Functions: Notation, Domain, Range and Composition

1. 1 A function is a rule that maps each input to exactly one output. Machine metaphor: input → [rule] → output. f(x) = 2x + 1. 'f of x' means apply the rule to x. f(3) = 2(3)+1 = 7. Students find f(0), f(-2), f(1/2).
2. 2 Domain = the set of valid inputs. Range = the set of resulting outputs. For f(x) = √x, domain is x ≥ 0 (cannot square root a negative). Range is f(x) ≥ 0. For f(x) = 1/x, domain is x ≠ 0. Students state domain and range for three functions.
3. 3 'If f(x) = 3x − 2 and f(a) = 13, find a.' Set up: 3a − 2 = 13. Solve: a = 5. 'If g(x) = x² + 1 and g(b) = 17, find b.' Two solutions: b = ±4. Discuss: why two answers here?
4. 4 f(x) = 2x + 1, g(x) = x². fg(x) means 'apply g first, then f.' fg(3) = f(g(3)) = f(9) = 2(9)+1 = 19. gf(3) = g(f(3)) = g(7) = 49. 'Note: fg ≠ gf in general.' Students find fg(x) algebraically: fg(x) = f(x²) = 2x²+1.
5. 5 The inverse f⁻¹(x) reverses the function. If f(x) = 2x+1, find f⁻¹(x). Method: write y = 2x+1, swap x and y: x = 2y+1, rearrange for y: y = (x−1)/2. So f⁻¹(x) = (x−1)/2. Verify: f(f⁻¹(x)) = x.
6. 6 Graph f(x) and f⁻¹(x). They are reflections of each other in the line y = x. Verify for f(x) = 2x+1 and f⁻¹(x) = (x−1)/2. Discuss why functions with restricted domains may have inverses where unrestricted ones do not.
7. 7 If fg(x) = 6x² + 1 and g(x) = x², find f(x). Work backwards: f must turn x² into 6x²+1, so f(x) = 6x+1. Verify: fg(x) = f(x²) = 6x²+1. ✓

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