Functions, Equations, and Scientific Models

Motivation

This chapter assumes arithmetic with signed numbers and fractions, and the distributive law. By the end, you should be able to state the domain of a formula, distinguish equivalent equations from merely necessary conditions, compose and invert suitable functions, and build a dimensionally consistent elementary model. These skills prepare for limits in calculus and for systems in linear algebra.

A formula alone does not tell the whole mathematical story. The expression $1/(x-2)$ is meaningful at $x=3$ but not at $x=2$. A square root may introduce a restriction even when it disappears after squaring. A scientific model adds another restriction: an algebraically valid negative concentration is not a physically admissible answer. We shall keep these three questions separate: Is the expression defined? Is the equation satisfied? Is the model applicable?

1.1 Real numbers, intervals, and absolute value

The real line is ordered. An interval records a range without listing its elements. For example,

$$[a,b]=\{x\in\mathbb R:a\le x\le b\},\qquad (a,b]=\{x\in\mathbb R:a<x\le b\}.$$

The symbols $\infty$ and $-\infty$ are not real endpoints, so $(a,\infty)$ uses a round bracket at infinity. An inequality may specify a union of intervals rather than one interval.

The absolute value of $x$ is its distance from zero:

$$|x|=\begin{cases}x,&x\ge0,\\-x,&x<0.\end{cases}$$

Proposition

Consequently, for $r\ge0$, $|x-a|\le r$ is equivalent to $a-r\le x\le a+r$. For $r>0$, $|x-a|>r$ means $x<a-r$ or $x>a+r$. The word “or” is essential: a point outside a neighbourhood need not be on both sides.

Proof

Put $z=x-a$. If $z\ge0$, the inequality $|z|\le r$ is $z\le r$, and $-r\le z$ already holds. If $z<0$, it is $-z\le r$, or $z\ge-r$, and $z\le r$ already holds. Thus, for $r\ge0$, the two cases give exactly $-r\le z\le r$. Taking the complementary cases gives $|z|>r$ exactly when $z<-r$ or $z>r$. Translating by $a$ proves both assertions, including the equality case $r=0$ in the first assertion.

Worked example

Worked example 1: a measurement tolerance. A sensor is certified when its reading $T$ differs from $25$ by at most $0.4$. The condition is $|T-25|\le0.4$, hence $T\in[24.6,25.4]$. The endpoints are allowed because “at most” includes equality. If the units are degrees Celsius, every number in this inequality is understood in those same units.

1.2 Functions and domains

Definition

A function $f:D\to C$ assigns exactly one value $f(x)\in C$ to each input $x\in D$. The set $D$ is its domain, $C$ its codomain, and $f(D)=\{f(x):x\in D\}$ its range. A codomain is a declared target; it need not equal the range.

For a real-valued formula, the natural real domain is the set of real inputs for which its operations are defined. In this chapter:

• a denominator must be nonzero;
• the radicand of a real square root must be nonnegative;
• the argument of a real logarithm must be positive.

Intersect restrictions; do not choose only the most conspicuous one.

Worked example

Worked example 2: two restrictions at once. For

$$f(x)=\frac{\sqrt{x+1}}{x-2},$$

we require $x\ge-1$ and $x\ne2$. Thus the natural domain is $[-1,2)\cup(2,\infty)$. The formula does not become defined at $2$ merely because points arbitrarily close to $2$ are permitted.

Common mistake

Two functions are equal only when their domains agree and their values agree on that domain. The formulas $(x^2-1)/(x-1)$ and $x+1$ agree when $x\ne1$, but as natural-domain functions they are not equal: the second is defined at $1$, the first is not. Cancelling a factor simplifies values where cancellation is allowed; it does not restore a deleted input.

A piecewise rule is still a function if it assigns one unambiguous value at every permitted input. The branches should not leave an unintended gap or give conflicting values where their conditions overlap.

1.3 Solving equations without losing logical control

An equation is a statement about an input. Its solution set consists of the inputs in its stated domain for which the two sides agree. Two equations are equivalent on a domain if they have the same solution set there.

Adding the same expression to both sides preserves equivalence wherever that expression is defined. Multiplication by a known nonzero expression also preserves equivalence. Multiplication by an expression that might vanish can create extra solutions. Division by an expression that might vanish can remove solutions. Squaring both sides preserves every genuine solution, but the resulting equation may have new ones because $u^2=v^2$ allows $u=-v$ as well as $u=v$.

Worked example

Worked example 3: a radical equation. Solve $\sqrt{x+1}=x-1$ over the reals. The left side is nonnegative, so any solution must satisfy $x\ge1$. Squaring gives

$$x+1=(x-1)^2\iff x^2-3x=0\iff x=0\text{ or }x=3.$$

The candidate $0$ violates $x\ge1$. Substituting $3$ in the original equation gives $2=2$. The solution set is $\{3\}$. Writing only the squared equation and both roots would not solve the original problem.

Worked example

Worked example 4: a rational inequality. Solve $(x-1)/(x+2)\ge0$. The critical numbers are $-2$, where the expression is undefined, and $1$, where it is zero. On $(-\infty,-2)$ the numerator and denominator are both negative, so the quotient is positive. On $(-2,1)$ their signs differ. On $(1,\infty)$ both are positive. Include $1$ and exclude $-2$:

$$(-\infty,-2)\cup[1,\infty).$$

Common mistake

Multiplying the original inequality by $x+2$ without splitting according to its sign would be unsafe: multiplication by a negative number reverses the inequality.

1.4 Composition and inverse functions

For $f:D\to C$ and a function $g$ whose domain contains the appropriate outputs of $f$, the composition is $(g\circ f)(x)=g(f(x))$. Its domain is precisely

$$\{x\in D:f(x)\in\operatorname{dom}(g)\}.$$

Worked example

Worked example 5: composition order. Let $f(x)=x^2-1$ on $\mathbb R$ and $g(u)=\sqrt u$ on $[0,\infty)$. Then

$$g(f(x))=\sqrt{x^2-1},\quad\operatorname{dom}(g\circ f)=(-\infty,-1]\cup[1,\infty).$$

In the reverse order, $f(g(x))=x-1$ on $[0,\infty)$. Although the simplified formula $x-1$ is meaningful on all of $\mathbb R$, the composition is not. The original inner function still controls its domain.

Proposition

A function is injective if different inputs have different outputs: $f(x_1)=f(x_2)$ implies $x_1=x_2$. An injective function has an inverse from its range to its domain. The inverse $f^{-1}$ satisfies $f^{-1}(f(x))=x$ and $f(f^{-1}(y))=y$ on the respective domains. The notation $f^{-1}$ does not mean $1/f$.

Proof

Let $f:D\to C$ be injective. For each $y\in f(D)$, the definition of range gives an $x\in D$ with $f(x)=y$, and injectivity makes that $x$ unique. Define $g(y)$ to be this unique input. Then $g:f(D)\to D$ is a function and $g(f(x))=x$ for every $x\in D$. Also $f(g(y))=y$ for every $y\in f(D)$ by its definition. Thus $g=f^{-1}$ has the claimed domain and both inverse identities. Conversely, a function $g$ satisfying $g\circ f=\operatorname{id}_D$ forces injectivity: apply $g$ to $f(x_1)=f(x_2)$. No surjectivity onto the originally declared codomain $C$ is needed when the inverse is defined only on $f(D)$.

The function $x\mapsto x^2$ on $\mathbb R$ is not injective because $2$ and $-2$ have the same square. On $[0,\infty)$ it is injective and its inverse is $y\mapsto\sqrt y$. A restriction of domain is a mathematical choice, not an algebraic trick.

Worked example

Worked example 6: calibrating an instrument. A dimensionless calibration rule is $y=3x/(1+x)$ for $x\ge0$. Its denominator is positive. If two inputs give the same output, cross multiplication gives $3x_1(1+x_2)=3x_2(1+x_1)$ and hence $x_1=x_2$. Thus it is injective. Solving for the input gives

$$y+yx=3x\implies x=\frac{y}{3-y}.$$

For $x\ge0$, $0\le y<3$; conversely every $y$ in that interval produces a nonnegative $x$. The inverse is therefore defined on $[0,3)$, not on all real numbers except $3$.

1.5 Exponentials and logarithms as inverse models

For a base $a>0$ with $a\ne1$, the exponential $a^x$ is positive and injective on $\mathbb R$. Its inverse is $\log_a x$, defined for $x>0$. In particular, $\ln x$ uses the base $e$. The laws

$$\ln(uv)=\ln u+\ln v,\qquad \ln(u^p)=p\ln u$$

hold for positive $u,v$ and real $p$. There is no corresponding law $\ln(u+v)=\ln u+\ln v$.

A non-calculus construction of positive real powers, their positive range and their monotonicity is given in the real-power foundation. Chapter 9 derives the logarithm laws from that construction; the present introductory section uses those results.

In a decay model $C(t)=C_0e^{-kt}$, assume $C_0>0$, $k>0$, and $t\ge0$. To find the time at which the amount is a fraction $r$ of its initial amount, where $0<r\le1$, solve $e^{-kt}=r$. Taking logarithms yields $t=-\ln r/k$. Since $\ln r\le0$, the answer is nonnegative as required. The quantity $kt$ must be dimensionless; if time is measured in hours, $k$ has units per hour.

1.6 A model is a statement with assumptions

Proof

Consider two salt solutions mixed without loss or reaction. Let their concentrations be $c_1,c_2$ in grams per litre and their volumes be $V_1,V_2>0$ in litres. If volumes are additive, conservation of salt gives

$$c=\frac{c_1V_1+c_2V_2}{V_1+V_2}.$$

The numerator is a mass and the denominator a volume. When $c_1\le c_2$, the result lies in $[c_1,c_2]$, since it is a weighted average with positive weights summing to one. This bound is a useful error check.

Proof

The weighted-average bound can be checked directly, without a graphical assumption. For $V_1,V_2>0$ and $c_1\le c_2$,

$$c-c_1=\frac{(c_2-c_1)V_2}{V_1+V_2}\ge0,\qquad c_2-c=\frac{(c_2-c_1)V_1}{V_1+V_2}\ge0.$$

Hence $c_1\le c\le c_2$. If one volume is allowed to be zero but the total remains positive, the same proof holds and includes pure-solution endpoints. The algebraic bound holds for ordered real $c_1,c_2$; a physical concentration model additionally uses nonnegative concentrations.

Worked example

Worked example 7: designing a mixture. Mix a $2$ g/L solution and a $7$ g/L solution to obtain $10$ L at $5$ g/L. Let $v$ be the litres of the second solution. Then the first has volume $10-v$, and

$$2(10-v)+7v=5(10)\implies5v=30\implies v=6.$$

Use $6$ L of the second solution and $4$ L of the first. The admissibility condition $0\le v\le10$ is satisfied. A requested target above $7$ g/L would have no admissible solution under this model even if an unrestricted linear equation produced a number.

1.7 Graded exercises

Foundation

Checkpoint

Exercise 1

1. Write the solution of $|2x-3|\le5$ as an interval.

Checkpoint

Exercise 2

2. Find the natural real domain of $h(x)=\sqrt{4-x}/(x^2-1)$.

Checkpoint

Exercise 3

3. Solve $\sqrt{2x+3}=x$ over $\mathbb R$ and check every candidate.

Checkpoint

Exercise 4

4. Solve $(x+3)/(x-2)<0$.

Core

Checkpoint

Exercise 5

5. Let $f(x)=1/(x-1)$ and $g(u)=\sqrt{u+2}$. Find both compositions and their domains.

Checkpoint

Exercise 6

6. On $[0,\infty)$ let $F(x)=2x/(x+3)$. Prove that $F$ is injective, find its range, and give its inverse with its domain.

Checkpoint

Exercise 7

7. A quantity obeys $Q(t)=80e^{-0.2t}$ for $t\ge0$, with $t$ in hours. Find the exact time at which $Q=20$ and state the units of $0.2$.

Synthesis

Checkpoint

Exercise 8

8. A mixture uses solutions of $4$ g/L and $10$ g/L to make $15$ L at $8$ g/L. Find both volumes. Explain, without solving a second linear equation, why a target of $12$ g/L is impossible.

Checkpoint

Exercise 9

9. A student solves $x(x-2)=0$ by dividing by $x$ and obtaining $x=2$. Identify the precise lost case. Then explain why the cancellation $(x^2-4)/(x-2)=x+2$ is valid at $x=3$ but does not define the left side at $x=2$.

Checkpoint

Exercise 10

10. Suppose $c_1<c_2$ and $V>0$ is fixed. Let $c(v)=[c_1(V-v)+c_2v]/V$ for $0\le v\le V$. Prove algebraically that this function is injective and that its range is exactly $[c_1,c_2]$. Find its inverse. Interpret the endpoints.

1.8 Full solutions

Solution · Exercise 1

1. $-5\le2x-3\le5$, so $-2\le2x\le8$ and $x\in[-1,4]$.

Solution · Exercise 2

2. The square root requires $x\le4$ and the denominator requires $x\ne-1,1$. The domain is $(-\infty,-1)\cup(-1,1)\cup(1,4]$.

Solution · Exercise 3

3. A solution must have $x\ge0$. Squaring gives $2x+3=x^2$, or $(x-3)(x+1)=0$. The candidates are $3$ and $-1$; only $3$ is nonnegative, and $\sqrt9=3$. The solution is $x=3$.

Solution · Exercise 4

4. The quotient is negative exactly when numerator and denominator have opposite signs, which occurs for $-3<x<2$. Neither endpoint is allowed: at $-3$ the quotient is zero, and at $2$ it is undefined.

Solution · Exercise 5

5. $(g\circ f)(x)=\sqrt{1/(x-1)+2}=\sqrt{(2x-1)/(x-1)}$. Require $x\ne1$ and $(2x-1)/(x-1)\ge0$. A sign table gives $(-\infty,1/2]\cup(1,\infty)$. In the reverse order, $(f\circ g)(x)=1/(\sqrt{x+2}-1)$; require $x\ge-2$ and $x\ne-1$. Its domain is $[-2,-1)\cup(-1,\infty)$.

Solution · Exercise 6

6. Equality of outputs gives $x_1(x_2+3)=x_2(x_1+3)$, so $3x_1=3x_2$. Also $F(x)=2-6/(x+3)$ lies in $[0,2)$ for $x\ge0$. For any $y\in[0,2)$, solving gives $x=3y/(2-y)\ge0$, so every such output occurs. Thus $F^{-1}(y)=3y/(2-y)$ on $[0,2)$.

Solution · Exercise 7

7. $e^{-0.2t}=1/4$, so $t=\ln4/0.2=5\ln4$ hours. The coefficient has units h$^{-1}$; its product with $t$ has no units.

Solution · Exercise 8

8. If $v$ is the volume at $10$ g/L, then $4(15-v)+10v=8(15)$. Hence $6v=60$, so $v=10$ L and the other volume is $5$ L. Every admissible mixture is a weighted average of $4$ and $10$ and cannot have concentration $12$.

Solution · Exercise 9

9. Division assumes $x\ne0$ and loses the genuine solution $x=0$. A correct case split gives $x=0$ or, if $x\ne0$, $x-2=0$. The rational cancellation is valid only for $x\ne2$, because it divides numerator and denominator by $x-2$. At $3$ this factor is nonzero. At $2$ the original quotient contains division by zero even though the simplified formula has the value $4$.

Solution · Exercise 10

10. Write $c(v)=c_1+(c_2-c_1)v/V$. If $c(v_1)=c(v_2)$, then $(c_2-c_1)(v_1-v_2)/V=0$; the nonzero coefficient forces $v_1=v_2$. For $v\in[0,V]$, the fraction $v/V$ lies in $[0,1]$, so $c(v)\in[c_1,c_2]$. Conversely $v=V(c-c_1)/(c_2-c_1)$ lies in $[0,V]$ whenever $c\in[c_1,c_2]$ and produces that concentration. This is the inverse. The endpoints mean using only the first or only the second solution.

Reading a model's domain from a graph

A graph has an input axis and an output axis, and each axis needs its own unit or dimensionless-ratio label. Before reading a height, decide whether the corresponding horizontal input belongs to the model domain. A dashed continuation can show what an algebraic formula would produce outside that domain without declaring those values scientifically admissible. The ends of a plotting window are not automatically endpoints of the mathematical domain.

For a horizontal tolerance boundary, first identify which part of the admissible curve lies on the permitted side. Then justify the crossings and their inclusion using the formula and the strict or non-strict comparison. This avoids treating a visually approximate crossing as an exact measured fact. The restricted parabola in Chapter 3's graphical workshop G1 makes these three sets visible separately: the algebraic curve, the operating domain and the inputs meeting the tolerance.

Summary

• A function includes a domain, not merely a formula. Domain restrictions survive simplification and composition.
• State the assumptions that make an equation transformation reversible. Check candidates after squaring.
• An inverse exists on the range when the original function is injective. It is not a reciprocal.
• Exponential models require dimensionless exponents; logarithms invert them only on positive arguments.
• An admissible scientific answer satisfies both the equation and the model's constraints.

Prerequisite map. Interval and sign reasoning supports trigonometric domains and later limits. Composition and inverse functions support logarithmic and trigonometric inverse rules. Conservation equations lead to linear systems and matrix representation.