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Operations on Functions Section 1-8

Operations on Functions Section 1-8. Objectives. I can add, subtract, multiply, and divide functions I can find the domains of newly formed functions. Chapter 1 Section 8 Arithmetic of Functions. NOTATION. f(x) + g(x) means. Add the f(x) function to the g(x) function.

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Operations on Functions Section 1-8

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  1. Operations on FunctionsSection 1-8

  2. Objectives • I can add, subtract, multiply, and divide functions • I can find the domains of newly formed functions

  3. Chapter 1 Section 8Arithmetic of Functions NOTATION f(x) + g(x) means Add the f(x) function to the g(x) function.

  4. Chapter 1 Section 8Arithmetic of Functions NOTATION f(x) - g(x) means Subtract the g(x) function from the f(x) function.

  5. Chapter 1 Section 8Arithmetic of Functions NOTATION (fg)(x) means f(x) g(x) Multiply f(x) times g(x)

  6. Chapter 1 Section 8Arithmetic of Functions NOTATION (f/g)(x) means f(x) g(x) Divide f(x) by g(x)

  7. Let f(x) = 3x + 4 & g(x) = 2x – 3 find: • f(x) + g(x) • f(x) – g(x) • f(x)g(x) d) f(x)/g(x) 5x + 1 x + 7 6x2 – x - 12

  8. Operations on Functions: Addition: (f + g)(x) = f(x) + g(x) Example: If f(x) = 3x – 6 and g(x) = -3x2 + 3x + 4 then (f + g)(x) = 3x –6 + -3x2 + 3x + 4 = -3x2 + 6x –2 Subtraction: (f – g)(x) = f(x) – g(x) Example: Use the same functions as above. Then (f - g)(x) = 3x – 6 – (-3x2 + 3x + 4 ) (f – g)(x) = 3x2 - 10

  9. Operations on Functions: Multiplication: (fg)(x) = f(x)g(x) Example: If f(x) = 3x – 6 and g(x) = -3x2 + 3x + 4 then (fg)(x) = (3x –6)(-3x2 + 3x + 4) = -9x3 + 9x2 + 12x + 18x2 – 18x – 24 = -9x3 + 27x2 – 6x - 24 Division: (f/g)(x) = f(x)/g(x) Example: Use the same functions as above. Then (f/g)(x) = (3x – 6)/(-3x2 + 3x + 4) except for x – values that make denominator 0

  10. Finding the domains of our new functions • Finding the domains of these new functions is a little complicated. • The first step is to find the domain of each function. • The second step is to find the intersection of the domains. This intersection is the domain of the new function*. * Unless the operation is division in which case you must also exclude x-values that make the denominator zero

  11. Lets do an example; Let a) Find (f + g)(x) and b) it’s domain. (f + g)(x) = Step 1: The domain of f(x) is [-2,). The domain of g(x) is [0,). Step 2: [-2,)  [0,) = [0,). Since this is not a quotient I don’t have to worry about division by zero.

  12. Lets do an example; Let a) Find (f - g)(x) and b) it’s domain. (f - g)(x) = Step 1: The domain of f(x) is (- ,). The domain of g(x) is (- ,). Step 2: (- ,)  (- ,) = (- ,). Since this is not a quotient I don’t have to worry about division by zero.

  13. Example; Let a) Find (f · g)(x) and b) it’s domain. (f · g)(x) = Step 1: The domain of f(x) is (- ,). The domain of g(x) is (- ,). Step 2: (- ,)  (- ,) = (- ,). Since this is not a quotient I don’t have to worry about division by zero.

  14. Another example: • Find (f /g)(x) and b) it’s domain. Step 1: Domain of f(x) is [0,). Domain of g(x) is [10, ) Step 2: [0,)  [10, ) = [10, )* * Division so we can’t plug in 10 so Domain is (10, ).

  15. Example • Given f(x) = x – 5 and g(x) = x2 -1, find (f+g)(x), (f - g)(x), (fg)(x) and (f/g)(x) • (f+g)(x) = f(x) + g(x) = (x – 5)+(x2 –1) • =x2 + x - 6 • Domain (f+g)(x): (- , ) • (f – g)(x) = f(x) – g(x) = (x – 5) – (x2 –1 ) • = x – 5 – x2 + 1 • = -x2 + x – 4 • Domain (f – g)(x) = (- , )

  16. Example • Given f(x) = x – 5 and g(x) = x2 -1, find (f - g)(x), (fg)(x) and (f/g)(x) • (fg)(x) = f(x)g(x) = (x – 5)(x2 –1) • = x3 – x – 5x2 + 5 • = x3 – 5x2 – x + 5 • Domain (fg)(x) : (- , ) • (f/g)(x)=f(x)/g(x) = (x – 5)/(x2 – 1) • Domain (f/g)(x): {All real except x 1 or –1} • (- , -1) U (-1,1) U (1, )

  17. Homework • WS 2-1

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