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📐 Pre-Calculus Study Notebook

20 Key Problems · Self-Study Edition · Write & Learn ✏️

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📘 UNIT 1 — Functions & Their Graphs Ch. 1

VERTICAL LINE TEST → If a vertical line hits the graph MORE than once → NOT a function!
DOMAIN = all valid x-values | RANGE = all valid y-values

✏️ EXAMPLE

Is \( f(x) = x^2 \) a function? → YES ✓ (each x gives exactly one y)
Is \( x^2 + y^2 = 9 \) a function? → NO ✗ (one x can give two y values)

Q1.

What is the domain of \( f(x) = \dfrac{1}{\sqrt{x - 3}} \) ? ⚠️ COMMON TRAP

→ Remember: sqrt needs ≥ 0, denominator needs ≠ 0!

💡 EXPLANATION

For \( \sqrt{x-3} \): need \( x - 3 \geq 0 \) → \( x \geq 3 \)
BUT the denominator \( \sqrt{x-3} \neq 0 \) → \( x \neq 3 \)
Combining: \( x > 3 \) ✓
Trap: Many pick ≥ 3, forgetting denominator can't be 0!

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Q2.

If \( f(x) = 2x + 1 \) and \( g(x) = x^2 \), find \( (f \circ g)(3) \).

→ f∘g means f(g(x)). Do INSIDE first!

💡 EXPLANATION

\( (f \circ g)(3) = f(g(3)) \)
Step 1: \( g(3) = 3^2 = 9 \)
Step 2: \( f(9) = 2(9) + 1 = \mathbf{19} \) ✓
Trap: Don't do f(3) first! Always start INSIDE.

📗 UNIT 2 — Polynomial & Rational Functions Ch. 2-3

REMAINDER THEOREM: Divide \(f(x)\) by \((x-c)\) → remainder = \(f(c)\)
FACTOR THEOREM: \(f(c) = 0\) ↔ \((x-c)\) is a FACTOR
End Behavior: Even degree → same ends | Odd degree → opposite ends

✏️ EXAMPLE

\( f(x) = x^3 - 2x^2 - 5x + 6 \). Is \((x-1)\) a factor?
→ \( f(1) = 1 - 2 - 5 + 6 = 0 \) ✓ → YES, \((x-1)\) is a factor!

Q3.

When \( f(x) = x^3 - 3x^2 + 2x - 5 \) is divided by \( (x - 2) \), the remainder is:

💡 EXPLANATION

Remainder Theorem: plug in \(x = 2\)
\( f(2) = 8 - 12 + 4 - 5 = \mathbf{-5} \) ✓
No long division needed! Just substitute the zero of the divisor.

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Q4.

The polynomial \( p(x) = x^4 - 5x^2 + 4 \). How many real zeros does it have? ⚠️ TRICKY

→ Treat as a quadratic! Let u = x²

💡 EXPLANATION

Let \(u = x^2\): \( u^2 - 5u + 4 = (u-1)(u-4) = 0 \)
→ \(u = 1\) or \(u = 4\)
→ \(x^2 = 1 \Rightarrow x = \pm 1\) and \(x^2 = 4 \Rightarrow x = \pm 2\)
→ 4 real zeros: \(-2, -1, 1, 2\) ✓

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Q5.

Find the vertical asymptote(s) of \( f(x) = \dfrac{x+2}{x^2 - x - 6} \). ⚠️ HOLE vs ASYMPTOTE

→ Factor denominator first! Cancel = hole, leftover = asymptote

💡 EXPLANATION

Factor: \( x^2 - x - 6 = (x-3)(x+2) \)
\( f(x) = \dfrac{(x+2)}{(x-3)(x+2)} = \dfrac{1}{x-3} \) (with hole at \(x = -2\))
→ Vertical asymptote: \(x = 3\) only ✓
\(x = -2\) creates a HOLE (cancelled factor), NOT an asymptote!

📙 UNIT 3 — Exponential & Logarithmic Functions Ch. 4

LOG ↔ EXP: \( \log_b x = y \) ↔ \( b^y = x \)
LOG RULES: Product → +, Quotient → −, Power → ×
CHANGE OF BASE: \( \log_b x = \dfrac{\ln x}{\ln b} \)

✏️ EXAMPLE

Solve: \( 2^{x+1} = 16 \)
→ \( 2^{x+1} = 2^4 \) → \( x + 1 = 4 \) → \( x = 3 \) ✓

Q6.

Solve: \( \log_2(x+3) + \log_2(x-1) = 5 \) ⚠️ CHECK DOMAIN!

→ Combine logs using Product Rule, then convert to exponential

💡 EXPLANATION

\( \log_2[(x+3)(x-1)] = 5 \) → \( (x+3)(x-1) = 2^5 = 32 \)
\( x^2 + 2x - 3 = 32 \) → \( x^2 + 2x - 35 = 0 \) → \( (x+7)(x-5) = 0 \)
\(x = 5\) or \(x = -7\) — but check domain:
\(x = -7\): \(\log_2(-4)\) is undefined! → Reject
Answer: \(x = 5\) ✓

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Q7.

Simplify: \( \log_3 \dfrac{27\sqrt{3}}{9} \)

→ Convert everything to powers of 3!

💡 EXPLANATION

\( \dfrac{27\sqrt{3}}{9} = \dfrac{3^3 \cdot 3^{1/2}}{3^2} = 3^{3 + 1/2 - 2} = 3^{3/2} \)
\( \log_3 3^{3/2} = \mathbf{\dfrac{3}{2}} \) ✓
Key: Convert radicals to fractional exponents first!

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Q8.

A bacteria doubles every 3 hours. Starting with 500, how many after 12 hours?

→ Formula: \(A = A_0 \cdot 2^{t/d}\) where d = doubling time

💡 EXPLANATION

\( A = 500 \cdot 2^{12/3} = 500 \cdot 2^4 = 500 \cdot 16 = \mathbf{8{,}000} \) ✓
In 12 hours: doubles 4 times (3→6→9→12 hrs) → 500 × 16 = 8,000

📕 UNIT 4 — Trigonometry Ch. 5-6

SOH-CAH-TOA for right triangles
ALL STUDENTS TAKE CALCULUS (Quadrant signs: All, Sin, Tan, Cos = +)
Unit Circle: \((\cos\theta, \sin\theta)\) | Radius = 1

✏️ EXAMPLE — Special Angles

\(\sin 30° = \dfrac{1}{2}\) | \(\cos 45° = \dfrac{\sqrt{2}}{2}\) | \(\tan 60° = \sqrt{3}\)

Q9.

If \( \sin\theta = \dfrac{3}{5} \) and \(\theta\) is in Quadrant II, find \(\cos\theta\). ⚠️ SIGN TRAP

→ Use Pythagorean identity + check the quadrant sign!

💡 EXPLANATION

\( \cos^2\theta = 1 - \sin^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25} \)
→ \( |\cos\theta| = \dfrac{4}{5} \)
In QII: cos is NEGATIVE → \( \cos\theta = \mathbf{-\dfrac{4}{5}} \) ✓
In QII: sin(+), cos(−), tan(−)

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Q10.

The period of \( f(x) = 3\sin(2x - \pi) + 1 \) is: ⚠️ COMMON MISTAKE

→ Period = 2π ÷ |B| where B is coefficient of x

💡 EXPLANATION

For \(y = A\sin(Bx + C) + D\): Period \(= \dfrac{2\pi}{|B|}\)
Here \(B = 2\): Period \(= \dfrac{2\pi}{2} = \mathbf{\pi}\) ✓
The \(-\pi\) shifts the phase, NOT the period! Don't confuse C with B.

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Q11.

Simplify: \( \dfrac{\sin^2\theta + \cos^2\theta}{\tan\theta} \cdot \sec\theta \)

→ Pythagorean identity first, then convert to sin/cos

💡 EXPLANATION

\( \sin^2\theta + \cos^2\theta = 1 \) (Pythagorean identity)
\( = \dfrac{1}{\tan\theta} \cdot \sec\theta = \dfrac{\cos\theta}{\sin\theta} \cdot \dfrac{1}{\cos\theta} = \dfrac{1}{\sin\theta} = \mathbf{\csc\theta} \) ✓

🔴 UNIT 5 — Trig Identities & Equations Ch. 6-7

DOUBLE ANGLE: \(\sin 2\theta = 2\sin\theta\cos\theta\) | \(\cos 2\theta = \cos^2\theta - \sin^2\theta\)
RANGE of Inverse: \(\arcsin\): \(\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\) | \(\arccos\): \([0, \pi]\)

✏️ EXAMPLE

Solve \(2\sin x - 1 = 0\) on \([0, 2\pi)\):
→ \(\sin x = \dfrac{1}{2}\) → \(x = \dfrac{\pi}{6}\) or \(x = \dfrac{5\pi}{6}\)

Q12.

Solve: \( 2\cos^2 x - 3\cos x + 1 = 0 \) on \([0, 2\pi)\) ⚠️ QUADRATIC IN DISGUISE

→ Let u = cos x, solve quadratic, then find angles

💡 EXPLANATION

Factor: \((2\cos x - 1)(\cos x - 1) = 0\)
→ \(\cos x = \dfrac{1}{2}\) → \(x = \dfrac{\pi}{3}, \dfrac{5\pi}{3}\)
→ \(\cos x = 1\) → \(x = 0\)
All solutions: \(x = 0, \dfrac{\pi}{3}, \dfrac{5\pi}{3}\) ✓

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Q13.

\( \arcsin\!\left(\sin\!\left(\dfrac{5\pi}{6}\right)\right) = \) ? ⚠️ RANGE RESTRICTION TRAP

→ arcsin's output range is only [−π/2, π/2]!

💡 EXPLANATION

\(\sin\!\left(\dfrac{5\pi}{6}\right) = \dfrac{1}{2}\)
Now: \(\arcsin\!\left(\dfrac{1}{2}\right) = \dfrac{\pi}{6}\) ✓ (must be in \(\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\))
arcsin NEVER returns \(\dfrac{5\pi}{6}\) — that's outside its range!

📓 UNIT 6 — Sequences, Series & Conics Ch. 8-9

ARITHMETIC: \(a_n = a_1 + (n-1)d\) | \(S_n = \dfrac{n}{2}(a_1 + a_n)\)
GEOMETRIC: \(a_n = a_1 r^{n-1}\) | Infinite Sum: \(S = \dfrac{a_1}{1-r}\) if \(|r| < 1\)
CONIC HINT: \(x^2 + y^2\) → circle | \(x^2 - y^2\) → hyperbola

✏️ EXAMPLE — Geometric Series

Sum of \(1 + \dfrac{1}{2} + \dfrac{1}{4} + \cdots\) = \(\dfrac{1}{1 - \frac{1}{2}} = 2\) (infinite geometric, \(r = \dfrac{1}{2}\))

Q14.

Find the sum of the first 20 terms of the arithmetic sequence: 5, 8, 11, 14, ...

→ d = 3, use Sₙ formula

💡 EXPLANATION

\(a_1 = 5,\ d = 3,\ n = 20\)
\(a_{20} = 5 + 19(3) = 62\)
\(S_{20} = \dfrac{20}{2}(5 + 62) = 10 \times 67 = \mathbf{670}\) ✓

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Q15.

The infinite geometric series \( 3 + 1 + \dfrac{1}{3} + \cdots \) converges to:

💡 EXPLANATION

\(r = \dfrac{1}{3},\ a_1 = 3\)
\(S = \dfrac{a_1}{1-r} = \dfrac{3}{1 - \frac{1}{3}} = \dfrac{3}{\frac{2}{3}} = \dfrac{9}{2}\) ✓

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Q16.

Identify the conic: \( \dfrac{x^2}{9} - \dfrac{y^2}{4} = 1 \) ⚠️ ELLIPSE vs HYPERBOLA

→ Key: What sign is between the fractions?

💡 EXPLANATION

Minus sign between the two fraction terms → Hyperbola ✓
\(x^2\) term is positive → opens left and right (horizontal hyperbola)
Ellipse uses PLUS: \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\)

🔵 UNIT 7 — Intro to Limits & Vectors Ch. 10-11

LIMIT: "What value does f(x) approach as x→c?" (doesn't matter what f(c) is!)
DOT PRODUCT: \(\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta\)
PERPENDICULAR: \(\vec{u} \cdot \vec{v} = 0\)

Q17.

Evaluate: \( \displaystyle\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2} \) ⚠️ INDETERMINATE FORM

→ Factor! Never just substitute first when you get 0/0

💡 EXPLANATION

Direct sub: \(\dfrac{4-4}{2-2} = \dfrac{0}{0}\) → indeterminate!
Factor: \(\dfrac{(x+2)(x-2)}{x-2} = x + 2\)
\(\lim_{x\to 2}(x+2) = \mathbf{4}\) ✓
The function is UNDEFINED at x=2, but the LIMIT is 4. These are different things!

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Q18.

Vectors \(\vec{u} = \langle 3, 4 \rangle\) and \(\vec{v} = \langle -4, 3 \rangle\). Are they perpendicular?

→ Compute dot product. If = 0, they're perpendicular!

💡 EXPLANATION

\( \vec{u} \cdot \vec{v} = (3)(-4) + (4)(3) = -12 + 12 = \mathbf{0} \) ✓
Since the dot product equals 0 → Perpendicular!
Note: \(|\vec{u}| = |\vec{v}| = 5\). These are perpendicular unit-scaled vectors.

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Q19.

\( \displaystyle\lim_{x \to \infty} \dfrac{4x^3 - 2x + 1}{6x^3 + 5} = \) ?

→ Same degree? Ratio of leading coefficients!

💡 EXPLANATION

Both numerator and denominator are degree 3 (same highest power).
→ Limit = ratio of leading coefficients = \(\dfrac{4}{6} = \mathbf{\dfrac{2}{3}}\) ✓
Rules: Higher degree top → ∞ | Higher degree bottom → 0 | Same degree → ratio of coefficients

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Q20.

Which expression is equivalent to \( \cos^2\theta - \sin^2\theta \)? ⚠️ IDENTITY CHALLENGE

→ Double angle identities! You must memorize these.

💡 EXPLANATION

\(\cos 2\theta\) has three equivalent forms:
1. \(\cos^2\theta - \sin^2\theta\)
2. \(1 - 2\sin^2\theta\)
3. \(2\cos^2\theta - 1\)
So Both B and C are correct! ✓
This is a very common exam question — know all three forms of cos 2θ!

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