2 3 we compute p p1 x p p1 x p 5003 2x 15 5003

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Unformatted text preview: tack deconstructing G from an operational approach. Given an input x, the first step is to square x, then add 1, then divide the result into 2. We will assign each of these steps a function so as to write G as a composite of three functions: f , g and h. Our first function, f , is the function that squares its input, f (x) = x2 . The next function is the function that adds 1 to its input, g (x) = x + 1. Our last function takes its input and divides it into 2, 2 h(x) = x . The claim is that G = h ◦ g ◦ f . We find (h ◦ g ◦ f )(x) = h(g (f (x))) = h(g x2 ) = 2 2+1 = hx = G(x). x2 +1 √ 3. If we look H (x) = √x+1 with an eye towards building a complicated function from simpler x− 1 √ functions, we see the expression x is a simple piece of the larger function. If we define √ f (x)+1 f (x) = x, we have H (x) = f (x)−1 . If we want to decompose H = g ◦ f , then we can glean the formula from g (x) by looking at what is being done to f (x). We find g (x) = x+1 . We x...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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