# nutriamigos

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No topic   Arithmic Operations on Functions1. (f + g)(x)= f(x) + g(x)2. (f - g)(x)= f(x) - g(x)3. (f * g)(x)= f(x) * g(x)4. (f / g)(x)= f(x) / g(x) g diferente a 0.

Lesson 3-1

Lesson 3-2

Definition of g o f:f and g are functions. The composite of g with f, written g o f, is the function with rule:gf(x)=g(f (x))whose domain is the set of all x in the domain of f for which f(x) is in the domain of g.

fog(x)=f(g(x)) is different fromf*g(x)=f(x)*g(x)

Arithmic Operations On Sequences1. (U+V)n= Un +Vn2.(U-V)n= Un- Vn3.(U*V)n= Un* Vn4. (U/V)n= Un/Vn Vn differente a 0

LESSON 3-3

and g are inverse functions, written f = g-1 or g = f-1, if and only if f o g(x) = x for all x in the domain of g and g o f(x) = x for all x in the domain of f.

Una funcion es inversa si su funcion es 1 a 1. Esto quiere decir que para cada Y en la funcion original solo exista 1 valor en X.Por ejemplo:una funcion cuadratica no es una funcion 1 a 1 por lo cual la funcion inversa de esta no existe.

Lesson 3-4... The Logic of Equation-Solving

Addition property of equality:For all x in the intersection of the domains of functions f,g,h, f(x) = g(x) + h(x)Multiplication Property of Equality:For all x in the intersection of the domains of functions f,g and h, if h(x) is not = 0, then..f(x)=g(x) <-> f(x) * h(x) = g(x) * h(x)

Reversible steps: corresponds to justifications that are biconditional statements.Nonreversible steps: correspond to true conditional statements for which the converse is false.Equivalent Equations: equations related by reversible steps that have the same solution.

Function composition Property of EqualityLet h be a 1-1 function. Then, for all x in the domains of f and g for which f(x) and g(x) are in the domain of h, f(x) = g(x) <-> h(f(x)) = h(g(x))Revesible steps theoremLet f,g, and h be fucntions. Then for all x in the intersection of the domains of f, g, and h,1. f(x)=g(x) <-> f(x) + h(x) = g(x) + h(x)2. f(x)=g(x) <-> f(x) * h(x) = g(x) * h(x)3. If h is 1-1, then for all x in the domains of f and g for which f(x) and g(x) are in the domain of h,f(x) = g(x) <-> h(f(x)) = h(g(x))