0. Functions and Coordinate Preparation

Learning goals and assumed background

This preparatory chapter assumes secondary-school algebra and elementary geometry. It makes explicit the function language used throughout the engineering-calculus chapters: domains, compositions, inverse branches, exponential and trigonometric models, and polar coordinates. The aim is not to replace the assumed secondary-school background but to prevent hidden domain or unit errors from contaminating later calculus.

0.1 Domains, ranges, and transformations

A function $f:D\to C$ assigns one output in $C$ to each input in $D$. The range $f(D)$ may be smaller than the declared codomain $C$. Two formulas that agree wherever both are defined need not define the same function if their domains differ.

For real formulas, denominators cannot be zero, even roots require nonnegative radicands, and logarithms require positive arguments. Intersect all restrictions. In a model, restrict further for physically meaningful quantities.

Worked example 1: intersect the restrictions. For $f(x)=\sqrt{\ln x}/(x-2)$, require $x>0$, $\ln x\ge0$, and $x\ne2$. The first two combine to $x\ge1$, so the domain is $[1,2)\cup(2,\infty)$. The denominator restriction is not removed by the logarithm restriction.

From a known graph $y=f(x)$, the graph of $f(x-a)+b$ is shifted right by $a$ and up by $b$. The graph of $cf(x)$ scales output by $c$; negative $c$ also reflects vertically. For $f(kx)$ with $k\ne0$, input intervals are scaled by $1/k$ and reflected horizontally when $k<0$. An input transformation acts before the function, so it changes both shape and domain.

The composition $g\circ f$ has domain $\{x\in D:f(x)\in\operatorname{dom}g\}$. Its order matters: exponentiating and then squaring is not the same operation as squaring and then exponentiating, for example $e^{x^2}$ versus $e^{2x}$.

0.2 Polynomial, rational, and power functions

A polynomial is $a_0+a_1x+\cdots+a_nx^n$ for nonnegative integer powers. It is defined for every real input. A rational function is a quotient of polynomials and excludes zeros of its original denominator. A cancelled factor can reveal a removable hole but does not fill it unless the function is explicitly redefined.

For an integer $n<0$, $x^n=1/x^{-n}$ excludes zero. For a rational power such as $x^{1/2}$, use the real root's domain and principal nonnegative value. In particular, $\sqrt{x^2}=|x|$, not $x$ on all of $\mathbb R$. For an arbitrary real exponent $a$, the standard positive-input definition is $x^a=e^{a\ln x}$ for $x>0$; some rational exponents have additional real-domain extensions that must be specified separately.

Worked example 2: an algebraic identity with a domain. The quotient $(x^2-9)/(x-3)$ equals $x+3$ for $x\ne3$. Its natural domain excludes $3$ even though the simplified expression has no visible denominator. Later, a limit at the missing point can exist without the original function having a value there.

0.3 Exponentials and logarithms

For $a>0$, $a\ne1$, the function $a^x$ is positive and one-to-one; its inverse is $\log_a x$ on $x>0$. The natural logarithm uses base $e$. For positive inputs,

$$\ln(uv)=\ln u+\ln v,\quad\ln(u/v)=\ln u-\ln v,\quad\ln(u^r)=r\ln u.$$

There is no sum rule for $\ln(u+v)$. When expressing a dimensional quantity inside a logarithm, it is best interpreted relative to a fixed reference scale so the logarithm's argument is dimensionless.

Worked example 3: solve before approximating. Solve $5e^{-0.4t}=2$ for $t\ge0$. Dividing and taking logarithms gives $t=\ln(5/2)/0.4$. The result is positive because $5/2>1$. If $t$ is in seconds, the coefficient $0.4$ has units s$^{-1}$. A numerical decimal is optional; the logarithmic expression is exact.

For $y(t)=Ae^{kt}$ with real constants $A,k$ and $A\ne0$, the model never vanishes and has constant relative rate $y'(t)/y(t)=k$. If $A=0$, then $y\equiv0$ still satisfies $y'=ky$, but its relative quotient is undefined. An affine model $A+bt$ has constant absolute rate $b$. Choosing between these models is a modeling assumption, not an algebraic preference.

0.4 Trigonometric functions and radian models

An angle in radians is signed arc length divided by radius. The unit-circle point at angle $\theta$ is $(\cos\theta,\sin\theta)$, so $\sin^2\theta+\cos^2\theta=1$. Sine and cosine have period $2\pi$; tangent has period $\pi$ and is undefined where cosine is zero.

A model $s(t)=A\sin(\omega t+\phi)$ has amplitude $|A|$ and period $2\pi/|\omega|$ when $A\ne0$, $\omega\ne0$. The argument must be dimensionless. If $t$ is seconds, $\omega$ is recorded in rad/s. Using degrees in derivative formulas would introduce an additional conversion factor; the standard calculus rules assume radians.

Worked example 4: recover a period. For $s(t)=2\cos(5t-\pi/3)$, the amplitude is $2$ and the period $2\pi/5$. The value at zero is $1$. The first maximum at nonnegative time occurs when $5t-\pi/3=0$, namely $t=\pi/15$.

To solve $\sin\theta=c$, first require $|c|\le1$, then include all periodic solutions allowed by the requested interval. A principal inverse value is only one representative; it is not the entire solution set.

0.5 Inverse functions and derivative-ready branches

A function is injective when $f(x_1)=f(x_2)$ forces $x_1=x_2$. An injective function has an inverse from its range back to its domain. Periodic trigonometric functions need restricted domains before they can be inverted:

• $\arcsin:[-1,1]\to[-\pi/2,\pi/2]$;
• $\arccos:[-1,1]\to[0,\pi]$;
• $\arctan:\mathbb R\to(-\pi/2,\pi/2)$.

Thus $\arcsin(\sin\theta)=\theta$ only when $\theta$ is in the chosen sine branch. The notation $f^{-1}$ means inverse function, not reciprocal.

When an inverse is differentiable and $f'(x)\ne0$, differentiate $f(f^{-1}(y))=y$ to get

$$(f^{-1})'(y)=\frac1{f'(f^{-1}(y))}.$$

For $y=\arcsin x$, $\sin y=x$ and the branch gives $\cos y=\sqrt{1-x^2}>0$ for $|x|<1$. Hence $d(\arcsin x)/dx=1/\sqrt{1-x^2}$. At $x=\pm1$, the inverse remains defined but does not have this finite two-sided derivative. Likewise $d(\arccos x)/dx=-1/\sqrt{1-x^2}$ for $|x|<1$, and $d(\arctan x)/dx=1/(1+x^2)$ for every real $x$.

Worked example 5: branch check. $\arctan(\tan(3\pi/4))=-\pi/4$, because tangent is $-1$ at that angle and the principal inverse must return a value between $-\pi/2$ and $\pi/2$. Ignoring the branch would return an inadmissible inverse value.

0.6 Cartesian and polar coordinates

For $r\ge0$, polar coordinates satisfy $x=r\cos\theta$, $y=r\sin\theta$. A nonzero point has $r=\sqrt{x^2+y^2}$ and direction modulo $2\pi$. At the origin every angle gives the same point, so direction is not unique. Recovering an angle from $\tan\theta=y/x$ requires quadrant information and fails directly when $x=0$.

Worked example 6: a polar curve. The curve $r=2\cos\theta$ with $r\ge0$ and $-\pi/2\le\theta\le\pi/2$ satisfies $r^2=2r\cos\theta=2x$. Thus $x^2+y^2=2x$, or $(x-1)^2+y^2=1$. Conversely each point on this circle has nonnegative horizontal coordinate and a compatible polar representation in the stated interval (with the origin represented by an endpoint). The algebraic conversion and the angular range together establish the curve; squaring a polar equation without checking restrictions can introduce extra points.

Polar coordinates offer a different description of the same plane, not a different physical space. A later integration formula must include the appropriate geometric measure; $dr\,d\theta$ is not automatically an area element without a scale factor.

0.7 Graded exercises

Foundation

1. Find the natural real domain of $\ln(4-x^2)$.
2. Simplify $\sqrt{(x-2)^2}$ for all real $x$.
3. Solve $3e^{2t}=12$ exactly.

Core

4. Find the domain of $g(f(x))$ when $f(x)=x^2-4$ and $g(u)=\sqrt u$.
5. Give the amplitude, period and initial value of $v(t)=-4\sin(2t+\pi/2)$.
6. Solve $\cos\theta=1/2$ for $0\le\theta<2\pi$.
7. Evaluate $\arccos(\cos(5\pi/3))$ and explain the inverse range.
8. Find positive-radius polar coordinates for $(-1,\sqrt3)$, taking $0\le\theta<2\pi$.

Synthesis

9. Differentiate $y=\arctan(2x)$ using the inverse derivative and chain rule. Explain why the result has no real denominator singularity.
10. Convert $r=4\sin\theta$, $0\le\theta\le\pi$, to Cartesian form, and verify the geometric curve without discarding the nonnegative-radius restriction.

0.8 Full solutions

1. Require $4-x^2>0$, so $-2<x<2$.
2. The result is $|x-2|$, equal to $x-2$ for $x\ge2$ and $2-x$ for $x<2$.
3. $e^{2t}=4$, hence $t=(\ln4)/2=\ln2$.
4. Require $x^2-4\ge0$, giving $(-\infty,-2]\cup[2,\infty)$.
5. Amplitude is $4$, period is $\pi$, and the initial value is $-4\sin(\pi/2)=-4$.
6. The first-turn solutions are $\theta=\pi/3$ and $5\pi/3$.
7. Cosine at $5\pi/3$ is $1/2$, and $\arccos$ returns the angle in $[0,\pi]$, namely $\pi/3$.
8. Radius is $\sqrt{1+3}=2$. The point is in quadrant II, with cosine $-1/2$ and sine $\sqrt3/2$, so $\theta=2\pi/3$.
9. The derivative is $2/(1+4x^2)$. The denominator is at least one for every real $x$, so the derivative is defined everywhere, consistent with the real domain of $\arctan(2x)$.
10. Multiply by $r$: $r^2=4r\sin\theta=4y$, giving $x^2+(y-2)^2=4$. This is the circle of radius two centered at $(0,2)$. Every point on it has $y\ge0$ and can be represented with $0\le\theta\le\pi$ and $r\ge0$; the origin occurs at the angular endpoints. Thus no extra lower-half-plane points are included.

Summary and onward map

Preserve domains through simplification and composition. Powers, logarithms, trigonometric formulas and inverse branches have distinct restrictions. Exponential and periodic models need dimensionless arguments and clear time units. Polar coordinates need quadrant and radius conventions. These preparations support the derivative, limit and integration chapters without silently importing undefined inputs.