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The Timpson Lyceum · Calculus 1 (2024)

Review Functions, Function Operations, & Graphing - Lecture 2 (Calculus 1)

Overview

This lecture reviews the foundational language of functions that the rest of calculus is built on, working through sections 1.1 and 1.2 of the text. It defines what a function is, how to read off its domain and range, and how to recognize and graph the standard families of functions you will use all year: linear, power, polynomial, rational, exponential, logarithmic, and trigonometric. It also covers the tools for describing and building functions — piecewise definitions, even/odd symmetry, increasing/decreasing behavior, the algebraic combinations (sum, difference, product, quotient, composition), and the geometric transformations (shifts, stretches, compressions, reflections). The material is mostly review, so the goal is to surface any gaps and fix them before calculus proper begins.

Key Concepts

Function, domain, and range

Natural domain

Graph and the vertical line test

Piecewise functions

Increasing / decreasing

Even / odd symmetry

Equations of a line (all equivalent; choose by what you know)

Families of functions

Sine and cosine graphs

Combining functions

Transformations (for constant k,c>0k, c > 0)

Worked Example

Composition and its domain. Let f(x)=xf(x) = \sqrt{x} and g(x)=x+1g(x) = x + 1. Compute (f∘g)(x)(f \circ g)(x) and find its domain. The rule is (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)), evaluate gg first, then substitute into ff:

(f∘g)(x)=f(g(x))=f(x+1)=x+1domain of g: (−∞,∞),domain of f: [0,∞)need g(x)≥0: x+1≥0 ⟹ x≥−1domain of f∘g: [−1,∞)\begin{aligned} (f \circ g)(x) &= f\big(g(x)\big) = f(x+1) = \sqrt{x+1} \\ \text{domain of } g &: \ (-\infty, \infty), \qquad \text{domain of } f : \ [0, \infty) \\ \text{need } g(x) \ge 0 &: \ x + 1 \ge 0 \ \Longrightarrow\ x \ge -1 \\ \text{domain of } f \circ g &: \ [-1, \infty) \end{aligned}

The domain is not all reals: even though gg accepts every real number, only inputs with x≥−1x \ge -1 produce an output g(x)≥0g(x) \ge 0 that f= ⋅ f = \sqrt{\,\cdot\,} can accept.

Summary