Appendix A — Precalculus Review

Calculus is built on top of algebra, functions, exponentials, logarithms, trigonometry, and coordinate geometry. If those foundations are shaky, calculus feels harder than it really is — not because the calculus is hard, but because every problem secretly asks you to factor, simplify a rational expression, or recall a trig identity while you're also learning a brand-new idea like the limit or the derivative.

This appendix is a focused refresher. It will not reteach precalculus from scratch — for that, see the free resources in Section A.7 — but it collects the tools you will actually reach for, with worked examples and practice problems (answers hidden in expandable blocks so you can test yourself first). Throughout, look for the "Why this matters for calculus" notes: they tell you exactly where each topic returns.

A word of encouragement: nobody is born knowing this. The students who succeed in calculus are not the ones who never forgot precalculus — they are the ones who keep this kind of reference nearby and look things up without shame. Use it that way.

A.1 Algebra

Algebra is the language. Differentiation and integration both produce expressions you then have to simplify, and a surprising amount of "calculus difficulty" is really algebra rust.

Factoring

The core identities:

$$a^2 - b^2 = (a-b)(a+b)$$ $$a^2 + 2ab + b^2 = (a+b)^2, \qquad a^2 - 2ab + b^2 = (a-b)^2$$ $$a^3 + b^3 = (a+b)(a^2 - ab + b^2), \qquad a^3 - b^3 = (a-b)(a^2 + ab + b^2)$$

For a monic quadratic $x^2 + bx + c$, look for two numbers that multiply to $c$ and add to $b$.

Worked example. Factor $x^2 - 5x + 6$. Two numbers multiplying to $6$ and adding to $-5$ are $-2$ and $-3$. So $$x^2 - 5x + 6 = (x-2)(x-3).$$

Worked example. Factor $6x^2 + 7x - 3$ (non-monic). Multiply $a \cdot c = 6 \cdot (-3) = -18$; find two numbers multiplying to $-18$ and adding to $7$: those are $9$ and $-2$. Split the middle term and group: $$6x^2 + 9x - 2x - 3 = 3x(2x+3) - 1(2x+3) = (3x-1)(2x+3).$$

Why this matters for calculus. Limits of the form $0/0$ are almost always resolved by factoring and cancelling — e.g. $\lim_{x\to 2}\frac{x^2-4}{x-2} = \lim_{x\to 2}(x+2) = 4$. Factoring also reveals roots, which become critical points in optimization (Ch. 4) and intercepts in curve sketching.

Rational expressions

Treat them like fractions: factor everything first, cancel common factors, and find a common denominator before adding.

Worked example. Simplify $\dfrac{x^2 - 4}{x^2 - x - 2}$. Factor top and bottom: $$\frac{x^2-4}{x^2-x-2} = \frac{(x-2)(x+2)}{(x-2)(x+1)} = \frac{x+2}{x+1}, \quad x \neq 2.$$ (The restriction $x\neq 2$ matters — that's exactly the kind of "hole" a limit detects.)

Worked example. Combine $\dfrac{1}{x} - \dfrac{1}{x+h}$ over a common denominator: $$\frac{1}{x} - \frac{1}{x+h} = \frac{(x+h) - x}{x(x+h)} = \frac{h}{x(x+h)}.$$

Why this matters for calculus. That last manipulation is the difference quotient for $f(x)=1/x$. The whole definition of the derivative (Ch. 3) lives or dies on combining fractions cleanly.

Exponents and radicals

$$a^m a^n = a^{m+n}, \quad \frac{a^m}{a^n} = a^{m-n}, \quad (a^m)^n = a^{mn}, \quad a^0 = 1, \quad a^{-n} = \frac{1}{a^n}$$ $$(ab)^n = a^n b^n, \qquad a^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m$$

Worked example. Rewrite $\dfrac{\sqrt{x}}{x^2}$ as a single power: $\dfrac{x^{1/2}}{x^2} = x^{1/2 - 2} = x^{-3/2}$.

Why this matters for calculus. The power rule $\frac{d}{dx}x^n = nx^{n-1}$ (Ch. 3) only applies once you've rewritten roots and reciprocals as powers. You cannot differentiate $\sqrt{x}$ until you see it as $x^{1/2}$.

The quadratic formula

For $ax^2 + bx + c = 0$ with $a \neq 0$: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$ The discriminant $b^2 - 4ac$ tells you the number of real roots: positive → two, zero → one (repeated), negative → none (two complex).

Worked example. Solve $2x^2 - 5x + 1 = 0$. Here $a=2,\,b=-5,\,c=1$: $$x = \frac{5 \pm \sqrt{25 - 8}}{4} = \frac{5 \pm \sqrt{17}}{4}.$$

Completing the square

To rewrite $x^2 + bx + c$ as a perfect square plus a constant, add and subtract $(b/2)^2$: $$x^2 + bx + c = \left(x + \tfrac{b}{2}\right)^2 + c - \tfrac{b^2}{4}.$$

Worked example. $x^2 + 6x + 5 = (x+3)^2 - 9 + 5 = (x+3)^2 - 4$.

For a leading coefficient $a\neq 1$, factor it out first: $2x^2 + 8x + 1 = 2(x^2 + 4x) + 1 = 2[(x+2)^2 - 4] + 1 = 2(x+2)^2 - 7$.

Why this matters for calculus. Completing the square is the standard move for integrals with an irreducible quadratic under a root or in a denominator (Ch. 16, integration techniques and trig substitution) and for putting conics into standard form (Ch. 27). When you see $\int \frac{dx}{x^2+6x+13}$, your first instinct should be to write the denominator as $(x+3)^2 + 4$.

Inequalities and absolute value

Solving inequalities is like solving equations, with one rule: multiplying or dividing by a negative number flips the inequality sign. For products and quotients, find the values that make each factor zero, then test the sign on each interval.

Worked example. Solve $x^2 - x - 6 > 0$. Factor: $(x-3)(x+2) > 0$. The roots are $-2$ and $3$. Testing intervals, the product is positive when $x < -2$ or $x > 3$. Solution: $(-\infty,-2)\cup(3,\infty)$.

Absolute value measures distance from zero: $|x| = a$ (for $a\ge 0$) means $x = \pm a$, and $|x| < a$ means $-a < x < a$.

Worked example. Solve $|2x - 1| \le 5$: $-5 \le 2x - 1 \le 5 \Rightarrow -4 \le 2x \le 6 \Rightarrow -2 \le x \le 3$.

Why this matters for calculus. The formal $\varepsilon$–$\delta$ definition of a limit (Ch. 2) is stated entirely in absolute-value inequalities: $|f(x) - L| < \varepsilon$ whenever $0 < |x - a| < \delta$. Comfort with $|\cdot|$ inequalities makes that definition readable instead of terrifying.

Partial-fraction setup

You won't decompose fractions until integration (Ch. 16), but the setup is pure algebra, so practice it now. A proper rational function (degree of top < degree of bottom) splits according to its denominator's factors:

• Each distinct linear factor $(x - r)$ contributes $\dfrac{A}{x - r}$.
• A repeated factor $(x-r)^2$ contributes $\dfrac{A}{x-r} + \dfrac{B}{(x-r)^2}$.
• Each irreducible quadratic $(x^2 + px + q)$ contributes $\dfrac{Bx + C}{x^2 + px + q}$.

Worked example. Set up (don't solve) the decomposition of $\dfrac{3x+1}{(x-1)(x+2)}$: $$\frac{3x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}.$$ Multiplying through by the denominator gives $3x+1 = A(x+2) + B(x-1)$. Setting $x=1$: $4 = 3A$, so $A = 4/3$. Setting $x=-2$: $-5 = -3B$, so $B = 5/3$.

Practice. 1. Factor $x^3 - 8$. 2. Simplify $\dfrac{x^2 - 9}{x^2 + 6x + 9}$. 3. Complete the square: $x^2 - 10x + 21$. 4. Solve $\dfrac{x-1}{x+2} \ge 0$. 5. Set up the partial-fraction form of $\dfrac{x}{(x+1)(x^2+4)}$.

A.2 Functions

A function $f$ assigns to each input exactly one output. Calculus is the study of how functions change, so you must be fluent in describing, combining, and transforming them.

Domain and range

The domain is the set of allowable inputs; the range is the set of resulting outputs. The two domain rules you'll use constantly: you cannot divide by zero, and you cannot take an even root (or a logarithm) of a negative number.

Worked example. The domain of $f(x) = \dfrac{1}{\sqrt{x-3}}$ requires $x - 3 > 0$ (strictly positive, since it's under a root and in a denominator), so the domain is $(3, \infty)$.

Composition

$$(f \circ g)(x) = f(g(x)) \quad\text{— do } g \text{ first, then } f.$$

Worked example. If $f(x) = x^2 + 1$ and $g(x) = \sqrt{x}$, then $(f\circ g)(x) = (\sqrt{x})^2 + 1 = x + 1$ (for $x\ge 0$), while $(g\circ f)(x) = \sqrt{x^2+1}$. Order matters.

Why this matters for calculus. The chain rule (Ch. 7) differentiates compositions: $\frac{d}{dx}f(g(x)) = f'(g(x))\,g'(x)$. To use it, you must first recognize a function as a composition — spotting that $\sqrt{x^2+1}$ is "square-root of (inside)" is the whole game.

Inverses

$f^{-1}$ undoes $f$: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. An inverse exists only when $f$ is one-to-one (passes the horizontal line test). To find it, swap $x$ and $y$ and solve for $y$.

Worked example. For $f(x) = 2x + 3$: write $y = 2x+3$, swap to $x = 2y+3$, solve $y = \frac{x-3}{2}$. So $f^{-1}(x) = \frac{x-3}{2}$. The graph of $f^{-1}$ is the reflection of $f$ across the line $y = x$.

Why this matters for calculus. The natural log is the inverse of the exponential; the inverse trig functions (below) are inverses of restricted sine, cosine, tangent. The derivative of an inverse function (Ch. 7) relies on this relationship.

Even, odd, and piecewise

• Even: $f(-x) = f(x)$ (symmetric about the $y$-axis; e.g. $x^2$, $\cos x$).
• Odd: $f(-x) = -f(x)$ (symmetric about the origin; e.g. $x^3$, $\sin x$).

A piecewise function uses different rules on different parts of its domain: $$f(x) = \begin{cases} x^2 & x < 0 \\ x + 1 & x \ge 0 \end{cases}$$ Here $f(-2) = 4$ and $f(0) = 1$. The "jump" at $x=0$ is exactly the kind of place where you check one-sided limits and continuity.

Why this matters for calculus. Even/odd symmetry shortcuts integrals: $\int_{-a}^{a} (\text{odd})\,dx = 0$ (Ch. 13). Piecewise functions are the standard testbed for continuity and differentiability (Ch. 2–3).

Transformations

Starting from $y = f(x)$, with $c > 0$:

Note the counterintuitive part: changes inside the function (to the input $x$) act horizontally and in the opposite direction you'd expect — $f(x-c)$ shifts right, not left.

Practice. 1. Find the domain of $g(x) = \ln(4 - x^2)$. 2. With $f(x)=x^2$ and $g(x)=x-1$, compute $(f\circ g)(x)$ and $(g\circ f)(x)$. 3. Is $f(x) = x^3 - x$ even, odd, or neither? 4. Describe the transformations turning $y = x^2$ into $y = -2(x-3)^2 + 1$.

A.3 Exponentials and Logarithms

An exponential function has the form $f(x) = a^x$ with base $a > 0$, $a \neq 1$. A logarithm is its inverse: $\log_a x = y \iff a^y = x$. Logarithms answer "what exponent produces this number?"

The number $e$ and the natural log

The special base $e \approx 2.71828$ makes calculus clean: the natural log $\ln x = \log_e x$ is the logarithm calculus prefers because $\frac{d}{dx}e^x = e^x$ and $\frac{d}{dx}\ln x = \frac{1}{x}$.

Laws

$$\log_a(MN) = \log_a M + \log_a N \qquad \log_a\!\left(\frac{M}{N}\right) = \log_a M - \log_a N$$ $$\log_a(M^p) = p\,\log_a M \qquad \log_a 1 = 0 \qquad \log_a a = 1$$

Inverse relationships (the most useful pair in all of calculus): $$e^{\ln x} = x \;(x>0), \qquad \ln(e^x) = x \;(\text{all } x).$$

Solving equations

To solve an exponential equation, take a log of both sides; to solve a log equation, exponentiate both sides.

Worked example (exponential). Solve $3 \cdot 2^x = 48$. Divide: $2^x = 16$, so $x = 4$. If the answer isn't clean, use logs: $2^x = 20 \Rightarrow x = \dfrac{\ln 20}{\ln 2}$.

Worked example (log). Solve $\ln(x) + \ln(x - 3) = \ln 10$. Combine: $\ln[x(x-3)] = \ln 10 \Rightarrow x^2 - 3x - 10 = 0 \Rightarrow (x-5)(x+2)=0$. The candidate $x=-2$ is rejected (negative input to $\ln$), so $x = 5$.

Common pitfall. $\ln(a + b)$ does not equal $\ln a + \ln b$. The product law applies to a product inside the log, not a sum. Always check domain restrictions after solving log equations — extraneous solutions are routine.

Change of base

$$\log_a b = \frac{\ln b}{\ln a} = \frac{\log_{10} b}{\log_{10} a}.$$

Worked example. $\log_2 10 = \dfrac{\ln 10}{\ln 2} \approx \dfrac{2.3026}{0.6931} \approx 3.3219$.

Why this matters for calculus. Logarithmic differentiation (Ch. 7) uses the log laws to turn products and powers into sums before differentiating — it's how you differentiate things like $x^x$. Exponentials model growth and decay (Ch. 9, differential equations), and $\ln$ is the antiderivative of $1/x$ (Ch. 13).

Practice. 1. Solve $5^{x+1} = 125$. 2. Write $\ln\dfrac{x^2\sqrt{y}}{z^3}$ as a sum/difference of logs. 3. Solve $\log_3(x) + \log_3(x+6) = 3$. 4. Evaluate $\log_5 40$ using change of base (to 4 decimals).

A.4 Trigonometry

More calculus students lose points to rusty trig than to anything else. Invest here.

Radians versus degrees

Calculus uses radians, always. A radian is the angle subtending an arc equal to the radius, and a full circle is $2\pi$ radians $= 360°$. Convert with $$\text{radians} = \text{degrees}\cdot\frac{\pi}{180}, \qquad \text{degrees} = \text{radians}\cdot\frac{180}{\pi}.$$ So $90° = \pi/2$, $180° = \pi$, $30° = \pi/6$. (The clean derivative formulas $\frac{d}{dx}\sin x = \cos x$ hold only in radians.)

The unit circle and the six functions

On the unit circle, a point at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$. The other four functions are ratios: $$\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \csc\theta = \frac{1}{\sin\theta}.$$

Key values

(Tangent is undefined where $\cos\theta = 0$.) Memory aid: for $\theta = 0,\tfrac{\pi}{6},\tfrac{\pi}{4},\tfrac{\pi}{3},\tfrac{\pi}{2}$, the sines are $\tfrac{\sqrt{0}}{2},\tfrac{\sqrt{1}}{2},\tfrac{\sqrt{2}}{2},\tfrac{\sqrt{3}}{2},\tfrac{\sqrt{4}}{2}$ — and cosine runs the same list backward.

Fundamental identities

$$\sin^2\theta + \cos^2\theta = 1, \qquad 1 + \tan^2\theta = \sec^2\theta, \qquad 1 + \cot^2\theta = \csc^2\theta$$

Angle-sum identities: $$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$$ $$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$$

Double-angle identities (set $B = A$ above): $$\sin(2A) = 2\sin A\cos A$$ $$\cos(2A) = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$$

The last form rearranges into the power-reduction identities, which you'll use to integrate $\sin^2$ and $\cos^2$: $$\cos^2 A = \frac{1 + \cos 2A}{2}, \qquad \sin^2 A = \frac{1 - \cos 2A}{2}.$$

Why this matters for calculus. The Pythagorean identity is the engine of trigonometric substitution and the simplification of trig integrals (Ch. 16). You cannot integrate $\int \cos^2 x\,dx$ without the power-reduction identity, and you cannot finish a trig substitution like $x = \sin\theta$ without $\sin^2 + \cos^2 = 1$.

Graphs: period and amplitude

For $y = A\sin(Bx + C) + D$: - Amplitude $= |A|$ (vertical half-height), - Period $= \dfrac{2\pi}{|B|}$, - horizontal shift (phase) $= -C/B$, vertical shift $= D$.

Sine and cosine have period $2\pi$ and range $[-1,1]$; tangent has period $\pi$.

Worked example. $y = 3\sin(2x)$ has amplitude $3$ and period $\dfrac{2\pi}{2} = \pi$.

Inverse trig

Because sine, cosine, and tangent are not one-to-one, we restrict their domains before inverting:

Worked example. $\arcsin\!\left(\tfrac{1}{2}\right) = \tfrac{\pi}{6}$ (the angle in $[-\tfrac\pi2,\tfrac\pi2]$ whose sine is $\tfrac12$). Note $\arccos(-1) = \pi$.

Why this matters for calculus. Inverse trig functions are antiderivatives you'll memorize: $\int \frac{dx}{1+x^2} = \arctan x + C$ and $\int \frac{dx}{\sqrt{1-x^2}} = \arcsin x + C$ (Ch. 13). They also appear as the output of trig substitution (Ch. 16).

Practice. 1. Convert $135°$ to radians. 2. Evaluate $\cos\!\left(\tfrac{2\pi}{3}\right)$. 3. Use a double-angle identity to write $\sin(2x)$ if $\sin x = \tfrac{3}{5}$ and $\cos x = \tfrac{4}{5}$. 4. State the amplitude and period of $y = -2\cos\!\left(\tfrac{x}{3}\right)$. 5. Evaluate $\arctan(1)$.

A.5 Coordinate Geometry

Calculus is geometry made dynamic: derivatives are slopes, integrals are areas. You need the static geometry first.

Lines

$$\text{slope } m = \frac{y_2 - y_1}{x_2 - x_1}$$ - Slope-intercept: $y = mx + b$. - Point-slope: $y - y_1 = m(x - x_1)$.

Parallel lines have equal slopes; perpendicular lines have slopes that are negative reciprocals ($m_1 m_2 = -1$).

Worked example. The line through $(1, 2)$ and $(4, 11)$ has slope $m = \frac{11-2}{4-1} = 3$, so $y - 2 = 3(x-1)$, i.e. $y = 3x - 1$.

Why this matters for calculus. The derivative at a point is the slope of the tangent line (Ch. 3), and you'll write its equation using point-slope form constantly. The perpendicular slope gives the normal line.

Distance and midpoint

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, \qquad \text{midpoint} = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right).$$

Worked example. The distance between $(1, 2)$ and $(4, 6)$ is $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

Circles

$$(x - h)^2 + (y - k)^2 = r^2 \quad\text{— center } (h,k),\ \text{radius } r.$$ If a circle is given in expanded form, complete the square (Section A.1) to recover center and radius.

Worked example. Put $x^2 + y^2 - 6x + 4y - 3 = 0$ in standard form. Group and complete the square: $$(x^2 - 6x) + (y^2 + 4y) = 3 \;\Rightarrow\; (x-3)^2 - 9 + (y+2)^2 - 4 = 3 \;\Rightarrow\; (x-3)^2 + (y+2)^2 = 16.$$ Center $(3, -2)$, radius $4$.

Conics preview

The conic sections — parabolas, ellipses, hyperbolas — are studied in depth in Ch. 27. Standard forms (centered at the origin):

Completing the square is again the tool that converts a messy second-degree equation into one of these recognizable forms.

Practice. 1. Find the equation of the line through $(2, -1)$ perpendicular to $y = \tfrac{1}{2}x + 4$. 2. Find the distance between $(-2, 3)$ and $(4, -5)$. 3. Find the center and radius of $x^2 + y^2 + 8x - 2y + 8 = 0$.

A.6 Quick Reference: Where Each Topic Returns

A.7 Where to Go Deeper

If a section above moved too fast, these free resources are excellent and have hundreds of practice problems with worked solutions:

• Khan Academy — search for Algebra, Precalculus, and Trigonometry. Short videos plus instant-feedback exercises; ideal for targeted review of a single weak spot.
• OpenStax College Algebra — free, full textbook covering Sections A.1–A.3 and A.5 in depth (openstax.org).
• OpenStax Precalculus — free, full textbook with thorough trigonometry coverage (Section A.4) and a chapter on conics.
• OpenStax Algebra and Trigonometry — combines the above into one volume if you want a single reference.

Keep this appendix bookmarked. When a calculus problem stalls, ask first: is this actually a calculus difficulty, or an algebra/trig one? More often than you'd guess, it's the latter — and that's good news, because it means the fix is right here.