High School Mathematics · Grades 10–12

4Reference Functions

A handful of simple functions — affine, square, inverse, square root, cube — appear everywhere, alone or combined. Knowing their graphs 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.

4.1 Affine functions

Definition 4.1 (Affine function)

An affine function is a function of the form

f(x) = mx + p,

where $m$ and $p$ are fixed real numbers. Its 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 is linear: $f(x) = mx$ expresses proportionality.

Proposition 4.2 (Slope and variations)

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

For any two distinct inputs $u \neq v$ : $m = \dfrac{f(v) - f(u)}{v - u}$ (the slope is the change of the output per unit change of the input).
If $m > 0$ , $f$ is strictly increasing on $\R$ ; if $m < 0$ , strictly decreasing; 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), negative $m$ gives $f(u) > f(v)$ (decreasing). ∎

Two affine functions: increasing for m = 1/2 > 0, decreasing for m = -1 < 0. Moving one unit to the right changes y by m. — Two affine functions: increasing for $m = \frac12 > 0$ , decreasing for $m = -1 < 0$ . Moving one unit to the right changes $y$ by $m$ .

Example 4.3

Find the affine function whose graph passes through $A(1, 3)$ and $B(4, 9)$ . The slope 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 $f(x) = x^2$ , defined on $\R$ :

is strictly decreasing on $\intoc{-\infty}{0}$ and strictly increasing on $\intco{0}{+\infty}$ , with minimum $0$ at $x = 0$ ;
satisfies $f(-x) = f(x)$ for all $x$ : its 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 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. 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 across the vertical axis, are both on the graph. ∎

Remark 4.5 (Squares and order)

Because the square 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 $f(x) = \dfrac1x$ , defined for $x \neq 0$ :

is strictly decreasing on $\intoo{-\infty}{0}$ and strictly decreasing on $\intoo{0}{+\infty}$ ;
satisfies $f(-x) = -f(x)$ : its 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 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 on its whole domain. 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). — 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 $f(x) = \sqrt{x}$ , defined on $\intco{0}{+\infty}$ , is strictly increasing.

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 $f(x) = x^3$ , defined on $\R$ , is strictly increasing, and its graph is symmetric about the origin.

Proof. Admitted at this level. ∎

The square root (defined for x ≥ 0 only) and the cube, both increasing. — The square root (defined for $x \geq 0$ only) and the cube, both increasing.

Proposition 4.10 (Comparing $x$ , $x^2$ and $\sqrt x$ )

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 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. — 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 $f$ :

place $a$ and $b$ on an interval where the variations of $f$ are known;
if $f$ increases there, images are in the same order as the inputs; if $f$ decreases, the order is reversed;
if $a$ and $b$ are not in the same 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 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, give its slope, its $y$ -intercept, and say whether it is increasing or decreasing:

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

Solution

Solution of Exercise 4.1.

$f$ : slope $3$ , intercept $-2$ , increasing. $g$ : slope $-\frac12$ , intercept $4$ , decreasing. $h$ : slope $0$ , intercept $7$ , constant. $k$ : slope $-1$ , intercept $0$ , decreasing (and linear).

Exercise 4.2 ★

Find the affine function whose graph passes through $(2, 1)$ and $(5, 10)$ ; then the one through $(-1, 4)$ and $(3, -4)$ .

Solution

Solution of Exercise 4.2.

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

Through $(-1,4)$ and $(3,-4)$ : slope $m = \frac{-4 - 4}{3 - (-1)} = \frac{-8}{4} = -2$ ; 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

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 of the square function, solve $x^2 = 16$ , then $x^2 \leq 16$ , then $x^2 > 9$ .

Solution

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 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

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 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

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 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 of $x^2$ is not at an endpoint.)

Solution

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 contains $0$ , where the square reaches its minimum $0$ ; the maximum 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 of the distance $x$ , plot both lines, and find from which distance the first company is cheaper.

Solution

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

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 $x \neq 1$ .

Show that $f(x) = 2 + \dfrac{3}{x - 1}$ for all $x \neq 1$ .
Deduce the variations of $f$ on $\intoo{1}{+\infty}$ from those of the inverse function.

Solution

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, times the positive constant $3$ ), so $f(x) = 2 + \frac{3}{x-1}$ decreases: $f$ is strictly decreasing 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

Open a physics book and the same few graphs appear on every page: the parabola, the hyperbola, the root curve. Nature writes its laws with this chapter’s reference functions — and knowing their shapes (Proposition 4.4, Proposition 4.6, Proposition 4.8, Proposition 4.9) 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.

Compute the braking distances at $30$ , $50$ , $90$ and $130$ km/h.
Doubling the speed multiplies the braking distance by how much? Explain from the square function’s scaling, and check on $50 \to 100$ km/h.
Forensics: skid marks measure $50$ m. What speed does the formula convict the driver of (to the km/h)? Which reference function answered — and on which domain is the reading unambiguous (Proposition 4.8)?
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) do the two answers illustrate?
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.

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 $F(0.5)$ , $F(1)$ , $F(3)$ .
What do the variations of the inverse function (Proposition 4.6) 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 $d = 0$ ?
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)?
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.
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.)

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 $a = 9.54$ , $T = 29.4$ .

Compute $a^3$ and $T^2$ for the Earth, Mars and Jupiter. What did Kepler notice in 1618?
State the law as a function: $T = \sqrt{a^3} = a\sqrt a$ . Test it on Saturn.
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).
Which three reference functions 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.)
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 in a table of numbers?

Part IV — The tortoise and the hare.

Order $\sqrt x$ , $x$ and $x^2$ on $\intoo{0}{1}$ and on $\intoo{1}{+\infty}$ (Proposition 4.10), and verify the two orderings at $x = 0.25$ and $x = 4$ .
Solve $\sqrt x > x$ completely (for $x \geq 0$ , square soundly and finish with a sign table).
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 cross.
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?
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 carries the physical meaning (steepening rise, vertical asymptote, spreading sphere, scaling exponent, flattening growth). Conclusion: the reference functions are the alphabet; nature writes with them.

Solution

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’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 climbs strictly (Proposition 4.8: one positive preimage).

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 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 on $\intoo{0}{+\infty}$ (inverse 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, yielded a law of the universe seventy years before anyone knew why it held — recognizing the 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, five laws: nature’s alphabet indeed.