Pre-Calculus · Units 1–9 · Quick reference

Pre-Calculus

Grades 11–126 pages

Functions and transformations, inverses, polynomial, rational, exponential and log functions, conics, matrices, parametric equations, series, counting, limits and rates of change.

1Functions2Transformations3Inverses4Polynomials5Rational6Exp & log7Conics8Matrices9Parametric10Series & binomials11Counting12Limits13Rates

1Functions & Their Featuresone input, one output, and how the graph behaves

Function
each input has exactly one output
Domain / range
allowed inputs / outputs produced
Zero
an input with f(x) = 0 (an x-intercept)
y-intercept
the point (0, f(0))
Basics
vertical line test, evaluating, f + g: Algebra §10

Interval notation

  • [a, b] includes the ends; (a, b) leaves them out
  • ∞ always takes a parenthesis; ∪ joins pieces

EXx ≤ −1 or x > 3 → (−∞, −1] ∪ (3, ∞)

EX−2 < x ≤ 5 → (−2, 5]

Domain of √(x + 3)x − 2 Worked

x + 3 ≥ 0 → x ≥ −3even root
x ≠ 2denominator
domain [−3, 2) ∪ (2, ∞)
xy(−1, 2) max(1, −2) min
f(x) = x³ − 3x rises on (−∞, −1) and (1, ∞), falls on (−1, 1).

Increasing, decreasing, extrema

  • Increasing on an interval: x₁ < x₂ gives f(x₁) < f(x₂).
  • A local max (min) is a peak (valley) of nearby values.
  • Name intervals by their x-values, with open ends.

Symmetry

  • Even: f(−x) = f(x), mirror in the y-axis.
  • Odd: f(−x) = −f(x), half-turn about the origin.

EXx³ − 3x: f(−x) = −x³ + 3x = −f(x) → odd

Difference quotient

The slope between x and x + h; it becomes the derivative in §13.

f(x + h) − f(x)h, h ≠ 0

f(x) = x² − 3x Worked

f(x + h) − f(x) = 2xh + h² − 3hexpand, cancel
÷ h → 2x + h − 3

Piecewise Worked

f(x) = x² for x < 1; 3 − x for x ≥ 1
f(−2) = 4, f(1) = 2, f(4) = −1pick the piece first

Average rate of change Worked

f(x) = x³ − 3x on [0, 2]
f(2) − f(0)2 − 0 = 2 − 02 = 1secant slope

DR. ZIPS

Describe any graph in six moves: Domain, Range, Zeros, Intervals (up or down), Peaks and valleys, Symmetry.

2Parent Functions & Transformationsa few shapes, moved and stretched

Parent functions (graphs: Algebra §10)
ParentDomainRangeSymmetry
x, x³, ∛xall realsall realsodd
x², |x|all reals[0, ∞)even
√x[0, ∞)[0, ∞)none
1xx ≠ 0y ≠ 0odd
bˣall reals(0, ∞)none
logb x(0, ∞)all realsnone
Transformations of y = f(x)
ChangeEffect on the graph
f(x) + kup k (down if k < 0)
f(x − h)right h (left if h < 0)
−f(x)reflect over the x-axis
f(−x)reflect over the y-axis
a · f(x)vertical stretch by |a|
f(bx)horizontal shrink by 1b

Inside Opposite, Outside Obvious

Changes inside f( ) act on x and run backwards: f(x − 3) moves right, f(2x) shrinks. Changes outside do just what they say.

Point mapping for y = a·f(b(x − h)) + k

Move a few key points of the parent; draw through them.

(x, y) → (xb + h, ay + k)

g(x) = −2√(x + 3) + 1 Worked

left 3, stretch 2, flip, up 1
(0, 0) → (−3, 1); (1, 1) → (−2, −1); (4, 2) → (1, −3)
domain [−3, ∞), range (−∞, 1]
xy(−3, 1)(1, −3)y = √x
Dashed: √x. Solid: g(x) = −2√(x + 3) + 1.

Absolute value Worked

f(x) = |2x − 4| − 3
vertex (2, −3); zeros: |2x − 4| = 3 → x = 0.5, 3.5

Step function

Greatest integer ⟦x⟧: the largest integer ≤ x.

EX⟦2.7⟧ = 2, ⟦−1.2⟧ = −2

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Pre-CalculusPre-Calculus · Units 1–9 · continued

3Composition & Inversesfunctions in a chain, and undoing them

Composition

(f ∘ g)(x) = f(g(x)): apply g first. The domain keeps x in the domain of g and g(x) in the domain of f.

Basic examples: Algebra §10.

f(x) = √x, g(x) = x² − 4 Worked

(f ∘ g)(x) = √(x² − 4): domain (−∞, −2] ∪ [2, ∞)
(g ∘ f)(x) = x − 4: domain [0, ∞)not all reals

Decompose h(x) = (3x + 1)⁵ Worked

inner g(x) = 3x + 1, outer f(x) = x⁵h = f ∘ g

f AFTER g

Read f ∘ g as “f after g”: the function written nearest x acts first.

Operations & domain Worked

f(x) = √x, g(x) = x − 1
(fg)(x) = √xx − 1: domain [0, 1) ∪ (1, ∞)both domains, g ≠ 0

Inverse functions

  • f⁻¹ exists only if f is one-to-one (passes the horizontal line test).
  • f(f⁻¹(x)) = x and f⁻¹(f(x)) = x.
  • Domain and range trade places; the graphs mirror in y = x.

Inverse of f(x) = 2x + 1x − 3 Worked

x = 2y + 1y − 3 → xy − 3x = 2y + 1swap x and y
y(x − 2) = 3x + 1 → f⁻¹(x) = 3x + 1x − 2
check: f⁻¹(5) = 163, f(163) = 5 ✓

One-to-one? Worked

f(x) = x² − 4x: f(0) = f(4) = 0two inputs, one output
not one-to-one; restrict to x ≥ 2vertex at 2

Swap, Solve, Check

To invert: Swap x and y, Solve for y, Check that f(f⁻¹(x)) = x.

xy(0, 1)(1, 0)y = x
f(x) = ¼x³ + 1 (color) and f⁻¹(x) = ∛(4(x − 1)) (black): (a, b) on f ↔ (b, a) on f⁻¹.

Restrict, then invert Worked

f(x) = (x − 1)² + 2, x ≥ 1one-to-one half
f⁻¹(x) = 1 + √(x − 2), x ≥ 2

Inverse from a table Worked

f(−1) = 4, f(0) = 1, f(2) = −3, f(5) = −7
f⁻¹(4) = −1, f⁻¹(−3) = 2, f⁻¹(f(5)) = 5read backwards

4Polynomial Functionsend behavior, zeros, division (factoring: Algebra §4)

End behavior (aₙxⁿ decides)
Leading terma > 0a < 0
even degree↖ ↗ both up↙ ↘ both down
odd degree↙ ↗ down, up↖ ↘ up, down

Even arms Agree

Even degree: both arms point the same way. Odd degree: the arms split. The sign of the leading coefficient sets the right arm.

Zeros & multiplicity

  • Factor (x − c)ᵐ: odd m crosses the x-axis, even m touches and turns.
  • Degree n: at most n real zeros and at most n − 1 turning points.
xy−213
f(x) = ¼(x + 2)(x − 1)²(x − 3): crosses at −2 and 3, touches at 1; degree 4, a > 0.

Long division Worked

(2x³ − x² + 3x − 4) ÷ (x² + 1)
= 2x − 1, remainder x − 3
so 2x³ − x² + 3x − 4 = (x² + 1)(2x − 1) + (x − 3)check
322-363-119-26-60
Synthetic: (2x³ − 3x² − 11x + 6) ÷ (x − 3) = 2x² + 3x − 2, remainder 0, so zeros 3, ½, −2.

Remainder & factor theorems

  • P(c) is the remainder of P(x) ÷ (x − c).
  • P(c) = 0 exactly when (x − c) is a factor.

EXx⁴ − 3x² + 2x − 5 at 2: 16 − 12 + 4 − 5 = 3

Polynomial inequality Worked

x³ − 4x > 0 → x(x − 2)(x + 2) > 0factor
signs on (−∞, −2), (−2, 0), (0, 2), (2, ∞): − + − +test points
(−2, 0) ∪ (2, ∞)

Counting zeros

  • Fundamental Theorem of Algebra: degree n has exactly n complex zeros, counting multiplicity.
  • Conjugate pairs: with real coefficients, a + bi brings a − bi.
  • Descartes' rule: sign changes in P(x) give the most positive zeros (or fewer by 2s); use P(−x) for negatives.

All zeros of x³ − x² + 4x − 4 Worked

signs + − + −: 3 or 1 positive; P(−x): 0 negativeDescartes
x²(x − 1) + 4(x − 1) = (x − 1)(x² + 4)group
zeros 1, 2i, −2i

Build from zeros Worked

degree 3, zeros 2 and 1 + iso 1 − i too
(x − 2)(x² − 2x + 2) = x³ − 4x² + 6x − 4

Intermediate Value Theorem

If P is continuous and P(a), P(b) have opposite signs, P has a zero between a and b.

EXx³ − x − 1: P(1) = −1, P(2) = 5 → a zero in (1, 2), ≈ 1.3247

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Pre-CalculusPre-Calculus · Units 1–9 · continued

5Rational Functionsa polynomial over a polynomial

Happy Vets Help Injured Sparrows

Graph in order: Holes (canceled factors), Vertical asymptotes, Horizontal or slant asymptote, Intercepts, Sign chart.

Analyze f(x) = x² − x − 2x² − 4 Worked

= (x − 2)(x + 1)(x − 2)(x + 2)factor
hole at x = 2: 2 + 12 + 2 → (2, ¾)cancel, then plug in
VA x = −2; HA y = 1equal degrees
x-int −1; y-int ½

Rational equation Worked

2x − 1 + 1x + 1 = 4x² − 1
× (x − 1)(x + 1): 2(x + 1) + (x − 1) = 4 → x = 1
x = 1 makes a denominator 0: no solutionextraneous
xy(2, ¾)x = −2y = 1
The graph of the analyzed f: a hole, not an asymptote, at x = 2.

Horizontal asymptote (degrees n over d)

  • n < d: y = 0 · n = d: ratio of leading coefficients · n > d: none (BOBO: Algebra §10)
  • A graph may cross its HA; it never crosses a VA.

Slant asymptote

When n = d + 1, divide: the quotient line is the asymptote.

EXx² + 3x + 1x − 1 = x + 4 + 5x − 1 → y = x + 4

Near a vertical asymptote

  • Odd multiplicity: the two sides go opposite ways (±∞).
  • Even multiplicity: both sides go the same way.

Partial fractions

Split a proper fraction over its factored denominator: one constant per linear factor.

EX5x − 4x² − x − 2 = Ax − 2 + Bx + 1: x = 2 → A = 2; x = −1 → B = 3

Rational inequality x + 3x − 1 ≤ 0 Worked

critical points −3 (top = 0) and 1 (undefined)
test −4: +, 0: −, 2: + → [−3, 1)1 never included
-5-4-3-2-10123[−3, 1)

6Exponential & Logarithmic Functionsgrowth, decay and the exponents that measure them

Inverse pair (b > 0, b ≠ 1)
y = bˣy = logb x
domainall reals(0, ∞)
range(0, ∞)all reals
key point(0, 1)(1, 0)
asymptotey = 0x = 0
123123(0, 1)(1, 0)
eˣ (green) and ln x (black) mirror in y = x.

Laws (b, M, N > 0)

  • logb MN = logb M + logb N; logb MN = logb M − logb N
  • logb Mᵖ = p logb M; logb x = ln xln b
  • blogb x = x and logb bˣ = x

More examples: Algebra §10.

Logs step down

A log lowers each operation one step: × becomes +, ÷ becomes −, a power becomes a multiplier in front.

Exponential equations Worked

4e0.3t = 20 → e0.3t = 5 → t = ln 50.3 ≈ 5.365
3x + 1 = 2ˣ → (x + 1) ln 3 = x ln 2log both sides
x = ln 3ln 2 − ln 3 ≈ −2.710

Log equation Worked

log₂ x + log₂(x − 2) = 3 → x(x − 2) = 8combine
x² − 2x − 8 = 0 → x = 4 or −2
x = 4 (log₂(−2) is undefined)check

Quadratic in eˣ Worked

e2x − 5eˣ + 6 = 0 → (eˣ − 2)(eˣ − 3) = 0u = eˣ
x = ln 2 ≈ 0.693, ln 3 ≈ 1.099

I Love Solving Carefully

Isolate the power or log, Log (or exponentiate) both sides, Solve, Check every answer in the original.

Growth & decay

k > 0 growth, k < 0 decay. Doubling time ln 2k; half-life h: A = A₀(½)t/h.

A = A₀ekt

Find k Worked

5,000 grows to 8,000 in 6 years
k = ln 1.66 ≈ 0.0783; doubling ≈ 8.85 yr

Half-life Worked

200 g, half-life 8 days, after 20 days
200(½)20/8 ≈ 35.36 g

The number e

e = the limit of (1 + 1n)ⁿ ≈ 2.71828. Continuous growth: A = Pert.

EX$2,000 at 5% for 10 years: 2000e0.5 ≈ $3,297.44

Log scales

Each step of 1 is a factor of 10: pH = −log[H⁺], decibels, earthquake magnitude.

EX[H⁺] = 3.2 × 10⁻⁴ → pH ≈ 3.49

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Pre-CalculusPre-Calculus · Units 1–9 · continued

7Conic Sectionsfoci, eccentricity and general form (basic forms: Algebra §9)

F(0, 1)Vdirectrix y = −1parabola x² = 4y
FFc = 4b = 3a = 5ellipse x²/25 + y²/9 = 1
FFy = ⁴⁄₃xhyperbola x²/9 − y²/16 = 1
Foci and eccentricity (shift by (h, k) for other centers)
Conic (center or vertex at origin)Focus relationEccentricity eDefining distance
Parabola x² = 4pyfocus (0, p), directrix y = −pe = 1to focus = to directrix
Ellipse x²a² + y²b² = 1, a > bc² = a² − b², foci (±c, 0)e = ca < 1sum to foci = 2a
Hyperbola x²a² − y²b² = 1c² = a² + b², foci (±c, 0)e = ca > 1difference = 2a
Circleboth foci at the centere = 0to center = r

Ellipse from general form Worked

9x² + 25y² − 36x + 150y + 36 = 0
9(x − 2)² + 25(y + 3)² = 225complete squares
(x − 2)²25 + (y + 3)²9 = 1
center (2, −3), c = 4: foci (−2, −3), (6, −3), e = 0.8

Parabola y² = 12x Worked

4p = 12 → p = 3opens right
focus (3, 0), directrix x = −3

Hyperbola x²9 − y²16 = 1 Worked

c = √(9 + 16) = 5: foci (±5, 0)
e = 53; asymptotes y = ±43x

H is for Hypotenuse

In a hyperbola, c is the longest, like a hypotenuse: c² = a² + b². In an ellipse, a is the longest: c² = a² − b².

Classify Ax² + Bxy + Cy² + Dx + Ey + F = 0

  • B = 0: AC = 0 parabola; AC > 0 ellipse (circle if A = C); AC < 0 hyperbola.
  • Any B: B² − 4AC < 0 ellipse, = 0 parabola, > 0 hyperbola.

Earth’s orbit Worked

a ≈ 149.6 million km, e ≈ 0.0167nearly a circle
nearest a(1 − e) ≈ 147.1, farthest a(1 + e) ≈ 152.1 million km

8Systems & Matricessolve many equations at once (2×2 basics: Algebra §6)

Nonlinear system Worked

x² + y² = 25, y = x + 1
x² + (x + 1)² = 25 → x² + x − 12 = 0substitute
(3, 4) and (−4, −3)

Matrix arithmetic

  • Size m × n = rows × columns. Add same-size matrices entry by entry.
  • AB needs columns of A = rows of B; the result is rows(A) × cols(B).
  • Row of A times column of B, summed. In general AB ≠ BA.

EX1234201−1 = 4−210−4

2 × 2 determinant & inverse

For A = abcd, det A = ad − bc. A⁻¹ exists only if det A ≠ 0.

A⁻¹ = 1ad − bcd−b−ca

Swap, Negate, Divide

For a 2 × 2 inverse: Swap the main diagonal, Negate the other diagonal, Divide everything by the determinant.

Solve with A⁻¹ Worked

2x + y = 5, 3x + 2y = 8
det = 4 − 3 = 1, A⁻¹ = 2−1−32
xy = A⁻¹58 = 21x = 2, y = 1

Cramer's rule

x = DxD, y = DyD: replace that variable’s column with the constants.

EXsame system: D = 1, Dx = 10 − 8 = 2, Dy = 16 − 15 = 1 → (2, 1)

3 × 3 determinant (row 1)

a(ei − fh) − b(di − fg) + c(dh − eg)

EXthe coefficient matrix of the elimination system at right: 1(1 − 2) − 1(−2 − 1) + 1(4 + 1) = 7

Row operations

  • Swap two rows · multiply a row by a nonzero number · add a multiple of one row to another.
  • Aim for a staircase (row-echelon form), then back-substitute.

3 × 3 by elimination Worked

11162−11312−12augmented
R₂ − 2R₁: −3y − z = −9; R₃ − R₁: y − 2z = −4
y = 2z − 4 → −7z + 12 = −9 → z = 3
(1, 2, 3)check all three

No unique answer

  • A row 0 0 0 | c with c ≠ 0: no solution.
  • A row of all zeros: infinitely many (dependent).
  • det = 0 means one of these two.
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Pre-CalculusPre-Calculus · Units 1–9 · continued

9Parametric Equations & the Trig Toolkitx and y driven by a third variable

Parametric curves

x = f(t), y = g(t): each t gives a point, and increasing t gives the curve a direction.

  • Line from (x₁, y₁) to (x₂, y₂): x = x₁ + (x₂ − x₁)t, y = y₁ + (y₂ − y₁)t
  • Circle: x = h + r cos t, y = k + r sin t

Eliminate the parameter Worked

x = 2t + 1, y = t² → t = x − 12
y = (x − 1)²4, a parabola
x = 3 cos t, y = 2 sin t → x²9 + y²4 = 1cos² + sin² = 1

Line segment Worked

(1, 2) to (5, −6), 0 ≤ t ≤ 1
x = 1 + 4t, y = 2 − 8t; t = ½ → (3, −2)midpoint

Projectile motion

g ≈ 9.8 m/s² (32 ft/s²); air resistance ignored.

x = (v₀ cos θ)t

y = h₀ + (v₀ sin θ)t − ½gt²

xytop ≈ 15.2 m≈ 65.1 m
v₀ = 25 m/s at 40°, launched 2 m up.

That launch Worked

x = 19.15t, y = 2 + 16.07t − 4.9t²
top at t = 16.079.8 ≈ 1.64 s: height ≈ 15.2 m
y = 0 at t ≈ 3.40 s; range ≈ 65.1 mquadratic formula

Trig in Pre-Calculus (see the Trigonometry guide)

  • Radians and the unit circle: Trig §1, §3
  • Sinusoids y = A sin(B(x − C)) + D, period 2πB: Trig §6
  • Inverse trig and identities: Trig §7–9
  • Vectors, u · v = |u||v| cos θ: Trig §11
  • Polar graphs; De Moivre (r cis θ)ⁿ = rⁿ cis nθ: Trig §12–13

The puppeteer t

One hand (t) works two strings: x and y move separately, and the curve is where the puppet goes.

Direction of travel

x = cos t, y = sin t runs counterclockwise from (1, 0); x = sin t, y = cos t runs clockwise from (0, 1). Same circle, different motion.

10Sequences, Series, Induction & Binomialspatterns, sums, proofs and (a + b)ⁿ

Sequence
a list a₁, a₂, a₃, …
Series
the sum of its terms
Recursive
each term from the one before
Explicit
aₙ straight from n

Recursive to explicit Worked

a₁ = 3, aₙ = 2aₙ₋₁ − 1
3, 5, 9, 17, 33, … → aₙ = 2ⁿ + 1

Arithmetic & geometric (Algebra §11)

  • aₙ = a₁ + (n − 1)d; Sₙ = n(a₁ + aₙ)2
  • aₙ = a₁rⁿ⁻¹; Sₙ = a₁(1 − rⁿ)1 − r
  • Infinite geometric: S = a₁1 − r if |r| < 1, else no sum

Power sums

Sums run k = 1 to n. Constants and sums split: Σ(cak + bk) = cΣak + Σbk; Σc = cn.

Σk = n(n + 1)2

Σk² = n(n + 1)(2n + 1)6

Σk³ = n²(n + 1)²4

Use the formulas Worked

Σk=120 (3k − 2) = 3 · 210 − 2 · 20 = 590
Σk=110 k² = 10 · 11 · 216 = 385

Repeating decimal Worked

0.2727… = 27100 + 2710,000 + ⋯r = 1100
= 0.271 − 0.01 = 2799 = 311

Bouncing ball Worked

dropped 10 m, each bounce 60% as high
10 + 2 · 61 − 0.6 = 10 + 30 = 40 mup and down

Mathematical induction

  • Base: show the statement for n = 1.
  • Assume it holds for n = k.
  • Show it then holds for n = k + 1.

Prove 1 + 3 + ⋯ + (2n − 1) = n² Worked

n = 1: 1 = 1² ✓base
assume 1 + 3 + ⋯ + (2k − 1) = k²
add 2k + 1: k² + 2k + 1 = (k + 1)² ✓next odd number

BAS: the dominoes

Base knocks over the first domino; Assume one falls; Show it knocks over the next. Then they all fall.

n=01n=111n=2121n=31331n=414641n=515101051n=61615201561
Pascal's triangle: each entry is the sum of the two above; row n holds ₙCₖ.

Binomial theorem

Term k + 1 is ₙCₖ an−kbᵏ.

(a + b)ⁿ = Σk=0n ₙCₖ an−kbᵏ

Expand (2x − 1)⁴ Worked

row 4: 1 4 6 4 1
16x⁴ − 32x³ + 24x² − 8x + 1

One term Worked

x⁵ term of (x + 3)⁸: k = 3
₈C₃ · 3³ = 56 · 27 = 1,512x⁵

Down, up, total n

Powers of a count down from n, powers of b count up from 0, and each term’s powers add to n.

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Pre-CalculusPre-Calculus · Units 1–9 · continued

11Counting & Probabilityhow many ways, and how likely

Counting principle

m ways, then n ways: m · n ways in all.

EX3 shirts, 4 pants, 2 pairs of shoes: 24 outfits

Arrangements & selections

n! = n(n − 1)⋯1, 0! = 1. P counts ordered lists; C counts groups.

ₙPᵣ = n!(n − r)!

ₙCᵣ = n!r!(n − r)!

Order matters?Example (from 10 people)Count
yes: Ppresident, vice president, secretary₁₀P₃ = 720
no: Ca committee of 3₁₀C₃ = 120

P for Position, C for Committee

If swapping two people makes a new result (positions), use P. If not (a committee), use C.

Repeated letters Worked

arrangements of BANANA3 A, 2 N
6!3! 2! = 60

Probability rules

  • P(A) = favorable outcomesall equally likely outcomes; P(not A) = 1 − P(A)
  • P(A or B) = P(A) + P(B) − P(A and B)
  • Independent: P(A and B) = P(A) · P(B)

EXtwo dice, sum 7: 636 = 16; at least one 6: 1 − (56)² = 1136

Probability by counting Worked

6 red, 4 blue marbles; draw 2
P(both red) = ₆C₂₁₀C₂ = 1545 = 13

12Limits & Continuitywhere a function is heading

What a limit means

limx→c f(x) = L: f(x) gets as close to L as we like for x close enough to c, from both sides. The value f(c) does not matter.

x² − 9x − 3 near 3 → 6
x2.92.992.9993.0013.013.1
f(x)5.95.995.9996.0016.016.1
xy(3, 6)
y = x² − 9x − 3: a line with a hole at x = 3.

One-sided limits

The limit exists only if the left and right limits agree.

EXx + 1 for x < 2, x² for x ≥ 2: left 3, right 4 → no limit (a jump)

Please Fix Really Carefully

Plug in first. If you get 00: Factor, Rationalize, or Combine fractions, then plug in again.

Factor Worked

limx→3 x² − 9x − 3 = limx→3 (x + 3) = 6

Rationalize Worked

limx→0 √(x + 4) − 2x = limx→0 1√(x + 4) + 2 = 14multiply by the conjugate

Combine fractions Worked

limx→0 1/(x + 2) − 1/2x = limx→0 −12(x + 2) = −14

At infinity; a special limit

  • Divide by the highest power: limx→∞ 3x² − x2x² + 5 = 32 (the HA).
  • limx→0 sin xx = 1 (x in radians)

Continuity at x = c

f(c) is defined, limx→c f(x) exists, and the two are equal.

Breaks: removable (hole), jump, infinite (vertical asymptote).

D-E-E

Continuous means Defined, Exists, Equal: three checks, in that order.

13From Average to Instantaneous Ratethe first step into calculus

The derivative

Secant slope (average rate, §1) → tangent slope (instantaneous rate) as h → 0.

f′(x) = limh→0 f(x + h) − f(x)h

Power rule preview

ddx xⁿ = nxⁿ⁻¹, term by term (proved in Calculus).

EXddx(3x⁴ − 5x + 2) = 12x³ − 5

xy(1, 1)(3, 9)tangentsecant
y = x²: secant slope 4 from 1 to 3; tangent slope 2 at x = 1.

Zoom until straight

Zoom in on a smooth curve until it looks like a line. That line’s slope is the derivative.

From the definition Worked

f(x) = x²: (x + h)² − x²h = 2x + h → f′(x) = 2x
at x = 3: slope 6, tangent y = 6x − 9y − 9 = 6(x − 3)

Velocity Worked

height s(t) = −16t² + 64t (ft)
average on [0, 1]: 48 − 01 = 48 ft/s
s(1 + h) − s(1)h = 32 − 16h → 32 ft/s at t = 1instantaneous