Simplified Quantitative Formulas: Functions
- Function: A rule that assigns each input exactly one output. Notation: f(x).
- Domain: Set of all possible inputs (x-values).
- Range: Set of all possible outputs (f(x)-values).
- Types: Linear (f(x)=mx+b), Quadratic (f(x)=ax²+bx+c), Polynomial, Rational, Exponential, etc.
- Composite Function: (f∘g)(x) = f(g(x)).
- Inverse Function: f⁻¹(y) = x if f(x) = y.
- Key Properties: One-one, onto, bijection, even/odd, periodicity.
What do these mean? (Super Simple Explanations & Examples)
- Function: Like a vending machine: put in x, get out f(x).
- Example: f(x) = 2x+1. If x=3, f(3)=7. - Domain: All x you can use. Example: For f(x)=1/x, x≠0.
- Range: All possible answers. Example: For f(x)=x², range is all numbers ≥0.
- Composite: f(g(x)) means do g first, then f. Example: If f(x)=x+2, g(x)=3x, then f(g(2))=f(6)=8.
- Inverse: Reverses the function. Example: If f(x)=x+5, then f⁻¹(y)=y-5.
Basic Functions
A function is a relation between a set of inputs (domain) and a set of outputs (range) where each input is related to exactly one output.
Function Definition
A function f from set A to set B is a rule that assigns to each element x in A exactly one element y in B.
Notation: f: A → B
f(x) = y, where x ∈ A and y ∈ B
Example: f(x) = 2x + 1
Domain: All real numbers
Range: All real numbers
f(2) = 2(2) + 1 = 5
Example 1: Basic Function Operations
Q1.
Given f(x) = 3x - 2, find f(4) and f(-1)
Solution:
f(4) = 3(4) - 2 = 12 - 2 = 10
f(-1) = 3(-1) - 2 = -3 - 2 = -5
Q2.
If f(x) = x² + 2x - 3, find f(0) and f(2)
Solution:
f(0) = 0² + 2(0) - 3 = -3
f(2) = 2² + 2(2) - 3 = 4 + 4 - 3 = 5
Q3.
Given f(x) = √(x + 4), find the domain and f(5)
Solution:
Domain: x + 4 ≥ 0, so x ≥ -4
f(5) = √(5 + 4) = √9 = 3
Q4.
If f(x) = 1/(x - 2), find the domain and f(3)
Solution:
Domain: x - 2 ≠ 0, so x ≠ 2
f(3) = 1/(3 - 2) = 1/1 = 1
Linear Functions
Linear functions are functions of the form f(x) = mx + b, where m is the slope and b is the y-intercept.
Linear Function Properties
Linear functions have constant rate of change and graph as straight lines.
General form: f(x) = mx + b
m: slope (rate of change)
b: y-intercept (value when x = 0)
Example: f(x) = 2x + 3
Slope (m) = 2
Y-intercept (b) = 3
Rate of change = 2 units up for every 1 unit right
Example 1: Linear Function Analysis
Q1.
Find the slope and y-intercept of f(x) = -3x + 4
Solution:
Slope (m) = -3
Y-intercept (b) = 4
Q2.
Write the equation of a line with slope 2 passing through point (3, 5)
Solution:
Using point-slope form: y - y₁ = m(x - x₁)
y - 5 = 2(x - 3)
y = 2x - 6 + 5
y = 2x - 1
Q3.
Find the x-intercept of f(x) = 4x - 8
Solution:
Set f(x) = 0
4x - 8 = 0
4x = 8
x = 2
Q4.
Determine if the lines f(x) = 2x + 1 and g(x) = -0.5x + 4 are perpendicular
Solution:
For perpendicular lines, m₁ × m₂ = -1
2 × (-0.5) = -1
Therefore, the lines are perpendicular
Quadratic Functions
Quadratic functions are functions of the form f(x) = ax² + bx + c, where a ≠ 0.
Quadratic Function Properties
Quadratic functions graph as parabolas and have various important properties.
General form: f(x) = ax² + bx + c
Vertex form: f(x) = a(x - h)² + k
Vertex: (-b/2a, f(-b/2a))
Axis of symmetry: x = -b/2a
Example: f(x) = x² - 4x + 3
a = 1, b = -4, c = 3
Vertex: (2, -1)
Axis of symmetry: x = 2
Example 1: Quadratic Function Analysis
Q1.
Find the vertex of f(x) = x² - 6x + 8
Solution:
x-coordinate = -b/2a = -(-6)/2(1) = 3
y-coordinate = f(3) = 3² - 6(3) + 8 = -1
Vertex: (3, -1)
Q2.
Find the roots of f(x) = 2x² - 5x + 3
Solution:
Using quadratic formula: x = (-b ± √(b² - 4ac))/2a
x = (5 ± √(25 - 24))/4
x = (5 ± 1)/4
x = 1.5 or x = 1
Q3.
Write f(x) = x² + 4x - 5 in vertex form
Solution:
Complete the square:
f(x) = (x² + 4x + 4) - 4 - 5
f(x) = (x + 2)² - 9
Q4.
Find the maximum/minimum value of f(x) = -2x² + 8x - 3
Solution:
Since a < 0, the function has a maximum
x-coordinate of vertex = -b/2a = -8/-4 = 2
Maximum value = f(2) = -2(2)² + 8(2) - 3 = 5
Polynomial Functions
Polynomial functions are functions of the form f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀.
Polynomial Function Properties
Polynomial functions have various properties based on their degree and coefficients.
Degree: Highest power of x
Leading coefficient: Coefficient of highest power
End behavior: Determined by degree and leading coefficient
Roots: Solutions to f(x) = 0
Example: f(x) = 2x³ - 3x² + 4x - 1
Degree: 3
Leading coefficient: 2
End behavior: As x → ∞, f(x) → ∞; As x → -∞, f(x) → -∞
Example 1: Polynomial Function Analysis
Q1.
Find the degree and leading coefficient of f(x) = 3x⁴ - 2x² + 5x - 1
Solution:
Degree = 4 (highest power)
Leading coefficient = 3
Q2.
Describe the end behavior of f(x) = -x³ + 2x² - 3
Solution:
Degree is odd (3)
Leading coefficient is negative
As x → ∞, f(x) → -∞
As x → -∞, f(x) → ∞
Q3.
Find the roots of f(x) = x³ - 4x² + 4x
Solution:
Factor: f(x) = x(x² - 4x + 4)
f(x) = x(x - 2)²
Roots: x = 0, x = 2 (double root)
Q4.
Determine if x = 1 is a root of f(x) = x³ - 2x² + x - 1
Solution:
Use synthetic division or substitution
f(1) = 1³ - 2(1)² + 1 - 1 = -1
Since f(1) ≠ 0, x = 1 is not a root
Rational Functions
Rational functions are functions of the form f(x) = P(x)/Q(x), where P(x) and Q(x) are polynomials.
Rational Function Properties
Rational functions have various properties including asymptotes and holes.
Vertical asymptotes: Where Q(x) = 0
Horizontal asymptotes: Based on degrees of P(x) and Q(x)
Holes: Where P(x) and Q(x) have common factors
Example: f(x) = (x² - 4)/(x - 2)
Vertical asymptote: x = 2
Hole at x = 2 (after simplifying)
Horizontal asymptote: y = x + 2
Example 1: Rational Function Analysis
Q1.
Find the vertical asymptotes of f(x) = (x + 3)/(x² - 4)
Solution:
Set denominator = 0
x² - 4 = 0
x = ±2
Vertical asymptotes: x = 2 and x = -2
Q2.
Find the horizontal asymptote of f(x) = (2x² + 3)/(x² - 1)
Solution:
Degrees of numerator and denominator are equal
Horizontal asymptote: y = 2/1 = 2
Q3.
Find any holes in f(x) = (x² - 9)/(x - 3)
Solution:
Factor numerator: (x + 3)(x - 3)/(x - 3)
Hole at x = 3
After simplifying: f(x) = x + 3, x ≠ 3
Q4.
Find the domain of f(x) = (x + 2)/(x² - 5x + 6)
Solution:
Set denominator ≠ 0
x² - 5x + 6 ≠ 0
(x - 2)(x - 3) ≠ 0
Domain: x ≠ 2 and x ≠ 3
Advanced Concepts
Advanced topics in functions including composition, inverse functions, and transformations.
Advanced Function Concepts
Advanced function concepts include composition, inverse functions, and various transformations.
Composition: (f ∘ g)(x) = f(g(x))
Inverse: f⁻¹(f(x)) = x
Transformations:
- Vertical shift: f(x) ± k
- Horizontal shift: f(x ± h)
- Vertical stretch/compression: af(x)
- Horizontal stretch/compression: f(bx)
Example: f(x) = x², g(x) = x + 2
(f ∘ g)(x) = (x + 2)²
f⁻¹(x) = √x (for x ≥ 0)
f(x + 3) - 2 = (x + 3)² - 2
Example 1: Function Composition
Q1.
Find (f ∘ g)(x) if f(x) = 2x + 1 and g(x) = x²
Solution:
(f ∘ g)(x) = f(g(x))
= f(x²)
= 2(x²) + 1
= 2x² + 1
Q2.
Find the inverse of f(x) = 3x - 2
Solution:
y = 3x - 2
x = 3y - 2
x + 2 = 3y
y = (x + 2)/3
f⁻¹(x) = (x + 2)/3
Q3.
Describe the transformation of f(x) = x² to g(x) = (x - 2)² + 3
Solution:
Horizontal shift right 2 units
Vertical shift up 3 units
Q4.
Find (g ∘ f)(x) if f(x) = √x and g(x) = x² + 1
Solution:
(g ∘ f)(x) = g(f(x))
= g(√x)
= (√x)² + 1
= x + 1