# calculadoras

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**Precalculus10**

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Functions and Equations

3.1 = Arithmetic Operations on Functions

Definitions of Basic Arithmetic s, whOperations on Functions:Suppose f and g are real-valued functions defined on a set of S of real numbers , where S is the intersection of the domain of f and teh domain of g. Then:(f +g)(x) = f(x) + g(x)(f-g)(x)= f(x) - g(x)(f * g)= f(x) * g(x)(f/g) = f(x)/g(x) , if g(x) different to 0

3.2 = Functions Composition and Decomposition

Definition of g°f:Suppose that f and g are functions. The composite of g with f written f°g is the function with rule:g ° f(x)= g(f(x))Whose domain is the set of all x in the domain of f for wich f (x) is the domain of g.

3.3= Inverses of functions

Definition of Inverse Functions:f and g are inverse functions. Written f= g^-1 or g=f^-1 if and only if f °g (x) = x for all x in the domain of g and g°f(x) for all x in the domain of f.Definition of one to one Function:A function g is one to one if and only if there is one value of x for each value of y.One-to-one Function Inverse Theorem:A function has an inverse that is a fucntion if and olny if it is a one to one function.

Addition Property of Equality: For all x in the intersection of the domains of functions f, g and h , f(x)=g(x)<-->f(x)+h(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 different to 0, thenf(x)=g(x)<-->f(x)*h(x)=g(x)*h(x)Function Composition Property of equality:let 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 domains of h, f(x)=g(x)<-->h(f(x))=h(g(x))Reversible Step Theorem:Let f,g ad h be functions. Then for all x in the intersection of the domaind of funcions f,g,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))

3.4= The logic of equation solving

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