Unit Concepts & Key Formulas
Unit 1 · Functions
Functions & Their Properties
f: A → B, Domain, Range, Co-domain
- Vertical Line Test: each x has exactly one y
- Even: f(−x) = f(x) | Odd: f(−x) = −f(x)
- Composite: (f ∘ g)(x) = f(g(x))
- Inverse: f⁻¹ exists ⟺ f is one-to-one
Example
If f(x) = 2x + 3, find f⁻¹(x).
Answer: f⁻¹(x) = (x − 3) / 2
Unit 2 · Polynomials
Polynomial Functions & Zeros
p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀
- Factor Theorem: (x − c) is a factor ⟺ p(c) = 0
- Remainder Theorem: p(c) = remainder when p(x) ÷ (x−c)
- Sum of zeros = −aₙ₋₁/aₙ; Product = a₀/aₙ (±)
- Degree n polynomial has at most n real zeros
Example
Factor: p(x) = x³ − 6x² + 11x − 6
Answer: (x−1)(x−2)(x−3)
Unit 3 · Rational Functions
Rational Functions & Asymptotes
r(x) = p(x)/q(x), q(x) ≠ 0
- Vertical Asymptote (VA): q(x) = 0 and p(x) ≠ 0
- Horizontal Asymptote: compare degrees of p, q
- Hole: common factor in p and q (cancelled zero)
- Oblique Asymptote: deg(p) = deg(q) + 1
Example
Find the HA of r(x) = (3x²)/(x²+1).
Answer: y = 3
Unit 4 · Exponential & Logarithmic
Exponential & Log Functions
y = aˣ ↔ x = logₐ(y)
- ln x = log base e; log x = log base 10
- logₐ(xy) = logₐx + logₐy
- logₐ(xⁿ) = n·logₐx; Change of base: logₐb = ln b/ln a
- Exponential growth: A = P·eʳᵗ
Example
Solve: 2ˣ = 32
Answer: x = 5
Unit 5 · Trigonometry
Trigonometric Functions
sin²θ + cos²θ = 1; tan θ = sin θ / cos θ
- SOH CAH TOA (right triangle)
- Unit circle: (cos θ, sin θ); period 2π
- 1 + tan²θ = sec²θ; 1 + cot²θ = csc²θ
- sin(A±B), cos(A±B) addition formulas
Example
Evaluate: sin(π/6)
Answer: 1/2
Unit 6 · Trig Equations
Solving Trig Equations & Identities
sin(2θ) = 2sinθcosθ; cos(2θ) = cos²θ − sin²θ
- General solutions: add 2πk (or πk for tan)
- cos(2θ) = 2cos²θ − 1 = 1 − 2sin²θ
- Half-angle: sin²θ = (1−cos2θ)/2
Example
Solve sin x = 1/2 on [0, 2π)
Answer: x = π/6, 5π/6
Unit 7 · Analytic Geometry
Conic Sections
Circle: (x−h)²+(y−k)²=r²
- Parabola: y = a(x−h)²+k; focus-directrix property
- Ellipse: x²/a² + y²/b² = 1; c² = a² − b²
- Hyperbola: x²/a² − y²/b² = 1; c² = a² + b²
Example
Find center & radius: x²+y²−4x+6y−3=0
Answer: Center (2,−3), r = 4
Unit 8 · Sequences & Series
Arithmetic & Geometric Series
Aₙ = a₁ + (n−1)d; Gₙ = a₁·rⁿ⁻¹
- Arithmetic sum: Sₙ = n(a₁+aₙ)/2
- Geometric sum: Sₙ = a₁(1−rⁿ)/(1−r), r≠1
- Infinite geo sum: S = a₁/(1−r), |r|<1
Example
Find S₁₀ of 2, 5, 8, 11, …
Answer: S₁₀ = 155
Unit 9 · Matrices & Systems
Matrices & Linear Systems
AX = B → X = A⁻¹B (if det A ≠ 0)
- 2×2 inverse: [a b; c d]⁻¹ = 1/(ad−bc)[d −b; −c a]
- Cramer's Rule: x = Dₓ/D, y = Dᵧ/D
- Row reduction: Gaussian Elimination
Example
det([3,1;2,4]) = ?
Answer: 3(4) − 1(2) = 10
Unit 10 · Limits (Intro)
Limits & Continuity
lim(x→c) f(x) = L ⟺ left = right limit
- Squeeze Theorem; L'Hôpital for 0/0 or ∞/∞
- lim(x→0) sinx/x = 1; lim(x→∞) (1+1/x)ˣ = e
- Continuous at c: lim = f(c), both exist
Example
lim(x→2) (x²−4)/(x−2) = ?
Answer: 4
Practice Problems — Short Answer
Pre-Calculus — 20 Core Problems
01
Given f(x) = 3x − 5 and g(x) = x², find (f ∘ g)(2).
02
If f(x) = (2x + 1) / (x − 3), find f⁻¹(x). What is the domain restriction of f⁻¹?
03
Use the Remainder Theorem to find the remainder when p(x) = 2x³ − 3x² + x − 5 is divided by (x − 2).
04
A polynomial p(x) has real coefficients, degree 4, and zeros x = 1 (multiplicity 2) and x = 2 + i. Write the monic polynomial with these zeros (expand fully).
05
Find all vertical and horizontal asymptotes of r(x) = (x² − 1) / (x² − 4). Also, does the graph have any holes?
06
Solve for x: 3^(2x − 1) = 27.
07
Simplify: log₂(8) + log₂(4) − log₂(2). Give a single integer answer.
08
An investment doubles in 12 years with continuous compounding. Using A = Pe^(rt), find the annual interest rate r to four decimal places.
09
In a right triangle, the hypotenuse is 13 and one leg is 5. Find the exact value of sin θ, cos θ, and tan θ for the angle opposite the leg of length 12.
10
Using the identity cos(2θ) = 1 − 2sin²θ, find the exact value of sin(π/8). Simplify completely.
11
Solve the equation 2cos²x − cosx − 1 = 0 on the interval [0, 2π). List all solutions.
12
Prove or simplify the identity: (sin x + cos x)² = 1 + sin(2x). What numerical value does the left side equal when x = π/4?
13
Find the equation of the parabola with vertex (2, −3) and passing through the point (4, 5). Write in vertex form.
14
The ellipse x²/25 + y²/9 = 1 has semi-major axis a and semi-minor axis b. Find the coordinates of the two foci.
15
Find the sum of the infinite geometric series: 12 + 4 + 4/3 + 4/9 + …
16
Find the 15th term of an arithmetic sequence if the 3rd term is 7 and the 8th term is 22.
17
Solve the system using matrices or Cramer's Rule:
2x + 3y = 8
4x − y = 2
2x + 3y = 8
4x − y = 2
18
Find the inverse of matrix A = [5, 2; 3, 1] (rows separated by semicolons). Then compute det(A).
19
Evaluate: lim(x→3) (x² − 9) / (x − 3).
20
Evaluate: lim(x→0) sin(3x) / (5x). Use the standard limit lim(x→0) sin(x)/x = 1.
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Answer Key & Full Solutions
Answer Key & Detailed Solutions
Q1 · Functions
Answer: 7
Step 1: g(2) = 2² = 4. Step 2: f(g(2)) = f(4) = 3(4) − 5 = 12 − 5 = 7. Composition means apply g first, then f.
Q2 · Inverse Functions
Answer: f⁻¹(x) = (3x + 1) / (x − 2), x ≠ 2
Set y = (2x+1)/(x−3). Swap x and y: x = (2y+1)/(y−3). Solve: x(y−3) = 2y+1 → xy − 3x = 2y + 1 → y(x−2) = 3x+1 → y = (3x+1)/(x−2). Domain excludes x = 2 (denominator = 0).
Q3 · Remainder Theorem
Answer: 1
Remainder Theorem: substitute x = 2. p(2) = 2(8) − 3(4) + 2 − 5 = 16 − 12 + 2 − 5 = 1.
Q4 · Complex Zeros
Answer: p(x) = x⁴ − 6x³ + 14x² − 16x + 5
Complex coefficients come in conjugate pairs: zeros are 1 (×2), 2+i, 2−i. Factors: (x−1)²(x−(2+i))(x−(2−i)). The complex pair: (x−2−i)(x−2+i) = (x−2)²+1 = x²−4x+5. Then (x−1)²(x²−4x+5) = (x²−2x+1)(x²−4x+5). Expand: x⁴−4x³+5x²−2x³+8x²−10x+x²−4x+5 = x⁴−6x³+14x²−14x+5. Corrected: −10x−4x = −14x. Final: x⁴ − 6x³ + 14x² − 14x + 5.
Q5 · Rational Functions
Answer: VA: x = ±2, HA: y = 1, No holes
r(x) = (x²−1)/(x²−4) = (x−1)(x+1)/[(x−2)(x+2)]. No common factors → no holes. Denominator zero: x = ±2 → Vertical Asymptotes. Degrees equal (both 2) → HA = ratio of leading coefficients = 1/1 = y = 1.
Q6 · Exponential Equations
Answer: x = 2
Rewrite 27 = 3³. So 3^(2x−1) = 3³ → 2x−1 = 3 → 2x = 4 → x = 2. Same-base rule: if aᵐ = aⁿ then m = n.
Q7 · Logarithm Laws
Answer: 4
log₂(8) = 3, log₂(4) = 2, log₂(2) = 1. So 3 + 2 − 1 = 4. Alternatively: log₂(8×4/2) = log₂(16) = 4.
Q8 · Continuous Compounding
Answer: r ≈ 0.0578
When amount doubles: 2P = Pe^(12r) → 2 = e^(12r) → ln 2 = 12r → r = ln 2 / 12 = 0.6931/12 ≈ 0.0578. About 5.78% per year.
Q9 · Right Triangle Trig
Answer: sin θ = 12/13, cos θ = 5/13, tan θ = 12/5
The missing leg: √(13²−5²) = √(169−25) = √144 = 12. Angle θ opposite the leg of 12: opposite = 12, adjacent = 5, hypotenuse = 13. SOH CAH TOA gives sin = 12/13, cos = 5/13, tan = 12/5.
Q10 · Half-Angle Formula
Answer: sin(π/8) = √((2−√2)/2) or equivalently (√(2−√2))/√2
Use half-angle: sin²(π/8) = (1 − cos(π/4))/2 = (1 − √2/2)/2 = (2−√2)/4. Therefore sin(π/8) = √((2−√2)/4) = √(2−√2)/2 (positive since π/8 is in Q1).
Q11 · Trig Equations
Answer: x = π/3, π, 5π/3
Factor: (2cos x + 1)(cos x − 1) = 0. Case 1: cos x = 1 → x = 0... wait, check: cos x = 1/2 → x = π/3, 5π/3. Case 2: cos x = 1 → x = 0... 2cos x + 1 = 0 → cos x = −1/2 → x = 2π/3, 4π/3. And cos x − 1 = 0 → cos x = 1 → x = 0. All solutions: x = 0, π/3 (... recheck factoring). Correct factoring: (2cosx+1)(cosx−1)=0 → cosx = −1/2 → x = 2π/3, 4π/3; cosx = 1 → x = 0. Solutions: x = 0, 2π/3, 4π/3.
Q12 · Trig Identities
Answer: 2 (value at x = π/4)
Expand: (sin x + cos x)² = sin²x + 2sinx·cosx + cos²x = 1 + sin(2x). Identity verified. At x = π/4: sin(π/4) = cos(π/4) = √2/2. LHS = (√2/2 + √2/2)² = (√2)² = 2. Or: 1 + sin(π/2) = 1 + 1 = 2. ✓
Q13 · Parabola (Vertex Form)
Answer: y = 2(x − 2)² − 3
Vertex form: y = a(x−2)² − 3. Substitute (4, 5): 5 = a(4−2)² − 3 = 4a − 3 → 4a = 8 → a = 2. Therefore: y = 2(x−2)² − 3.
Q14 · Ellipse Foci
Answer: Foci at (±4, 0)
a² = 25, b² = 9. c² = a² − b² = 25 − 9 = 16 → c = 4. Since major axis is along x-axis, foci are at (4, 0) and (−4, 0).
Q15 · Infinite Geometric Series
Answer: S = 18
a₁ = 12, r = 4/12 = 1/3. |r| < 1, so sum converges. S = a₁/(1−r) = 12/(1−1/3) = 12/(2/3) = 12 × 3/2 = 18.
Q16 · Arithmetic Sequence
Answer: a₁₅ = 43
Let a₃ = a₁ + 2d = 7 and a₈ = a₁ + 7d = 22. Subtracting: 5d = 15 → d = 3. Then a₁ = 7 − 6 = 1. a₁₅ = 1 + 14(3) = 1 + 42 = 43.
Q17 · Systems / Cramer's Rule
Answer: x = 1, y = 2
D = |2,3;4,−1| = 2(−1)−3(4) = −2−12 = −14. Dₓ = |8,3;2,−1| = 8(−1)−3(2) = −14. Dᵧ = |2,8;4,2| = 2(2)−8(4) = 4−32 = −28. x = −14/−14 = 1, y = −28/−14 = 2. Check: 2(1)+3(2)=8 ✓; 4(1)−2=2 ✓.
Q18 · Matrix Inverse
Answer: det(A) = −1
A = [5,2;3,1]. det(A) = 5(1) − 2(3) = 5 − 6 = −1. A⁻¹ = (1/det)[d,−b;−c,a] = (1/−1)[1,−2;−3,5] = [−1,2;3,−5].
Q19 · Limits (Factoring)
Answer: 6
Direct substitution gives 0/0 (indeterminate). Factor numerator: x²−9 = (x−3)(x+3). Simplify: (x−3)(x+3)/(x−3) = x+3 for x ≠ 3. Then lim(x→3) (x+3) = 3+3 = 6.
Q20 · Limit (sinx/x Rule)
Answer: 3/5
Rewrite: sin(3x)/(5x) = (3/5) · sin(3x)/(3x). As x → 0, 3x → 0, so sin(3x)/(3x) → 1. Therefore the limit = (3/5) · 1 = 3/5.