βΈA function maps each input x to exactly one output f(x). Vertical Line Test confirms this graphically.
βΈDomain: all valid inputs. Avoid division by zero and square roots of negatives.
βΈTransformations: f(x β h) + k shifts right h, up k; βf(x) reflects over x-axis; f(βx) reflects over y-axis.
βΈComposition: (f β g)(x) = f(g(x)). Evaluate inner function first.
Even: f(βx) = f(x) | Odd: f(βx) = βf(x)
Inverse: f(fβ»ΒΉ(x)) = x (swap x and y, solve)
β¦ Worked Example
Given f(x) = 2x + 1 and g(x) = xΒ², find (f β g)(3).
Step 1: g(3) = 9
Step 2: f(9) = 2(9) + 1 = 19
Answer: 19
Functions Β· Domain
Q1What is the domain of f(x) = β(x β 3) / (x β 7)?
Functions Β· Composition
Q2If f(x) = xΒ² β 1 and g(x) = 2x + 3, what is (g β f)(2)?
UNIT 02Polynomial Functions
π Key Concepts to Memorize
βΈRemainder Theorem: Dividing p(x) by (x β c) gives remainder p(c).
βΈFactor Theorem: (x β c) is a factor of p(x) if and only if p(c) = 0.
βΈEnd behavior: determined by leading term. Even degree β same ends; Odd degree β opposite ends.
βΈMultiplicity: Even multiplicity β graph touches x-axis and bounces; Odd multiplicity β graph crosses.
Fundamental Theorem: Degree n polynomial has exactly n zeros (counting multiplicity, in β)
Rational Root Theorem: possible rational roots = Β±(factors of constant) / (factors of leading coeff)
β¦ Worked Example
Find all real zeros of p(x) = xΒ³ β 6xΒ² + 11x β 6.
βΈHorizontal asymptote rules: if deg(num) < deg(den) β y = 0; if degrees equal β y = ratio of leading coefficients; if deg(num) > deg(den) β no horizontal asymptote (oblique instead).
βΈHoles: occur where a common factor cancels from numerator and denominator.
Rational Functions Β· Asymptotes
Q5What is the horizontal asymptote of f(x) = (3xΒ² β 2) / (xΒ² + 5)?
UNIT 04Exponential & Logarithmic Functions
π Key Concepts to Memorize
βΈDefinition: log_b(x) = y means b^y = x. Always: base b > 0, b β 1, and x > 0.
βΈLog properties: Product β log(MN) = log M + log N; Quotient β log(M/N) = log M β log N; Power β log(M^p) = pΒ·log M.
βΈChange of base: log_b(x) = ln(x)/ln(b).
βΈNatural exponential: e β 2.718. Inverse of ln.
Compound Interest: A = P(1 + r/n)^(nt)
Continuous: A = Pe^(rt)
β¦ Worked Example
Solve: logβ(x + 3) = 4
Convert: x + 3 = 2β΄ = 16
Answer: x = 13
Logarithms Β· Properties
Q6Which expression is equal to logβ(8) + logβ(4)?
Q4β Answer: BEnd behavior of β3xβ΄ + 5xΒ² β 2
Leading term is β3xβ΄. Degree 4 is even β both ends behave the same. Leading coefficient β3 < 0 β both ends go to ββ. Therefore as x β Β±β, f(x) β ββ.
Q5β Answer: BHorizontal asymptote of (3xΒ²β2)/(xΒ²+5)
The degrees of numerator and denominator are both 2 (equal). The horizontal asymptote is the ratio of the leading coefficients: 3/1 = 3. So y = 3.
Q6β Answer: Blogβ(8) + logβ(4)
Product rule: logβ(8) + logβ(4) = logβ(8 Γ 4) = logβ(32). Since 2β΅ = 32, the answer is 5. Alternatively: logβ(8) = 3 and logβ(4) = 2, so 3 + 2 = 5.
Q7β Answer: BSolve 5^(2x) = 125
Write 125 = 5Β³. Then 5^(2x) = 5Β³ implies 2x = 3, so x = 3/2.
Q8β Answer: ASolve ln(x) + ln(xβ2) = ln(3)
Product rule: ln(x(xβ2)) = ln(3), so x(xβ2) = 3, giving xΒ² β 2x β 3 = 0, which factors as (xβ3)(x+1) = 0. Solutions: x = 3 or x = β1. But x = β1 makes ln(x) undefined. Therefore the only valid solution is x = 3.
Q9β Answer: Btan(225Β°)
225Β° is in Quadrant III (reference angle = 45Β°). In Q III, tangent is positive. tan(45Β°) = 1, so tan(225Β°) = +1. (In Q III, both sin and cos are negative, so their ratio tan is positive.)
Q10β Answer: BPeriod of 3sin(2x β Ο/4)
The general form is A sin(Bx + C). Here B = 2. Period = 2Ο / |B| = 2Ο / 2 = Ο.
We need the angle ΞΈ β [βΟ/2, Ο/2] where sin(ΞΈ) = ββ2/2. Since sin(Ο/4) = β2/2, and we need the negative value, the answer is ΞΈ = βΟ/4.
Q13β Answer: Acos(2ΞΈ) given sinΞΈ = 3/5, Quadrant II
In Q II: sin ΞΈ = 3/5, so cos ΞΈ = β4/5 (cos is negative in Q II; from cosΒ²ΞΈ = 1 β 9/25 = 16/25). cos(2ΞΈ) = cosΒ²ΞΈ β sinΒ²ΞΈ = 16/25 β 9/25 = 7/25.
Verification: 1 β 2sinΒ²ΞΈ = 1 β 2(9/25) = 1 β 18/25 = 7/25 β
Factor as a quadratic in cos x: (2cosx + 1)(cosx β 1) = 0.
Case 1: cosx = 1 β x = 0. (1 solution)
Case 2: cosx = β1/2 β x = 2Ο/3 or x = 4Ο/3. (2 solutions)
Total: 3 solutions: {0, 2Ο/3, 4Ο/3}.
Q15β Answer: BSum of 12 + 4 + 4/3 + Β·Β·Β·
First term aβ = 12. Common ratio r = 4/12 = 1/3. Since |r| = 1/3 < 1, the series converges. S = aβ/(1 β r) = 12/(1 β 1/3) = 12/(2/3) = 12 Γ 3/2 = 18.
Q16β Answer: BCoefficient of xΒ² in (x+3)β΅
The term with xΒ² in (x + 3)β΅ corresponds to k = 3 (since the power of x is 5 β k = 2, so k = 3).
Term = C(5,3) Β· xΒ² Β· 3Β³ = 10 Β· xΒ² Β· 27 = 270xΒ².
Coefficient = 270.
Q17β Answer: AFoci of xΒ²/25 + yΒ²/9 = 1
Here aΒ² = 25, bΒ² = 9. Since aΒ² > bΒ², major axis is along the x-axis. cΒ² = aΒ² β bΒ² = 25 β 9 = 16, so c = 4.
Foci are at (Β±4, 0).
Q18β Answer: AFocus of yΒ² = 12x
The form yΒ² = 4px is a rightward-opening parabola with vertex at origin. Here 4p = 12, so p = 3. The focus is at (3, 0).
Dot product: u Β· v = (3)(4) + (4)(β3) = 12 β 12 = 0.
Since the dot product equals zero, the vectors are perpendicular. The angle between them is 90Β°.
Q20β Answer: Diβ΄β·
Powers of i cycle with period 4: iΒΉ = i, iΒ² = β1, iΒ³ = βi, iβ΄ = 1.
Divide: 47 Γ· 4 = 11 remainder 3. So iβ΄β· = iΒ³ = βi.