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    FUNCTION

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    Function, in mathematical terms is mapping every member of a set (called a domain) to members of other himpunanyang (named as kodomain). This term is different than the same word that is used daily, such as "the equipment is functioning properly." The concept of function is one of the basic concepts of mathematics and every quantitative science. The term "function", "mapping", "map", "transformation", and "operator" is typically used synonymously.
    Summary
    A. Relationships
    Rules that connects each member of the set A to B is called Relation from A to B.
    Written: R: A → B.
    Terms:
    Set A is called Domain = Region of Origin
    Set B is called Kodomain = Regions Kawan
    Range = Local Results
    B. Declare Relation
         Relationships can be expressed in three ways, namely:
             1. Diagram Arrow
             2. Sequential Pair Association
             3. Graph Cartesian
    C. The Cartesian Product
           If x y ε ε A and B, then the Cartesian product A to B is a set of sequential pairs      (x, y).
    Posted: AxB = {(x, y) І yε xε A and B}
    example:
    A = {a, b, c}
    B = {1, 2}
    then using tables A x B is obtained:
    A x B
    1
    2
    a
    (a, 1)
    (a, 2)
    b
    (b, 1)
    (b, 2)
    c
    (c, 1)
    (c, 2)
    A x B = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}
    The properties:
    1. A x B ≠ B × A
    2. n (A x B) = n (B x A)
    D. Mapping (Function)
    Mapping is a special relationship that pairs each member of the set A with exactly one member of the set B.



    composition Function
    The composition of the function is a merger of two functions in sequence operation that will produce a new function.
    The composition of the two functions     f(x) and g(x)  denoted by the symbol (f \circ g)(x)  or (g \circ f)(x) .
    Where
    (f\circ g)(x)=f(g(x))
    (g\circ f)(x)=g(f(x))
    Personality Composition Functions
    • (g \circ f)(x) \neq (f \circ g)(x)
    • (f\circ (g\circ h))(x)=((f\circ g)\circ h)(x)
    example:
    given function:
    {\color{Red} f(x)=2x+1}
    {\color{Blue} g(x)=3x^2}
    {\color{DarkGreen} h(x)=\frac{1}{x+4}}
    1. (f\circ g)(x) = ….?
    * function g (x) is substituted into the function f (x)
    \begin{align*}(f\circ g)(x)&=&{\color{Red} f}({\color{Blue} g(x)})\\&=&{\color{Red} f}({\color{Blue} 3x^2})\\&=&{\color{Red} 2(}{\color{Blue} 3x^2}{\color{Red} )+1}\\(f\circ g)(x)&=&6x^2+1 \end{align*}
    2. (g\circ h)(x) = ….?
    * function h (x) is substituted into the function g (x)
    \begin{align*}(g\circ h)(x)&=&{\color{Blue} g}({\color{DarkGreen} h(x)})\\&=&{\color{Blue} g}({\color{DarkGreen} \frac{1}{x+4}})\\&=&{\color{Blue} 3}\left ({\color{DarkGreen} \frac{1}{x+4}} \right )^{\color{Blue} 2}\\&=&3\left (\frac{1}{x^2+8x+16} \right )\\(g\circ h)(x)&=&\frac{3}{x^2+8x+16} \end{align*}
    3.(h\circ g\circ f)(x) =…?
    *the function f (x) is substituted prior to the function g (x) nah, the new result is substituted into the function h (x), note the color represents the substitution ... .ok!
    \begin{align*}(h\circ g\circ f)(x)&=&{\color{DarkGreen} h}({\color{Blue} g}({\color{Red} f(x)}))\\&=&{\color{DarkGreen} h}({\color{Blue} g}({\color{Red} 2x+1}))\\&=&{\color{DarkGreen} h}({\color{Blue} 3}({\color{Red} 2x+1})^{\color{Blue} 2})\\&=&{\color{DarkGreen} h}(3(4x^2+4x+1))\\&=&{\color{DarkGreen} h}(12x^2+12x+3)\\&=&\frac{{\color{DarkGreen} 1}}{\left (12x^2+12x+3 \right ){\color{DarkGreen} +4}}\\&=&\frac{1}{12x^2+12x+7}\end{align*}
    How the above example ??? it's pretty obvious, is not it ??? !!
    Be careful in mensubtitusikan yes ....

    Looking for one function if the composition of unknown function
    1. Looking for g (x) if f (x) and (f \ circ g) (x) is known
    sample questions and discussion:
    Unknown (f \ circ g) (x) = 19-6x and {\ color {Red} f (x) = 3x + 1} define the function {\ color {Blue} g (x)}!
    replied:
    \begin{align*}(f\circ g)(x)&=&19-6x\\{\color{Red} f}({\color{Blue} g(x)})&=&19-6x\\{\color{Red} 3(}{\color{Blue} g(x)}{\color{Red} )+1}&=&19-6x\\{\color{Red} 3(}{\color{Blue} g(x)}{\color{Red} )}&=&19-6x{\color{Red} -1}\\{\color{Blue} g(x)}&=&\frac{18-6x}{3}\\{\color{Blue} g(x)}&=&6-2x \end{align*}
    2. Looking for f (x) if g (x) and (f \ circ g) (x) is known
    sample questions and discussion:
    Unknown (f \ circ g) (x) = 2x + 1 and {\ color {Blue} (g) (x) = x + 3} set {\ color {Red} f (x)}!
    replied:
    \begin{align*}(f\circ g)(x)&=&2x+1\\f({\color{Blue} g(x)})&=&2x+1\\f({\color{Blue} x+3})&=&2x+1\end{align*}
    We suppose the first:
    \begin{align*}{\color{Blue} x+3}&=&{\color{DarkGreen} y}\\x&=&{\color{DarkGreen} y-3}\end{align*}
    Substituting back to the function:
    \begin{align*}f({\color{Blue} x+3})&=&2x+1\\f({\color{DarkGreen} y})&=&2({\color{DarkGreen} y-3})+1\\f({\color{DarkGreen} y})&=&2y-6+1\\f({\color{DarkGreen} y})&=&2y-5\\f(x)&=&2x-5\end{align*}
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