Function Concepts
Complete Notes & Formulae for Twelfth Grade (Precalculus)
1. Domain and Range
Definitions:
Domain:
The set of all possible input values (x-values) for which the function is defined
Range:
The set of all possible output values (y-values) that result from the domain
Domain Restrictions:
Common restrictions to watch for:
1. Division by Zero:
Denominator cannot equal zero
Example: \( f(x) = \frac{1}{x-3} \) → Domain: \( x \neq 3 \) or \( (-\infty, 3) \cup (3, \infty) \)
2. Even Roots (Square Root, 4th Root, etc.):
Expression under even root must be ≥ 0
Example: \( f(x) = \sqrt{x-2} \) → Domain: \( x \geq 2 \) or \( [2, \infty) \)
3. Logarithms:
Argument must be > 0
Example: \( f(x) = \ln(x) \) → Domain: \( x > 0 \) or \( (0, \infty) \)
Notation Types:
| Interval Notation | Inequality Notation | Meaning |
|---|---|---|
| [a, b] | \( a \leq x \leq b \) | Includes endpoints |
| (a, b) | \( a < x < b \) | Excludes endpoints |
| [a, b) | \( a \leq x < b \) | Includes a, excludes b |
| \( (-\infty, \infty) \) | All real numbers | No restrictions |
2. Identify Functions
Definition of a Function:
A relation where each input (x-value) corresponds to exactly ONE output (y-value)
\[ \text{For every } x \text{ in the domain, there is exactly one } y \text{ in the range} \]
Vertical Line Test:
Rule:
If any vertical line intersects the graph at MORE THAN ONE POINT, it is NOT a function
✓ IS a Function:
• Lines: \( y = 2x + 3 \)
• Parabolas opening up/down: \( y = x^2 \)
• Any graph that passes vertical line test
✗ NOT a Function:
• Circles: \( x^2 + y^2 = 25 \)
• Sideways parabolas: \( x = y^2 \)
• Any graph that fails vertical line test
3. Evaluate Functions
Function Notation:
\[ f(x) = \text{expression in terms of } x \]
Read as: "f of x" or "f at x"
How to Evaluate:
Steps:
1. Replace every x in the function with the given value
2. Use parentheses when substituting
3. Simplify following order of operations
Examples:
Given: \( f(x) = 3x^2 - 2x + 1 \), find \( f(2) \)
\( f(2) = 3(2)^2 - 2(2) + 1 \)
\( f(2) = 3(4) - 4 + 1 \)
\( f(2) = 12 - 4 + 1 = 9 \)
Given: \( f(x) = x^2 + 5 \), find \( f(a+1) \)
\( f(a+1) = (a+1)^2 + 5 \)
\( f(a+1) = a^2 + 2a + 1 + 5 \)
\( f(a+1) = a^2 + 2a + 6 \)
4. Find Values Using Function Graphs
Reading Graphs:
To find f(a) from a graph:
1. Locate x = a on the horizontal axis
2. Draw a vertical line up/down to the curve
3. Read the y-coordinate of that point
4. That y-value is f(a)
To find x when f(x) = b:
1. Locate y = b on the vertical axis
2. Draw a horizontal line to the curve
3. Read the x-coordinate(s) of intersection point(s)
5. Complete a Table for a Function Graph
Process:
1. For each x-value in the table, locate it on the graph
2. Find the corresponding y-value (height of the curve)
3. Fill in the y-value in the table
4. Repeat for all x-values
Key Skills:
• Accurately reading coordinates from graphs
• Understanding the relationship between x and f(x)
• Interpolating between plotted points when necessary
6. Add, Subtract, Multiply, and Divide Functions
Function Operations:
Addition:
\[ (f + g)(x) = f(x) + g(x) \]
Subtraction:
\[ (f - g)(x) = f(x) - g(x) \]
Multiplication:
\[ (f \cdot g)(x) = f(x) \cdot g(x) \]
Division:
\[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}, \quad g(x) \neq 0 \]
Example:
Given: \( f(x) = 2x + 3 \) and \( g(x) = x^2 - 1 \)
\( (f + g)(x) = (2x + 3) + (x^2 - 1) = x^2 + 2x + 2 \)
\( (f - g)(x) = (2x + 3) - (x^2 - 1) = -x^2 + 2x + 4 \)
\( (f \cdot g)(x) = (2x + 3)(x^2 - 1) = 2x^3 + 3x^2 - 2x - 3 \)
\( \left(\frac{f}{g}\right)(x) = \frac{2x + 3}{x^2 - 1}, \quad x \neq \pm 1 \)
Domain of Combined Functions:
• For +, −, ×: Domain is the intersection of the domains of f and g
• For ÷: Domain is intersection of domains, but also exclude where g(x) = 0
7. Composition of Functions
Definition:
Composition is applying one function to the output of another function
\[ (f \circ g)(x) = f(g(x)) \]
Read as: "f composed with g of x" or "f of g of x"
How to Evaluate:
Steps for \( (f \circ g)(x) \):
1. Start with the inner function: Evaluate g(x)
2. Take the result from step 1
3. Substitute it into the outer function f
4. Simplify
Examples:
Given: \( f(x) = 2x + 1 \) and \( g(x) = x^2 \), find \( (f \circ g)(x) \)
\( (f \circ g)(x) = f(g(x)) \)
\( = f(x^2) \)
\( = 2(x^2) + 1 \)
\( = 2x^2 + 1 \)
Using same functions, find \( (g \circ f)(x) \)
\( (g \circ f)(x) = g(f(x)) \)
\( = g(2x + 1) \)
\( = (2x + 1)^2 \)
\( = 4x^2 + 4x + 1 \)
⚠️ Note: \( f \circ g \neq g \circ f \) (composition is NOT commutative!)
Domain of Composition:
• Domain of \( (f \circ g)(x) \): All x in domain of g such that g(x) is in domain of f
• Check both: x must work in g, and g(x) must work in f
8. Quick Reference Summary
Key Concepts:
Domain: Set of all possible x-values
Range: Set of all possible y-values
Vertical Line Test: If any vertical line hits graph more than once → NOT a function
Function Operations:
\( (f + g)(x) = f(x) + g(x) \)
\( (f - g)(x) = f(x) - g(x) \)
\( (f \cdot g)(x) = f(x) \cdot g(x) \)
\( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \), \( g(x) \neq 0 \)
Composition: \( (f \circ g)(x) = f(g(x)) \)
Work from inside out!
📚 Study Tips
✓ Always check domain restrictions (division by zero, square roots, logs)
✓ Use vertical line test to quickly identify functions from graphs
✓ When evaluating functions, use parentheses for substitution
✓ For composition, work from the inside function outward
✓ Remember: Composition is NOT commutative (f∘g ≠ g∘f)