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1.5 notation and algebra of functions

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  • 1. Notation and Algebra of Functions
  • 2. Functions are procedures that assign a unique output to each (valid) input. Notation and Algebra of Functions
  • 3. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions
  • 4. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions name of the function
  • 5. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions name of the function the input box
  • 6. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions name of the function the input box the defining formula
  • 7. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions name of the function the input box the defining formula the output
  • 8. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions The input box “ ( ) ” holds the input to be evaluated by the formula.
  • 9. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions The input box “ ( ) ” holds the input to be evaluated by the formula. Hence f (2) means to replace x by the input (2) in the formula,
  • 10. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions The input box “ ( ) ” holds the input to be evaluated by the formula. Hence f (2) means to replace x by the input (2) in the formula, so f (2) = (2) 2 – 2 (2) + 3
  • 11. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions The input box “ ( ) ” holds the input to be evaluated by the formula. Hence f (2) means to replace x by the input (2) in the formula, so f (2) = (2) 2 – 2 (2) + 3 = 3 = y.
  • 12. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions The input may be other mathematics expressions. The input box “ ( ) ” holds the input to be evaluated by the formula. Hence f (2) means to replace x by the input (2) in the formula, so f (2) = (2) 2 – 2 (2) + 3 = 3 = y.
  • 13. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions The input may be other mathematics expressions. Often in such problems we are to simplify the outputs algebraically. The input box “ ( ) ” holds the input to be evaluated by the formula. Hence f (2) means to replace x by the input (2) in the formula, so f (2) = (2) 2 – 2 (2) + 3 = 3 = y.
  • 14. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions The input may be other mathematics expressions. Often in such problems we are to simplify the outputs algebraically. The input box “ ( ) ” holds the input to be evaluated by the formula. Hence f (2) means to replace x by the input (2) in the formula, so f (2) = (2) 2 – 2 (2) + 3 = 3 = y. (The Square–Formula) (a ± b) 2 = a 2 ± 2ab + b 2 We write down the Square–Formula as a reminder below.
  • 15. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions The input may be other mathematics expressions. Often in such problems we are to simplify the outputs algebraically. The input box “ ( ) ” holds the input to be evaluated by the formula. Hence f (2) means to replace x by the input (2) in the formula, so f (2) = (2) 2 – 2 (2) + 3 = 3 = y. (The Square–Formula) (a ± b) 2 = a 2 ± 2ab + b 2 We write down the Square–Formula as a reminder below. Example A. Given f(x) = x 2 – 2x + 3, simplify the following. a. f (2a) = b. f (a + b) =
  • 16. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions The input may be other mathematics expressions. Often in such problems we are to simplify the outputs algebraically. The input box “ ( ) ” holds the input to be evaluated by the formula. Hence f (2) means to replace x by the input (2) in the formula, so f (2) = (2) 2 – 2 (2) + 3 = 3 = y. (The Square–Formula) (a ± b) 2 = a 2 ± 2ab + b 2 We write down the Square–Formula as a reminder below. Example A. Given f(x) = x 2 – 2x + 3, simplify the following. a. f (2a) = (2a) 2 – 2 (2a) + 3 b. f (a + b) =
  • 17. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions The input may be other mathematics expressions. Often in such problems we are to simplify the outputs algebraically. The input box “ ( ) ” holds the input to be evaluated by the formula. Hence f (2) means to replace x by the input (2) in the formula, so f (2) = (2) 2 – 2 (2) + 3 = 3 = y. (The Square–Formula) (a ± b) 2 = a 2 ± 2ab + b 2 We write down the Square–Formula as a reminder below. Example A. Given f(x) = x 2 – 2x + 3, simplify the following. a. f (2a) = (2a) 2 – 2 (2a) + 3 = 4a 2 – 4a + 3 b. f (a + b) =
  • 18. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions The input may be other mathematics expressions. Often in such problems we are to simplify the outputs algebraically. The input box “ ( ) ” holds the input to be evaluated by the formula. Hence f (2) means to replace x by the input (2) in the formula, so f (2) = (2) 2 – 2 (2) + 3 = 3 = y. (The Square–Formula) (a ± b) 2 = a 2 ± 2ab + b 2 We write down the Square–Formula as a reminder below. Example A. Given f(x) = x 2 – 2x + 3, simplify the following. a. f (2a) = (2a) 2 – 2 (2a) + 3 = 4a 2 – 4a + 3 b. f (a + b) = (a + b) 2 – 2 (a + b) + 3
  • 19. Functions are procedures that assign a unique output to each (valid) input. Most mathematics functions are given in mathematics formulas such as f (x) = x 2 – 2 x + 3 = y . Notation and Algebra of Functions The input may be other mathematics expressions. Often in such problems we are to simplify the outputs algebraically. The input box “ ( ) ” holds the input to be evaluated by the formula. Hence f (2) means to replace x by the input (2) in the formula, so f (2) = (2) 2 – 2 (2) + 3 = 3 = y. (The Square–Formula) (a ± b) 2 = a 2 ± 2ab + b 2 We write down the Square–Formula as a reminder below. Example A. Given f(x) = x 2 – 2x + 3, simplify the following. a. f (2a) = (2a) 2 – 2 (2a) + 3 = 4a 2 – 4a + 3 b. f (a + b) = (a + b) 2 – 2 (a + b) + 3 = a 2 + 2ab + b 2 – 2a – 2b + 3
  • 20. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions
  • 21. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g.
  • 22. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3
  • 23. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7
  • 24. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3)
  • 25. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1
  • 26. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1 f * g = (3x – 4) * (x 2 – 2x – 3)
  • 27. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1 f * g = (3x – 4) * (x 2 – 2x – 3) = 3x 3 – 10x 2 – x + 12
  • 28. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1 f * g = (3x – 4) * (x 2 – 2x – 3) = 3x 3 – 10x 2 – x + 12 f/g = (3x – 4)/(x 2 – 2x – 3)
  • 29. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1 f * g = (3x – 4) * (x 2 – 2x – 3) = 3x 3 – 10x 2 – x + 12 f/g = (3x – 4)/(x 2 – 2x – 3) Algebraic expressions may be formed with functions notations.
  • 30. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1 f * g = (3x – 4) * (x 2 – 2x – 3) = 3x 3 – 10x 2 – x + 12 f/g = (3x – 4)/(x 2 – 2x – 3) Algebraic expressions may be formed with functions notations. Example C. Let f(x) = x 2 – 2x – 3, g(x) = 2x + 1. a. Simplify f(2) – g(–3)
  • 31. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1 f * g = (3x – 4) * (x 2 – 2x – 3) = 3x 3 – 10x 2 – x + 12 f/g = (3x – 4)/(x 2 – 2x – 3) Algebraic expressions may be formed with functions notations. Example C. Let f(x) = x 2 – 2x – 3, g(x) = 2x + 1. a. Simplify f(2) – g(–3) We find f(2) = (2) 2 – 2 * (2) – 3
  • 32. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1 f * g = (3x – 4) * (x 2 – 2x – 3) = 3x 3 – 10x 2 – x + 12 f/g = (3x – 4)/(x 2 – 2x – 3) Algebraic expressions may be formed with functions notations. Example C. Let f(x) = x 2 – 2x – 3, g(x) = 2x + 1. a. Simplify f(2) – g(–3) We find f(2) = (2) 2 – 2 * (2) – 3 = 4 – 4 – 3
  • 33. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1 f * g = (3x – 4) * (x 2 – 2x – 3) = 3x 3 – 10x 2 – x + 12 f/g = (3x – 4)/(x 2 – 2x – 3) Algebraic expressions may be formed with functions notations. Example C. Let f(x) = x 2 – 2x – 3, g(x) = 2x + 1. a. Simplify f(2) – g(–3) We find f(2) = (2) 2 – 2 * (2) – 3 = 4 – 4 – 3 = – 3
  • 34. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1 f * g = (3x – 4) * (x 2 – 2x – 3) = 3x 3 – 10x 2 – x + 12 f/g = (3x – 4)/(x 2 – 2x – 3) Algebraic expressions may be formed with functions notations. Example C. Let f(x) = x 2 – 2x – 3, g(x) = 2x + 1. a. Simplify f(2) – g(–3) We find f(2) = (2) 2 – 2 * (2) – 3 = 4 – 4 – 3 = – 3 g(–3) = 2(–3) +1
  • 35. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1 f * g = (3x – 4) * (x 2 – 2x – 3) = 3x 3 – 10x 2 – x + 12 f/g = (3x – 4)/(x 2 – 2x – 3) Algebraic expressions may be formed with functions notations. Example C. Let f(x) = x 2 – 2x – 3, g(x) = 2x + 1. a. Simplify f(2) – g(–3) We find f(2) = (2) 2 – 2 * (2) – 3 = 4 – 4 – 3 = – 3 g(–3) = 2(–3) +1 = –6 + 1 = –5
  • 36. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1 f * g = (3x – 4) * (x 2 – 2x – 3) = 3x 3 – 10x 2 – x + 12 f/g = (3x – 4)/(x 2 – 2x – 3) Algebraic expressions may be formed with functions notations. Example C. Let f(x) = x 2 – 2x – 3, g(x) = 2x + 1. a. Simplify f(2) – g(–3) We find f(2) = (2) 2 – 2 * (2) – 3 = 4 – 4 – 3 = – 3 Hence f(2) – g(-3) = – 3 – (–5) g(–3) = 2(–3) +1 = –6 + 1 = –5
  • 37. We may form new functions by +, – , * , and / functions. Notation and Algebra of Functions Example B. Let f(x) = 3x – 4, g(x) = x 2 – 2x – 3, find f + g, f – g, f * g, and f/g. f + g = 3x – 4 + x 2 – 2x – 3 = x 2 + x – 7 f – g = 3x – 4 – (x 2 – 2x – 3) = 3x – 4 – x 2 + 2x + 3 = –x 2 + 5x – 1 f * g = (3x – 4) * (x 2 – 2x – 3) = 3x 3 – 10x 2 – x + 12 f/g = (3x – 4)/(x 2 – 2x – 3) Algebraic expressions may be formed with functions notations. Example C. Let f(x) = x 2 – 2x – 3, g(x) = 2x + 1. a. Simplify f(2) – g(–3) We find f(2) = (2) 2 – 2 * (2) – 3 = 4 – 4 – 3 = – 3 Hence f(2) – g(-3) = – 3 – (–5) = –3 + 5 = 2 g(–3) = 2(–3) +1 = –6 + 1 = –5
  • 38. b. Simplify f(x + h ) – f(x) Notation and Algebra of Functions
  • 39. b. Simplify f(x + h ) – f(x) First, we calculate f(x + h ) Notation and Algebra of Functions
  • 40. b. Simplify f(x + h ) – f(x) First, we calculate f (x + h ) = (x+h) 2 – 2 (x+h) – 3 Notation and Algebra of Functions
  • 41. b. Simplify f(x + h ) – f(x) First, we calculate f(x + h ) = (x+h) 2 – 2(x+h) – 3 = x 2 + 2xh + h 2 – 2x – 2h – 3 Notation and Algebra of Functions
  • 42. b. Simplify f(x + h ) – f(x) First, we calculate f(x + h ) = (x+h) 2 – 2(x+h) – 3 = x 2 + 2xh + h 2 – 2x – 2h – 3 Hence f(x+h) – f(x) = x 2 + 2xh + h 2 – 2x – 2h – 3 – (x 2 – 2x – 3) Notation and Algebra of Functions
  • 43. b. Simplify f(x + h ) – f(x) First, we calculate f(x + h ) = (x+h) 2 – 2(x+h) – 3 = x 2 + 2xh + h 2 – 2x – 2h – 3 Hence f(x+h) – f(x) = x 2 + 2xh + h 2 – 2x – 2h – 3 – (x 2 – 2x – 3) = x 2 + 2xh + h 2 – 2x – 2h – 3 – x 2 + 2x + 3 Notation and Algebra of Functions
  • 44. b. Simplify f(x + h ) – f(x) First, we calculate f(x + h ) = (x+h) 2 – 2(x+h) – 3 = x 2 + 2xh + h 2 – 2x – 2h – 3 Hence f(x+h) – f(x) = x 2 + 2xh + h 2 – 2x – 2h – 3 – (x 2 – 2x – 3) = x 2 + 2xh + h 2 – 2x – 2h – 3 – x 2 + 2x + 3 = 2xh + h 2 – 2h Notation and Algebra of Functions
  • 45. b. Simplify f(x + h ) – f(x) First, we calculate f(x + h ) = (x+h) 2 – 2(x+h) – 3 = x 2 + 2xh + h 2 – 2x – 2h – 3 Hence f(x+h) – f(x) = x 2 + 2xh + h 2 – 2x – 2h – 3 – (x 2 – 2x – 3) = x 2 + 2xh + h 2 – 2x – 2h – 3 – x 2 + 2x + 3 = 2xh + h 2 – 2h Notation and Algebra of Functions Composition of Functions
  • 46. b. Simplify f(x + h ) – f(x) First, we calculate f(x + h ) = (x+h) 2 – 2(x+h) – 3 = x 2 + 2xh + h 2 – 2x – 2h – 3 Hence f(x+h) – f(x) = x 2 + 2xh + h 2 – 2x – 2h – 3 – (x 2 – 2x – 3) = x 2 + 2xh + h 2 – 2x – 2h – 3 – x 2 + 2x + 3 = 2xh + h 2 – 2h Notation and Algebra of Functions Composition of Functions When one function is used as the input of another function, this operation is called composition .
  • 47. b. Simplify f(x + h ) – f(x) First, we calculate f(x + h ) = (x+h) 2 – 2(x+h) – 3 = x 2 + 2xh + h 2 – 2x – 2h – 3 Hence f(x+h) – f(x) = x 2 + 2xh + h 2 – 2x – 2h – 3 – (x 2 – 2x – 3) = x 2 + 2xh + h 2 – 2x – 2h – 3 – x 2 + 2x + 3 = 2xh + h 2 – 2h Notation and Algebra of Functions Composition of Functions When one function is used as the input of another function, this operation is called composition . Given f(x) and g(x), we define (f ○ g) (x) ≡ f ( g(x) ).
  • 48. b. Simplify f(x + h ) – f(x) First, we calculate f(x + h ) = (x+h) 2 – 2(x+h) – 3 = x 2 + 2xh + h 2 – 2x – 2h – 3 Hence f(x+h) – f(x) = x 2 + 2xh + h 2 – 2x – 2h – 3 – (x 2 – 2x – 3) = x 2 + 2xh + h 2 – 2x – 2h – 3 – x 2 + 2x + 3 = 2xh + h 2 – 2h Notation and Algebra of Functions Composition of Functions When one function is used as the input of another function, this operation is called composition . Given f(x) and g(x), we define (f ○ g) (x) ≡ f ( g(x) ). (It's read as f "circled" g).
  • 49. b. Simplify f(x + h ) – f(x) First, we calculate f(x + h ) = (x+h) 2 – 2(x+h) – 3 = x 2 + 2xh + h 2 – 2x – 2h – 3 Hence f(x+h) – f(x) = x 2 + 2xh + h 2 – 2x – 2h – 3 – (x 2 – 2x – 3) = x 2 + 2xh + h 2 – 2x – 2h – 3 – x 2 + 2x + 3 = 2xh + h 2 – 2h Notation and Algebra of Functions Composition of Functions When one function is used as the input of another function, this operation is called composition . Given f(x) and g(x), we define (f ○ g) (x) ≡ f ( g(x) ). (It's read as f "circled" g). In other words, the input for f is g(x) .
  • 50. b. Simplify f(x + h ) – f(x) First, we calculate f(x + h ) = (x+h) 2 – 2(x+h) – 3 = x 2 + 2xh + h 2 – 2x – 2h – 3 Hence f(x+h) – f(x) = x 2 + 2xh + h 2 – 2x – 2h – 3 – (x 2 – 2x – 3) = x 2 + 2xh + h 2 – 2x – 2h – 3 – x 2 + 2x + 3 = 2xh + h 2 – 2h Notation and Algebra of Functions Composition of Functions When one function is used as the input of another function, this operation is called composition . Given f(x) and g(x), we define (f ○ g) (x) ≡ f ( g(x) ). (It's read as f "circled" g). In other words, the input for f is g(x) . Similarly, we define (g ○ f) (x) ≡ g ( f(x) ).
  • 51. Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). Notation and Algebra of Functions
  • 52. Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). (f ○ g)(3) = f( g(3) ) Notation and Algebra of Functions
  • 53. Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). (f ○ g)(3) = f( g(3) ) Since g(3) = –4(3) + 3 Notation and Algebra of Functions
  • 54. Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). (f ○ g)(3) = f( g(3) ) Since g(3) = –4(3) + 3 = –9 Notation and Algebra of Functions
  • 55. Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). (f ○ g)(3) = f( g(3) ) Since g(3) = –4(3) + 3 = –9 Hence f( g(3) ) = f (–9) Notation and Algebra of Functions
  • 56. Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). (f ○ g)(3) = f( g(3) ) Since g(3) = –4(3) + 3 = –9 Hence f( g(3) ) = f (–9) = 3 (–9) – 5 = –32 Notation and Algebra of Functions
  • 57. Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). (f ○ g)(3) = f( g(3) ) Since g(3) = –4(3) + 3 = –9 Hence f( g(3) ) = f (–9) = 3 (–9) – 5 = –32 (g ○ f)(3) = g( f(3) ) Notation and Algebra of Functions
  • 58. Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). (f ○ g)(3) = f( g(3) ) Since g(3) = –4(3) + 3 = –9 Hence f( g(3) ) = f (–9) = 3 (–9) – 5 = –32 (g ○ f)(3) = g( f(3) ) Since f(3) = 3(3) – 5 Notation and Algebra of Functions
  • 59. Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). (f ○ g)(3) = f( g(3) ) Since g(3) = –4(3) + 3 = –9 Hence f( g(3) ) = f (–9) = 3 (–9) – 5 = –32 (g ○ f)(3) = g( f(3) ) Since f(3) = 3(3) – 5 = 4 Notation and Algebra of Functions
  • 60. Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). (f ○ g)(3) = f( g(3) ) Since g(3) = –4(3) + 3 = –9 Hence f( g(3) ) = f (–9) = 3 (–9) – 5 = –32 (g ○ f)(3) = g( f(3) ) Since f(3) = 3(3) – 5 = 4 Hence g( f(3) ) =g (4) Notation and Algebra of Functions
  • 61. Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). (f ○ g)(3) = f( g(3) ) Since g(3) = –4(3) + 3 = –9 Hence f( g(3) ) = f (–9) = 3 (–9) – 5 = –32 (g ○ f)(3) = g( f(3) ) Since f(3) = 3(3) – 5 = 4 Hence g( f(3) ) =g (4) = –4(4) + 3 Notation and Algebra of Functions
  • 62. Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). (f ○ g)(3) = f( g(3) ) Since g(3) = –4(3) + 3 = –9 Hence f( g(3) ) = f (–9) = 3 (–9) – 5 = –32 (g ○ f)(3) = g( f(3) ) Since f(3) = 3(3) – 5 = 4 Hence g( f(3) ) =g (4) = –4(4) + 3 = –13 Notation and Algebra of Functions
  • 63. Note: They are different! Example C. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify (f ○ g)(3), (g ○ f)(3). (f ○ g)(3) = f( g(3) ) Since g(3) = –4(3) + 3 = –9 Hence f( g(3) ) = f (–9) = 3 (–9) – 5 = –32 (g ○ f)(3) = g( f(3) ) Since f(3) = 3(3) – 5 = 4 Hence g( f(3) ) =g (4) = –4(4) + 3 = –13 Notation and Algebra of Functions
  • 64. f(x) = 2 – 3x g(x) = –2x 2 + 3x – 1 h(x) = Exercise A. Simplify the following expressions with the given functions. Notation and Algebra of Functions 2x – 1 x – 2 8. f(2a) 12. 2h(a) 9. g(2a) 11. h(2a) 10. 2g(a) 13. f(3 + b) 14. g(3 + b) 15. h(3 + b) 1. f(2) + f(3) 5. f(0) + g(0) + h(0) 2. 2f(3) 4. [h(2)] 2 3. 2g(0) + g(1) 6. [f(3)] 2 – [g(3)] 2 7. h(1) / h(–1 ) 16. f(3 + b) – f(b) 17. g(3 + b) – g(b) 18. h(3 + b) – h(b) 19. f(3 + b) – f(3 – b) 20. g(3 + b) – g(3 – b) 24. (f ○ g)(2) 25. (g ○ f)(–2) Exercise B. Simplify the following compositions. 26. (g ○ g)(–1) 27. (h ○ g)(3) 28. (f ○ h)(1) 29. (f ○ f)(1) 30. (f ○ g)(x) 31. (g ○ f)(x) 32. (f ○ f)(x) 33. (h ○ f)(x) 34. (f ○ h)(x) 21. g(x) + 3f(x) 22. 2g(x) + [f(x)] 2 23. g(x) / h(x)