Classifications and combinations of functions

in Popular STEM25 days ago

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The modern notion we have today of what a function is is due to the efforts of many mathematicians of the 17th and 18th centuries. One example is Leonhard Euler, to whom we owe the function notation: y = f(x).

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Already by the end of the 18th century, mathematicians and scientists had come to the conclusion that a large number of real-life phenomena could be represented by mathematical models, constructed from a collection of functions called: “elementary functions.

One can have a specific classification of these elementary functions:

[1] Algebraic functions.
[2] Trigonometric functions.
[3] Exponential and logarithmic functions.

Algebraic functions are all those expressed algebraically, such as polynomials, radicals and rationals.

Trigonometric functions are those derived from the right triangle, such as sine, cosine, tangent, secant, cotangent, etc.

In this opportunity I want to explain in a generalized way the algebraic functions, for this case the most common type of algebraic functions is the polynomial function, whose general form is:

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Where n is the positive integer that determines the degree of the polynomial function. The constantsimage.png are called coefficients, whereimage.png the dominant coefficient and image.png the constant term.

Although subscripts are often used for the coefficients of polynomial functions in general, the following simpler forms are often used for that of lower degrees:

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Definition of composite function

Let f and g be two functions. The function given by (f or g)(x) = f(g(x)) is called a composite function of f with g. The domain of f or g is the set of all x in the domain of g such that g(x) is in the domain of f.

It is necessary to take into account that the composite function of f with g may not be equal to the composite function of g with f.

Example of function composition

Given f(x) = 2x-3 and g(x) = cosx, find each of the composite functions:

[1] f o g
[2] g o f

Solution:

[1] (f o g)(x) = f(g(x))

(f o g)(x) = f(cosx)

(f o g)(x) = 2(cosx)-3

(f o g)(x) = 2cosx-3

[2] (g o f)(x) = g(f(x))

(g o f)(x) = g(2x-3)

(g o f)(x) = cos(2x-3)

This shows us that:

(𝑓𝑜𝑔)(𝑥) ≠ (𝑔𝑜𝑓)(𝑥)

Bibliographic Reference

Calculus with Analytic Geometry by Ron Larson, Robert, P. Hostetler and Bruce H. Edwards. Volume I. Eighth Edition. McGraw Hill. Año 2006