• [ended]Mathematics Tutoring Session #12[Mapping and Function(Wed. 2rd march 2016, 4:00 - 5:00pm)]

    Mathematics

    Emmanuel Iwara(Tutor)
    02-03-2016 13:04:00 +0000

    This session has ended for today

    Good day all

    The concept of Mapping and Function is the most fundamental in mathematics often tested on JAMB and other exams. so lets take a quick look into this topics.

    Mapping - Given two non-empty set A and B, if there is a rule, which assigned an element x∈A a unique element y∈B,then such rule is called mapping.

    The set A is called the Domain of the mapping, while the set B is called the co-domain of the mapping.

    e.g see (Image (1) below) for Domain and Co-domain.

    From (Image 1), the set { 2, 5, 4, 1} is the Domain, while the set {3, 6,10, 20} is the Co-domain.

    3 is the image of 2, 20 is the image of 1, 10 the image of 4 and 6 the image of 5.

    The collcetion of all the images is what is called the Range.

    Example 1.Determine the domain of the mapping f: x → 2x-3.If C = {-3, -1, 5} is the range and f is defined on D.

    Solution

    2x -3 is the formula connecting the parent set (Domain) and the Co-domain (range)

    Now by assumption what will be the value of (x) that when substitute into (2x-3) will give -3, -1, 5. then those numbers are elements of the domain.

    Observe that when x = 0,f: x → 2x-3 = 2(0)-3 = -3, and when x = 1,f: x → 2x-3 = -1, and when x = 4, f: x → 2x-3 = 2(4)-3 = 5.

    . Hence the domain of the mapping are {0, 1, 4} that gives the co-domain {-3,-1,5}.

    One -One- Mapping,- let f: x →Y be the mapping that establishes that the correspondence between the sets X and Y.The mapping f is called a One to One if different elements in the domain X have different images in the co-domain Y.

    e.g The mapping which associates each University in Nigeria with its Vice - Chancellor.

    From (Image 1) (3 is the image of 2) and (6 is the image of 5)

    Onto -Mapping -Let f: x →Y be mapping from the set X to the set Y.The mapping f is called Onto -mapping if every element of the co-domain is an image of atleast one element in the domain.

    e.g (See Image 2 below), the range is equall to the co-domain.

    Composite Mapping - Let f: x →Z and g: Z→Y be two mappings such that the co-domain of f is the domain of g.

    The mapping gof is called composite mapping.

    Example 2. Let the mapping f and g on the set of real numbers be defined by f(x) = x2 + 2 and g(x) = 2x + 1.

    Find;

    (i) gof

    (ii) fog

    Solution

    i) gof = g[f(x)] = 2( x2 + 2) + 1.

    That is putting the value of f(x) into (x) in g(x)

    = (2x2 + 4) + 1.

    = 2x2 + 5.

    ii) fog = f[g(x)] = (2x + 1)2 + 2.

    That is putting the value of g(x) into (x) in f(x),which give us.

    = 42+ 4x + 1 + 2

    = 42+ 4x + 3.

    Example 3. If f and g are mappings defined over the set of real numbers by f(x) = (x+3) and g(x) = 2x2 + 3 respectively, find the value of x for which fog = gof

    Solution

    fog = f[g(x)] = (2x2 + 3 + 3)

    That is putting the value of g(x) into (x) in f(x)

    = 2x2 + 6

    gof = g[f(x)] = 2(x+3)2 + 3.

    That is putting the value of g(x) into (x) in f(x)

    [2(x+3)(x+3)] + 3

    = 2[x2 + 6x + 9] + 3

    = 2x2 + 12x + 18 + 3

    Then fog = gof will be

    2x2 + 12x + 18 + 3 = 2x2 + 6

    Collecting terms

    12x + 21 = 6

    12x = -15

    x = -15/12.

    Inverse Functions - A function f has an inverse, if it is both ono to one and onto.

    ie the domain of f is the range of f1 and the domain of f1 is the range of f.

    Example 4.The function f over the set of real numbers is defined by f(x) = (1/3)x + 2. Find f1.

    Solution

    Let y = (1/3)x + 2

    y-2 = (1/3)x

    x = 3(y-2)

    ∴ x = f′(y) = 3(y-2)

    replacing y by x

    f′(x) = 3(x-2)

    Now see if you can try the following past Jamb and other exams questions.


    1.The function f and g on the set of real numbers are defined by f(x) = 3x-1 and g(x) = 5x + 2 respectively.

    Find;

    (a) fog

    (b) gof

    2.If f and g are mappings defined over the set of real numbers by f(x) = (x-1) and g(x) = 2x2 - 2 respectively, find the value of x for which fog = gof

    3.Given the functions.

    f(x) = 3x2 - x + 5.

    g(x) = 5x + 3.

    Simplify f(x) + 3g(x).

    4.If f(x) = (x-1)/(x2 -1), find f(1-x)

    5.Determine the inverse function of f(x) = 2x +3.

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