - d(90)$ . Which property of the parabola ( [Proposition 4.4](#prop-g10-reffunc-square) ) do the two answers illustrate?
5. Real stopping adds reaction time — about one second, during which the car covers $0.28v$ meters: $D(v) = 0.28v + \left(\frac{v}{10}\right)^2$ . Compute $D(50)$ and $D(130)$ . Which of the two terms — affine or square — dominates at city speed, and which on the highway?

**Part II — The hyperbolas of daily life.**

6. A lever balances when force $\times$ distance is the same on both sides. A $60$ kg child sits $2$ m from the pivot; the balancing force at distance $d$ on the other side is $F(d) = \frac{120}{d}$ (in kg-equivalents). Compute $F(0.5)$ , $F(1)$ , $F(3)$ .
7. What do the variations of the inverse [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) ( [Proposition 4.6](#prop-g10-reffunc-inverse) ) say about levers — and why does Archimedes’ boast (“give me a place to stand and I shall move the Earth”) hide in the hyperbola’s tail? What forbids $d = 0$ ?
8. Sound and light fade with the *square* of the distance: at $1$ m from a lamp the intensity is $100$ units; give it at $2$ , $3$ and $10$ m. Explain the exponent $2$ with a sphere: over what area has the light spread at distance $d$ (Archimedes’ tombstone problem, in the Middle School volume)?
9. A bus for the school trip costs $600$ euros, split equally among $n$ participants. Tabulate the cost per head for $n = 10, 20, 30$ , find how many participants bring it to $25$ euros or less, and state what the hyperbola’s shape promises — and refuses — as $n$ grows.
10. Producing $x$ posters costs $2x + 3$ euros ( $3$ euros of setup). Show that the *average* cost per poster is $a(x) = 2 + \frac3x$ , describe its variations, and interpret its horizontal asymptote economically. (Compare the algebra of [Exercise 4.10](#exo-g10-reffunc-10) .)

**Part III — Kepler’s harmony.** For each planet, let $a$ be its mean distance to the Sun (in astronomical units, Earth $= 1$) and $T$ its period (in years). Data: Mars $a = 1.52$, $T = 1.88$; Jupiter $a = 5.20$, $T = 11.86$; Saturn $a = 9.54$, $T = 29.4$.

11. Compute $a^3$ and $T^2$ for the Earth, Mars and Jupiter. What did Kepler notice in 1618?
12. State the law as a [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) : $T = \sqrt{a^3} = a\sqrt a$ . Test it on Saturn.
13. Two predictions: an asteroid orbits at $a = 4$ AU — find its period (the numbers come out whole); a comet returns every $27$ years — find its mean distance (look for a perfect cube).
14. Which three reference [functions](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) does the law braid together? And its scaling rule: when $a$ is multiplied by $4$ , what happens to $T$ ? (Check with the asteroid against the Earth.)
15. Kepler read this law in Tycho Brahe’s data seventy years before Newton’s gravitation explained it. In one sentence: what does this episode say about the power of recognizing a reference [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) in a table of numbers?

**Part IV — The tortoise and the hare.**

16. Order $\sqrt x$ , $x$ and $x^2$ on $\intoo{0}{1}$ and on $\intoo{1}{+\infty}$ ( [Proposition 4.10](#prop-g10-reffunc-compare) ), and verify the two orderings at $x = 0.25$ and $x = 4$ .
17. Solve $\sqrt x > x$ completely (for $x \geq 0$ , square soundly and finish with a [sign table](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-signtable) ).
18. Bring in the cube: order $x$ , $x^2$ , $x^3$ at $x = 0.5$ and at $x = 2$ , and find all points where the square and the cube [functions](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) cross.
19. Two savings schemes pay, after $t$ years, $100\sqrt t$ euros (scheme A) or $10 t^2$ euros (scheme B). Show that B overtakes A when $t^3 = 100$ , and give the crossover time to the month. Moral about roots against powers in the long run?
20. Finale: for each law of this problem — braking ( $v^2$ ), the lever ( $1/d$ ), the lamp ( $1/d^2$ ), Kepler ( $a^{3/2}$ ), scheme A ( $\sqrt t$ ) — say in one clause which feature of its reference [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) carries the physical meaning (steepening rise, vertical asymptote, spreading sphere, scaling exponent, flattening growth). Conclusion: the reference [functions](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) are the alphabet; nature writes with them.

**Solution of Problem 4.1.**

**1.** $d(30) = 3^2 = 9$ m; $d(50) = 25$ m; $d(90) =
81$ m; $d(130) = 169$ m.

**2.** Doubling $v$ doubles $\frac v{10}$ and multiplies its square by $4$: $d(100) = 100$ m $= 4 \times d(50)$. Twice the speed, four times the distance — the square [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function)’s scaling.

**3.** $\left(\frac{v}{10}\right)^2 = 50$ gives $\frac{v}{10} = \sqrt{50}$, so $v = 10\sqrt{50} \approx 71$ km/h. The square root answered; the reading is unambiguous because speeds are positive, and on $\intco{0}{+\infty}$ the square [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) climbs strictly ([Proposition 4.8](#prop-g10-reffunc-sqrt): one positive [preimage](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function)).

**4.** $d(31) - d(30) = 9.61 - 9 = 0.61$ m, while $d(91) - d(90) = 82.81 - 81 = 1.81$ m: the same $+1$ km/h costs three times more distance at high speed. The parabola not only rises, it *steepens* — its rate of climb grows with $x$.

**5.** $D(50) = 14 + 25 = 39$ m; $D(130) = 36.4 + 169 =
205.4$ m. In town the affine (reaction) term is the bigger share; on the highway the square term crushes it — affine grows steadily, squares run away.

**6.** $F(0.5) = 240$, $F(1) = 120$, $F(3) = 40$.

**7.** The inverse [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) decreases: the further from the pivot, the smaller the needed force — with a long enough lever arm ($d$ huge), any force, however small, balances any load: Archimedes’ boast lives in the hyperbola’s tail, which approaches $0$ without reaching it. And $d = 0$ is the forbidden value: no arm, no lever — the vertical asymptote.

**8.** $\frac{100}{4} = 25$ at $2$ m, $\frac{100}{9} \approx 11$ at $3$ m, $1$ at $10$ m. At distance $d$ the light has spread over a sphere of area $4\pi d^2$ (Archimedes’ tombstone problem, in the Middle School volume): the same energy divided by an area growing like $d^2$ — hence the inverse *square*.

**9.** $60$, $30$, $20$ euros per head for $n = 10, 20,
30$. For $\frac{600}{n} \leq 25$: $n \geq 24$ participants. The hyperbola promises ever-cheaper shares as $n$ grows — and refuses ever to reach $0$: the bus is never free.

**10.** $a(x) = \frac{2x + 3}{x} = 2 + \frac3x$: [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoo{0}{+\infty}$ (inverse [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) shifted), approaching the asymptote $y = 2$. Economically: spreading the fixed $3$ euros over more posters pushes the unit cost down towards the incompressible $2$ euros of materials — economies of scale, with a hard floor.

**11.** Earth: $a^3 = 1$, $T^2 = 1$. Mars: $1.52^3 \approx 3.51$ and $1.88^2 \approx 3.53$. Jupiter: $5.20^3 \approx 140.6$ and $11.86^2 \approx 140.7$. Kepler’s notice: $T^2 = a^3$ for every planet — one law for the whole sky.

**12.** $T = \sqrt{a^3}$: for Saturn $\sqrt{9.54^3} = \sqrt{868} \approx 29.5$ years, against the observed $29.4$: the law holds to a tenth.

**13.** Asteroid: $T = \sqrt{4^3} = \sqrt{64} = 8$ years. Comet: $a^3 = 27^2 = 729 = 9^3$, so $a = 9$ AU.

**14.** Square ($T^2$), cube ($a^3$) and square root (to extract $T$). Scaling: multiplying $a$ by $4$ multiplies $T$ by $4\sqrt4 = 8$ — as the asteroid confirms ($a$ four times Earth’s, $T$ eight times).

**15.** A table of numbers, matched against the shapes of a few reference [functions](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function), yielded a law of the universe seventy years before anyone knew *why* it held — recognizing the [function](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) is half of science.

**16.** On $\intoo{0}{1}$: $x^2 < x < \sqrt x$; on $\intoo{1}{+\infty}$: $\sqrt x < x < x^2$. Checks: at $x = 0.25$: $0.0625 < 0.25 < 0.5$; at $x = 4$: $2 < 4 < 16$.

**17.** For $x \geq 0$, both sides are $\geq 0$, so $\sqrt x > x \iff x > x^2 \iff x(1 - x) > 0 \iff
x \in \intoo{0}{1}$.

**18.** At $x = 0.5$: $x^3 = 0.125 < x^2 = 0.25 < x =
0.5$. At $x = 2$: $2 < 4 < 8$: the order reverses. Crossings of $x^2$ and $x^3$: $x^3 - x^2 = x^2(x - 1) = 0$: at $x = 0$ and $x = 1$.

**19.** $10t^2 = 100\sqrt t$ gives $t^2 = 10\sqrt t$; squaring, $t^4 = 100t$, so $t^3 = 100$ and $t = \sqrt[3]{100} \approx 4.64$ years — about $4$ years and $8$ months. (Both schemes then pay about $215$ euros.) Moral: roots sprint early, powers win every marathon.

**20.** Braking: the parabola’s *steepening* is the danger of speed. Lever: the *vertical asymptote* at $0$ and the long tail are the mechanic’s advantage. Lamp: the $d^2$ of the *spreading sphere* sets the fading. Kepler: the *exponent* $\frac32$ is the solar system’s tempo. Savings A: the root’s *flattening* is the slow saver’s fate. Five [graphs](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph), five laws: nature’s alphabet indeed.