Stewart Calculus · 8th Edition

Chapter 1: Functions & Models

Every key formula from Stewart Calculus Chapter 1, in one searchable page. Click a card to study it — worked examples included.

1.1  ·  FUNCTION BASICS
X Y ✓ each x → exactly one y X Y ✗ one x → two y’s
✓ passes: one crossing ✗ fails: two crossings
Symmetry
Even: f(−x) = f(x) · Odd: f(−x) = −f(x)
even · y-axis odd · origin
KEY NOTES
  • Test f(−x) to check even/odd — don’t trust your eyes alone.
WORKED EXAMPLE
Find the domain of f(x) = x + 2/(x − 1).
Need x + 2 ≥ 0 and x ≠ 1 ⇒ [−2, 1) ∪ (1, ∞).
1.2  ·  ALGEBRAIC FUNCTIONS
rise run m = rise / run x = −b/2a vertex
Linear
f(x) = mx + b,   m = y2 − y1x2 − x1
x² x³ y = 1/x
Power · Rational · Root
f(x) = xa · f(x) = P(x)/Q(x), Q(x) ≠ 0
WORKED EXAMPLE
Line through (2, −1) and (4, 5): m = (5 − (−1))/(4 − 2) = 3.
⇒ y + 1 = 3(x − 2), i.e. y = 3x − 7.
QUICK CHECKS
x − 3 → shifts RIGHT 3 f(x) f(x − 3)
  • Domain first: denominators ≠ 0, even roots need radicand ≥ 0, logs need argument > 0.
  • Not sure it’s a function? Run the Vertical Line Test.
1.2  ·  TRANSCENDENTAL FUNCTIONS
0 π 2π sin x cos x tan x −π/2 π/2
Trigonometric
sin x, cos x: period 2π, range [−1, 1]
sin²x + cos²x = 1 · tan x = sin xcos x, period π, asymptotes x = π/2 + nπ
Exponential
(0, 1) y = b^x b > 1 grows (1, 0) y = log_b x mirror across y = x
f(x) = bx (b>0, b≠1): domain ℝ, range (0, ∞), through (0, 1)
Logarithm
y = logb x ⇔ by = x; domain (0, ∞) · ln x = loge x
Laws of logarithms
log xy = log x + log y · log(x/y) = log x − log y
log xr = r log x
KEY NOTES
  • eln x = x (x>0) and ln(ex) = x — they undo each other.
1.3  ·  TRANSFORMATIONS
+2 f(x) → f(x) + 2 −f(x) reflect over x-axis
Shifts
y = f(x) + c up c · y = f(x) − c down c
y = f(x − c) right c · y = f(x + c) left c
Stretches
y = c·f(x): vertical ×c · y = f(cx): horizontal ×1/c
Reflections
y = −f(x) over the x-axis · y = f(−x) over the y-axis
WORKED EXAMPLE
From y = x2: shift right 2, reflect over the x-axis, shift up 1.
y = x2 → y = (x − 2)2 → y = −(x − 2)2 + 1.
1.3  ·  COMBINING & COMPOSITION
x g g(x) f f(g(x)) work inside → out
WORKED EXAMPLE
f(x) = √x, g(x) = x + 1:
(f∘g)(x) = √(x+1), x ≥ −1 · (g∘f)(x) = √x + 1, x ≥ 0
⇒ f∘g ≠ g∘f — order matters!
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