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Precalc 3.5 - Combinations of Functions

(quo) : f(x). (2x - 3). =. g(x). (x 2 - 1). Precalc 3.5 - Combinations of Functions. ex: f(x) = (2x - 3) g(x) = (x 2 - 1). (sum) : f(x) + g(x) = (2x - 3) + (x 2 - 1) = x 2 + 2x - 4. (diff) : f(x) - g(x) = (2x - 3) - (x 2 - 1) = -x 2 + 2x - 2.

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Precalc 3.5 - Combinations of Functions

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  1. (quo) : f(x) (2x - 3) = g(x) (x2 - 1) Precalc 3.5 - Combinations of Functions ex: f(x) = (2x - 3) g(x) = (x2 - 1) (sum) : f(x) + g(x) = (2x - 3) + (x2 - 1) = x2 + 2x - 4 (diff) : f(x) - g(x) = (2x - 3) - (x2 - 1) = -x2 + 2x - 2 (prod) : f(x)•g(x) = (2x - 3)(x2 - 1) = 2x3 - 3x2 - 2x + 3 (x + 1)

  2. ex: f(x) = (2x + 1) g(x) = (x2 + 2x - 1) • Find (f + g)(x) • Evaluate sum at x = 2 (f + g)(x) = f(x) + g(x) = 2x + 1 + x2 + 2x - 1 = x2 + 4x (f + g)(2) = (2)2 + 4(2) = 4 + 8 = 12 • Find (f - g)(x) • Evaluate difference at x = 2 (f - g)(x) = f(x) - g(x) = 2x + 1 - (x2 + 2x - 1) = 2x + 1 - x2 - 2x + 1 = -x2 + 2 (f - g)(2) = -(2)2 + 2 = -4 + 2 = -2

  3. 0 g(x) = x f 1 f(x) = 0 x g x  0 x > 0 0 (f + g) Mixed Domains ex: Find the Domain of (f + g): x > 0

  4. 0 0 f f f(x) = x -2 0 2 -2 0 2 g g g(x) = 4 - x2 -2 0 2 -2 0 2 f/g f/g x > 0 x f(x) = g(x) 4 - x2 -2 < x < 2 g(x) x -2 < x < 2 = f(x) x > 0 4 - x2 ex: Find the Domain of f/g and g/f: 0< x < 2 0 < x < 2

  5. (f g) f(x + 1) = (x + 1)2 = x2 + 2x + 1 3.6 continued - Compositions ex: f(x) = x2 and g(x) = x + 1 Notation: (f o g)(x) - say: “f-circle-g of x” f(x) = x2 f [g(x)] = [g(x)]2 Take function g… (x + 1) And put it into function f…

  6. (f o g)(x) = f [g(x)] = g(x) = x - 1 = 1 = 1 = 2 - 1 0 Not in Domain (x > 1) ex: Find (f o g)(x) for f(x) = x and g(x) = (x - 1) Domain: x > 1 [1, ) ex: Find (f o g)(2) ex: Find (f o g)(0)

  7. ex: For f(x) = (x + 2) and g(x) = (4 - x2) - Find (f o g)(x) and (g o f)(x) (f o g)(x) = f [g(x)] = f(4 - x2) = (4 - x2 + 2) = (- x2 + 6) (g o f)(x) = g [f(x)] = g(x + 2) = [4 – (x + 2) 2] = 4 – (x2 + 4x + 4) = 4 - x2 - 4x - 4 = - x2 - 4x

  8. Domains of Compositions 1 x x = ex: f (x) = g (x) = x + 2 x - 3 x - 3 What is the domain of (fog)(x) ? 1 -2 = -2 (fog)(x) = f [ g(x) ] = x 1 + 2 -2(x - 3) = x x - 3 -2x + 6 = x x ≠ 3 x x ≠ 3 (solve) 6 = 3x ≠ -2 x - 3 x ≠ 2 x = 2 Domain of (fog)(x) = (x ≠ 2, 3)

  9. Domains of Compositions (Summary / Quick Tips) f [ g(x) ] 1) Any restrictions on the inside function will stay 2 is BAD for f! 2) Anything bad for f that the inside function might produce has to be solved for and excluded… 1 ex: f (x) = x - 2 x ≠ 2 g (x) ≠ 2 9 - x ≠ 2 g (x) = 9 - x x < 9 x ≠ 5 So, for (fog)(x) … x < 9, ≠ 5

  10. Decomposing Composites ex: Break h(x) = (3x – 5)3 into 2 functions f and g such that (fog)(x) = h(x) (3x – 5)3 f (x) = x3 outside function g(x) = 3x - 5 inside function (fog)(x) = (3x – 5)3 = f(x) = h(x) = g(x)

  11. Decomposing Composites 1 ex: Break h(x) = into 2 functions f and g such that (fog)(x) = h(x) (x – 2)2 1 1 f (x) = x2 (x – 2) 2 g(x) = x - 2 outside function 1 (fog)(x) = inside function (x – 2)2 = f(x) = g(x) = h(x)

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