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L4.2 – Operations on Functions Creating new functions by combining functions: 1) Arithmetically 2) Using Composition What is the domain of the new function?

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Operations on Functions: Arithmetic (1/4) Functions combined arithmetically create a new function. SUM (f + g)(x) = f(x) + g(x) DIFFERENCE (f – g)(x) = f(x) – g(x) PRODUCT (f. g)(x) = f(x). g(x) QUOTIENT The domain of the new function is the intersection of the domains of the original functions. Notice that the outputs (values of the functions) are arithmetically combined. (f + g)(x) is two functions running in parallel on the same input. vs. f(a + b) is one function with two inputs (combined prior to running the function) f g x x + f(x) g(x) f(x) + g(x) Arithmetic Combinations Operate in Parallel Any domain restrictions of either f or g, apply to their arithmetic combination.

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Arithmetic Operations on Functions: (2/4) (A) Arithmetic operations can be performed By creating the new function and then evaluating Ex 1: h(x) = 3x 2 + 4x, j(x) = x – 8. Find (h + j)(x) = Then evaluate: (h + j)(2) = 3(2) 2 + 5(2) − 8 = 14 Numerically, by combining the output values of the orig’l fcns. From Ex 1 above: (h + j)(2) = Find (f + g)(–1) (f – g)(2) (f. g)(2) For what values of x is (f – g)(x) > 0? = f( – 1) + g( – 1) 3x 2 + 5x – 8 = f(2) – g(2) = 4 – 4 = 0 = f(2) ∙ g(2) = 4. 4 =16 f is above g for {x| –2 < x < 2} = = 8 h(2) + j(2) = 20 + (-6) = 14 Ex 2:

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Arithmetic Operations on Functions (3/4) (A) The difference function (f – g)(x) can be used to solve the equation f(x) = g(x), because f(x) = g(x) f(x) – g(x) = 0. So to find intersections between f(x) = 2x 2 and g(x) = x + 1, you can –Graph the two functions independently and use [F5]Math → Intersection Or –Graph the difference function (f – g)(x) = 2x 2 – (x + 1) and use [F5]Math → Zero –Either way you get x = −0.5, x = 1 and you can find corresponding y values (−.5,.5) (1, 2) (−.5, 0)(1, 0)

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Operations on Functions: Arithmetic (4/4) (A) Let f(x) = 4x 2 – 2x, g(x) = 2x, h(x) =, j(x) = Find each new function and state its’ domain. 1. (f + g)(x) 2. (f. g)(x) (j – h)(x) 4x 2 8x 3 – 4x 2 2x – 1, x 0 Domain is all Reals Hint: Recall that the domain of the new function is the intersection of the domains of the original functions. Domains: f: all R g: all R h: j: So what are the domains of the original functions? Ready for the answers?

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Operations on Functions: Composition (1/3) Functions can be combined using composition. The output of one function is sent as input into the other. (f o g)(x) = f(g(x)) x g(x) f(g(x)) Note that (f o g)(x) is not the same as (g o f)(x). The domain of (f o g)(x) is the subset of the domain of g which produces output which is in the domain of f. In other words: All x in D omain of g and g(x) in D omain of f g f f o g fg x g(x) (f ▫ g)(x) Composition Operates in Sequence

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Operations on Functions: Composition (2/3) (A) 1.Let f(x) = x 4 – 3x 2 and g(x) = Find (f o g)(x) and give the domain of the composite. 2. Find the domain of (f o g)(x), if and x 2 – 7x + 10, Exclude x = 1 from domain of (f o g)(x), since g(x) runs 1 st.. Also exclude x = -1., since this becomes input to f(x). D of f o g is

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Operations on Functions: Composition (3/3) 1.Sometimes, (f o g)(x) = (g o f)(x). For ex, let f(x) = 2x + 6 and g(x) = Show that (f o g)(x) = (g o f)(x) for all x. 2.You can also evaluate numerically, ≈ for arithmetic combos. Let f(x) = x + 2 and g(x) = x 2 + x – 1. Find f(g(3)) and f(g(x)). Find f(f(4)) and f(f(x)). 3.More complex functions can be rewritten as compositions. Let f(x) = x 3, g(x) =, and h(x) = x – 4, and j(x) = 2x. Rewrite k(x) = and l(x) = (2x – 4) 3 as compositions of the above functions. k(x) = g(f(h(x))l(x) = f(h(j(x)) f(g(3)) = f(11) = 13; f(g(x)) = x 2 + x + 1 f(f(4)) = f(6) = 8; f(f(x)) = x + 4 Do on board…

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WARMUP (A) Let and Evaluate each expression. Give answers in simplest form Ready for answers?

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L4.1 Homework Questions?

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