Unit 01

Functions & Their Properties

Domain · Range · Composition · Inverses · Even/Odd
📐 Core Concepts
  • A function f maps each input x to exactly one output f(x). Domain = valid inputs; Range = outputs produced.
  • Vertical Line Test: a graph is a function if every vertical line intersects it at most once.
  • Even: f(−x) = f(x) (symmetric about y-axis). Odd: f(−x) = −f(x) (symmetric about origin).
  • Composition: (f ∘ g)(x) = f(g(x)). Order matters — f ∘ g ≠ g ∘ f in general.
  • Inverse f⁻¹ exists iff f is one-to-one (passes Horizontal Line Test). If (a, b) is on f, then (b, a) is on f⁻¹.
★ Must Memorize
  • Domain restrictions: denominator ≠ 0; radicand ≥ 0 (even root); argument of log > 0
  • (f ∘ g)⁻¹ = g⁻¹ ∘ f⁻¹
  • To find inverse: swap x and y, then solve for y
  • f(f⁻¹(x)) = x and f⁻¹(f(x)) = x (on appropriate domains)
✏️ Worked Example
Composition & Inverse

Let f(x) = 2x − 1 and g(x) = x² + 3. Find (f ∘ g)(x) and f⁻¹(x).

(f ∘ g)(x) = f(g(x)) = f(x²+3) = 2(x²+3) − 1 = 2x² + 5 For inverse: y = 2x−1 → x = (y+1)/2 → f⁻¹(x) = (x+1)/2

▸ (f ∘ g)(x) = 2x² + 5  |  f⁻¹(x) = (x+1)/2

Unit 02

Polynomials & Rational Functions

Zeros · End Behavior · Asymptotes · Factor Theorem
📐 Core Concepts
  • Degree n polynomial has at most n real zeros and n−1 turning points.
  • Factor Theorem: (x−c) is a factor of p(x) if and only if p(c) = 0.
  • Rational Root Theorem: possible rational zeros = ±(factors of constant) / (factors of leading coefficient).
  • Vertical Asymptote at zeros of denominator (where numerator ≠ 0).
  • Horizontal Asymptote: deg(num) < deg(den) → y=0; equal degrees → ratio of leading coefficients; num > den → oblique asymptote.
★ Must Memorize
  • End behavior: even degree → both ends same; odd degree → opposite directions
  • Multiplicity: odd → crosses x-axis; even → bounces off x-axis
  • Remainder Theorem: p(c) = remainder when p(x) ÷ (x−c)
  • Complex zeros come in conjugate pairs for real polynomials
✏️ Worked Example
Finding All Zeros

Find all zeros of p(x) = x³ − 6x² + 11x − 6.

Test x=1: 1 − 6 + 11 − 6 = 0 ✓ Factor: (x−1)(x²−5x+6) = (x−1)(x−2)(x−3)

▸ x = 1, 2, 3

Unit 03

Exponential & Logarithmic Functions

Properties · Equations · Growth & Decay Models
📐 Core Concepts
  • Exponential: f(x) = a · bˣ, b > 0, b ≠ 1. Growth if b > 1; decay if 0 < b < 1.
  • Compound Interest: A = P(1 + r/n)ⁿᵗ. Continuous: A = Peʳᵗ. e ≈ 2.71828.
  • Logarithm definition: log_b(x) = y ⟺ bʸ = x. Common log = log₁₀. Natural log = ln = log_e.
  • Change of Base: log_b(x) = ln(x)/ln(b) = log(x)/log(b).
★ Must Memorize
  • log(xy) = log x + log y  |  log(x/y) = log x − log y  |  log(xʳ) = r·log x
  • b^(log_b x) = x   and   log_b(bˣ) = x  (inverse relationships)
  • Exponential growth/decay: N(t) = N₀ · e^(kt), k > 0 growth, k < 0 decay
  • Half-life T₁/₂ = ln 2 / |k|
✏️ Worked Example
Logarithmic Equation

Solve: log₂(x+3) + log₂(x−1) = 5.

Combine: log₂[(x+3)(x−1)] = 5 (x+3)(x−1) = 2⁵ = 32 x² + 2x − 3 = 32 → x² + 2x − 35 = 0 (x+7)(x−5) = 0 → x = −7 (rejected, log undefined) or x = 5

▸ x = 5

Unit 04

Trigonometric Functions

Unit Circle · Identities · Equations · Graphs · Law of Sines/Cosines
📐 Core Concepts
  • Unit Circle: sin θ = y, cos θ = x, tan θ = y/x at point (x,y) on unit circle.
  • ASTC: All Students Take Calculus — Q1: all +; Q2: sin +; Q3: tan +; Q4: cos +.
  • Graph: y = A·sin(Bx + C) + D. |Amplitude| = |A|, Period = 2π/|B|, Phase shift = −C/B.
  • Sum/Difference: sin(A±B) = sinA cosB ± cosA sinB; cos(A±B) = cosA cosB ∓ sinA sinB.
  • Law of Sines: a/sinA = b/sinB = c/sinC  |  Law of Cosines: a² = b² + c² − 2bc cosA.
★ Must Memorize
  • sin²θ + cos²θ = 1  |  1 + tan²θ = sec²θ  |  1 + cot²θ = csc²θ
  • sin(2θ) = 2 sinθ cosθ  |  cos(2θ) = cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1
  • Key values: sin 30°=½, sin 45°=√2/2, sin 60°=√3/2 (cosines in reverse)
  • Radian–degree: 180° = π rad. Arc length s = rθ. Sector area = ½r²θ.
✏️ Worked Example
Identity Verification

Verify: tan x + cot x = 1/(sin x cos x).

LHS = sin x/cos x + cos x/sin x = (sin²x + cos²x) / (sin x cos x) = 1 / (sin x cos x) = RHS ✓

▸ Identity verified using Pythagorean identity

Unit 05

Sequences, Series & Conic Sections

Arithmetic · Geometric · Binomial Theorem · Parabola · Ellipse · Hyperbola
📐 Sequences & Series
  • Arithmetic: aₙ = a₁ + (n−1)d.   Sₙ = n/2 · (a₁ + aₙ).
  • Geometric: aₙ = a₁ · rⁿ⁻¹.   Sₙ = a₁(1−rⁿ)/(1−r), r ≠ 1.
  • Infinite Geometric (|r| < 1): S∞ = a₁/(1−r).
  • Binomial Theorem: (a+b)ⁿ = Σ C(n,k) aⁿ⁻ᵏ bᵏ, k = 0 to n.
📐 Conic Sections
  • Parabola (vertical): (x−h)² = 4p(y−k). Focus: (h, k+p). Directrix: y = k−p.
  • Ellipse: (x−h)²/a² + (y−k)²/b² = 1 (a>b). c² = a²−b². Foci at (h±c, k).
  • Hyperbola: (x−h)²/a² − (y−k)²/b² = 1. c² = a²+b². Asymptotes: y−k = ±(b/a)(x−h).
  • Circle: (x−h)² + (y−k)² = r². Center (h,k), radius r.
★ Must Memorize
  • Arithmetic: constant difference d. Geometric: constant ratio r.
  • Infinite geometric sum requires |r| < 1
  • Eccentricity e = c/a: e < 1 → ellipse; e = 1 → parabola; e > 1 → hyperbola
  • Ellipse: larger denominator → major axis direction
✏️ Worked Example
Infinite Geometric Series

Find the sum: 12 + 4 + 4/3 + ···

a₁ = 12, r = 4/12 = 1/3. Since |1/3| < 1: S∞ = 12 / (1 − 1/3) = 12 / (2/3) = 18

▸ S∞ = 18

Unit 06

Limits, Vectors & Matrices

Continuity · Dot & Cross Product · Matrix Operations
📐 Limits & Continuity
  • lim_{x→a} f(x) = L: f(x) can be made arbitrarily close to L as x → a.
  • Continuity at a: (1) f(a) defined; (2) limit exists; (3) limit = f(a).
  • Standard limits: lim_{x→0} (sin x)/x = 1; lim_{x→∞} (1+1/x)ˣ = e.
  • Indeterminate forms (0/0, ∞/∞): factor, rationalize, or L'Hôpital's Rule.
📐 Vectors & Matrices
  • Magnitude: |v| = √(a²+b²+c²). Unit vector: v̂ = v/|v|.
  • Dot product: u·v = u₁v₁+u₂v₂+u₃v₃. Angle: cos θ = (u·v)/(|u||v|). Perpendicular: u·v = 0.
  • 2×2 Inverse: if A=[[a,b],[c,d]], then A⁻¹ = (1/det A)[[d,−b],[−c,a]] where det A = ad−bc.
  • Unique solution to Ax=b iff det(A) ≠ 0.
★ Must Memorize
  • f continuous at a ⟺ lim_{x→a} f(x) = f(a)
  • det([[a,b],[c,d]]) = ad − bc
  • Perpendicular vectors: u · v = 0; Parallel: u × v = 0
  • Matrix multiplication: (AB)_{ij} = Σ A_{ik}B_{kj}
✏️ Worked Example
Limit by Factoring

Evaluate: lim_{x→3} (x² − 9)/(x − 3).

Factor: (x−3)(x+3)/(x−3) = x + 3 (x ≠ 3) As x → 3: 3 + 3 = 6

▸ Limit = 6

Practice Exam · 20 Problems

Subjective Format  ·  Show All Work  ·  AP / SAT / IB Exam Style

Instructions: Solve each problem showing all steps clearly. Box or underline your final answer. Partial credit may be awarded for correct reasoning. Full solutions appear after Question 20.
1
Functions
Let f(x) = 3x + 2 and g(x) = x² − 1. Find (g ∘ f)(x) and simplify completely. Then state the domain of (g ∘ f)(x).
Work Space
2
Functions
Find the inverse function of f(x) = (2x − 5)/(x + 1). State the domain of f⁻¹(x) and verify your answer by computing f(f⁻¹(x)).
Work Space
3
Polynomials
A polynomial p(x) = x⁴ − 5x³ + 5x² + 5x − 6 has a known zero at x = 1. Use the Factor Theorem and polynomial division to find all remaining zeros.
Work Space
4
Rational Functions
Determine all vertical asymptotes, horizontal asymptotes, and holes of: f(x) = (3x² − 12) / (x² − x − 6). Describe the graph's behavior near each asymptote.
Work Space
5
Exponential
Solve the equation: 2^(3x−1) = 5^(x+2). Express your answer in exact logarithmic form, then give a decimal approximation rounded to four decimal places.
Work Space
6
Logarithms
Solve the system of equations:
    log(x) + log(y) = 3
    log(x) − log(y) = 1
Find the exact values of x and y.
Work Space
7
Trigonometry
Without a calculator, evaluate sin(195°). Show all work using a sum or difference identity. Express your answer in simplified radical form.
Work Space
8
Trigonometry
Prove the trigonometric identity: cos⁴(x) − sin⁴(x) = cos(2x). Show each algebraic step clearly, working from one side only.
Work Space
9
Trig Equations
Solve for all solutions in [0, 2π):   2cos²(x) − cos(x) − 1 = 0. Give answers in exact radian form.
Work Space
10
Applied Trig
A 40-foot ladder leans against a wall with its foot 10 feet from the base. Find: (a) the angle the ladder makes with the ground (to the nearest tenth of a degree), and (b) the height the ladder reaches on the wall.
Work Space
11
Sequences
Find the 50th term and the sum of the first 50 terms of the arithmetic sequence: 7, 13, 19, 25, ···
Work Space
12
Sequences
A geometric sequence has its 3rd term equal to 20 and common ratio r = 5/2. Find the first term a₁, write the general term aₙ, and find the sum of the first 6 terms.
Work Space
13
Binomial Theorem
Expand (2x − 3)⁵ completely using the Binomial Theorem. Identify the coefficient of the x³ term.
Work Space
14
Conic Sections
Write the equation of the parabola with vertex (2, −1) that passes through (4, 7). State whether it opens up or down, and find the focus and directrix.
Work Space
15
Conic Sections
Find the equation of the ellipse centered at the origin with major axis along the x-axis, vertex at (5, 0), and focus at (3, 0). Find the eccentricity and directrices.
Work Space
16
Limits
Evaluate without L'Hôpital's Rule:   lim_{x→4} (x² − 16) / (√x − 2). Show the algebraic manipulation required.
Work Space
17
Continuity
Determine whether f is continuous at x = 2, where:
    f(x) = (x² − 4)/(x − 2)   if x ≠ 2
    f(x) = 5                      if x = 2
Justify using the formal definition of continuity.
Work Space
18
Vectors
Given u = ⟨3, −1, 2⟩ and v = ⟨1, 4, −2⟩, find: (a) u · v, (b) |u| and |v|, (c) the angle between u and v (degrees, 1 decimal), and (d) a unit vector in the direction of u.
Work Space
19
Matrices
Solve the matrix equation AX = B where A = [[2, 1], [5, 3]] and B = [[4], [7]]. Find A⁻¹ first, then compute X = A⁻¹B.
Work Space
20
Exp. Modeling
A radioactive substance decays by N(t) = N₀ e^(−kt). If 500 g initially reduces to 350 g after 10 years: (a) find exact k and approximate value, (b) find the half-life T₁/₂, and (c) find the amount remaining after 30 years.
Work Space

Answer Key & Full Solutions

Detailed step-by-step solutions for all 20 problems

Q1 (g ∘ f)(x) = 9x² + 12x + 3  |  Domain: all real numbers ℝ
(g ∘ f)(x) = g(f(x)) = g(3x+2) = (3x+2)² − 1
= 9x² + 12x + 4 − 1 = 9x² + 12x + 3
It is a polynomial — domain = (−∞, +∞)
Q2 f⁻¹(x) = (−x − 5)/(x − 2)  |  Domain: x ≠ 2
Set y = (2x−5)/(x+1), swap x and y: x = (2y−5)/(y+1)
x(y+1) = 2y−5 → xy + x = 2y − 5 → y(x−2) = −x−5
f⁻¹(x) = (−x−5)/(x−2). Verify: f(f⁻¹(x)) = x ✓
Q3 All zeros: x = 1, 2, 3, −1
p(1) = 1−5+5+5−6 = 0 ✓ → divide by (x−1) → x³−4x²+x+6
Test x=2: 8−16+2+6 = 0 ✓ → divide by (x−2) → x²−2x−3 = (x−3)(x+1)
Complete factorization: (x−1)(x−2)(x−3)(x+1) → zeros: 1, 2, 3, −1
Q4 VA: x = 3  |  HA: y = 3  |  Hole at x = −2
Factor: 3(x−2)(x+2) / [(x−3)(x+2)] → cancel (x+2): hole at x = −2
Simplified: 3(x−2)/(x−3) → VA at x = 3
Equal degrees → HA: y = 3/1 = 3. As x→3⁻: f→+∞; as x→3⁺: f→−∞
Q5 x = (2ln5 + ln2)/(3ln2 − ln5) ≈ 8.3234
Take ln both sides: (3x−1)ln 2 = (x+2)ln 5
3x·ln2 − ln2 = x·ln5 + 2·ln5 → x(3ln2−ln5) = 2ln5 + ln2
x = (2ln5 + ln2)/(3ln2 − ln5) ≈ (3.219+0.693)/(2.079−1.609) ≈ 3.912/0.470 ≈ 8.3234
Q6 x = 100, y = 10
Add equations: 2log(x) = 4 → log(x) = 2 → x = 100
Substitute: 2 + log(y) = 3 → log(y) = 1 → y = 10
Verify: log(100)+log(10)=2+1=3 ✓; log(100)−log(10)=2−1=1 ✓
Q7 sin(195°) = −(√6 + √2)/4
Write 195° = 150° + 45°
sin(150°+45°) = sin150°cos45° + cos150°sin45° = (1/2)(√2/2) + (−√3/2)(√2/2)
= √2/4 − √6/4 = −(√6 − √2)/4 = −(√6 + √2)/4 ... note: (√2−√6)/4 = −(√6−√2)/4
Q8 cos⁴x − sin⁴x = cos(2x) ✓ Proved
LHS = (cos²x + sin²x)(cos²x − sin²x)  [difference of squares]
= (1)(cos²x − sin²x)  [Pythagorean identity: sin²x+cos²x=1]
= cos(2x) = RHS ✓  [double angle identity]
Q9 x = 0, 2π/3, 4π/3
Let u = cos x: 2u²−u−1 = (2u+1)(u−1) = 0
u = −1/2 → cos x = −1/2 → x = 2π/3, 4π/3
u = 1 → cos x = 1 → x = 0. All solutions: {0, 2π/3, 4π/3}
Q10 (a) θ ≈ 75.5°  |  (b) h = 10√15 ≈ 38.7 ft
(a) cos θ = adj/hyp = 10/40 = 1/4 → θ = arccos(1/4) ≈ 75.5°
(b) h = √(40²−10²) = √(1600−100) = √1500 = 10√15 ≈ 38.7 ft
Q11 a₅₀ = 301  |  S₅₀ = 7,700
d = 13−7 = 6; aₙ = 7+(n−1)(6) = 6n+1; a₅₀ = 6(50)+1 = 301
S₅₀ = 50/2 · (7+301) = 25 × 308 = 7,700
Q12 a₁ = 16/5  |  aₙ = (16/5)(5/2)^(n−1)  |  S₆ ≈ 518.7
a₃ = a₁·r² = a₁·(5/2)² = (25/4)a₁ = 20 → a₁ = 80/25 = 16/5
aₙ = (16/5)·(5/2)^(n−1)
S₆ = (16/5)·[(5/2)⁶−1]/(5/2−1) = (16/5)·[(15625/64−1)]/(3/2) ≈ 518.7
Q13 Coefficient of x³ = 720  |  Full expansion: 32x⁵−240x⁴+720x³−1080x²+810x−243
x³ term: 5−k=3 → k=2; C(5,2)(2x)³(−3)² = 10·8x³·9 = 720x³
k=0: C(5,0)(2x)⁵(−3)⁰ = 32x⁵
k=1: C(5,1)(2x)⁴(−3)¹ = 5·16x⁴·(−3) = −240x⁴; k=3: −1080x²; k=4: 810x; k=5: −243
Q14 y = 2(x−2)²−1 (opens up)  |  Focus: (2, −7/8)  |  Directrix: y = −9/8
7 = a(4−2)²−1 → 8 = 4a → a = 2. Equation: y = 2(x−2)²−1 (opens up)
Rewrite: (x−2)² = (1/2)(y+1) → 4p = 1/2 → p = 1/8
Focus: (2, −1+1/8) = (2, −7/8). Directrix: y = −9/8
Q15 x²/25 + y²/16 = 1  |  e = 3/5  |  Directrices: x = ±25/3
a = 5, c = 3 → b² = 25−9 = 16. Equation: x²/25 + y²/16 = 1
Eccentricity: e = c/a = 3/5
Directrices: x = ±a/e = ±5/(3/5) = ±25/3
Q16 Limit = 32
Direct substitution gives 0/0. Rationalize by multiplying by (√x+2)/(√x+2):
(x²−16)(√x+2) / [(√x−2)(√x+2)] = (x²−16)(√x+2) / (x−4)
= (x−4)(x+4)(√x+2)/(x−4) = (x+4)(√x+2). As x→4: (8)(4) = 32
Q17 NOT continuous at x = 2 (removable discontinuity)
Step 1: f(2) = 5 ✓ (defined)
Step 2: lim_{x→2} (x²−4)/(x−2) = lim (x−2)(x+2)/(x−2) = lim(x+2) = 4
Step 3: limit (4) ≠ f(2) (5). Discontinuous at x=2 (removable). If f(2) were redefined as 4, it would be continuous.
Q18 (a) −5  (b) √14, √21  (c) ≈106.9°  (d) ⟨3/√14, −1/√14, 2/√14⟩
(a) u·v = 3(1)+(−1)(4)+2(−2) = 3−4−4 = −5
(b) |u|=√(9+1+4)=√14≈3.742; |v|=√(1+16+4)=√21≈4.583
(c) cosθ = −5/(√14·√21)= −5/√294 ≈ −0.2916 → θ ≈ 106.9°
(d) û = (1/√14)⟨3,−1,2⟩ = ⟨3√14/14, −√14/14, √14/7⟩
Q19 x₁ = 5, x₂ = −6
det(A) = 2·3−1·5 = 6−5 = 1 → A⁻¹ = [[3,−1],[−5,2]]
X = A⁻¹B = [[3,−1],[−5,2]]·[[4],[7]] = [[3·4+(−1)·7],[−5·4+2·7]] = [[5],[−6]]
Solution: x₁ = 5, x₂ = −6. Verify: 2(5)+(−6)=4 ✓; 5(5)+3(−6)=25−18=7 ✓
Q20 (a) k ≈ 0.03567 yr⁻¹  (b) T₁/₂ ≈ 19.43 yrs  (c) ≈ 171.5 g
(a) 350 = 500e^(−10k) → e^(−10k) = 0.7 → k = −ln(0.7)/10 ≈ 0.03567 yr⁻¹
(b) T₁/₂ = ln2/k = ln2/0.03567 ≈ 19.43 years
(c) N(30) = 500e^(−0.03567·30) = 500e^(−1.070) ≈ 500(0.3430) ≈ 171.5 grams