functions-Domain, Range

Functions

Graphical representation of function 
    Let ‘f’ be a mapping with domain D such that y = f(x) should assume single value for each x
    (i.e the straight line drawn parallel to y - axis in its domain should be cut at only one point)
Illustration - I
    Find whether y = f(x) = x³ forms a mapping or not ?
sol:\( \,y = f(x) = x^3 ,\,\,\forall n \in R \)

Clearly, the straight lines drawn parallel to y - axis cut y =x³ only at one point  
     \( \therefore \)y = f(x) forms a mapping 
     \( \therefore \)y = f(x)  is a function.

Illustration- II :
Find whether \( \frac{{x^2 }} {{a^2 }} + \frac{{y^2 }} {{b^2 }} = 1 \) forms a mapping or not
    Let us consider an ellipse \( \frac{{x^2 }} {{a^2 }} + \frac{{y^2 }} {{b^2 }} = 1 \)
\( \Rightarrow y = \pm \frac{b} {a}\sqrt {a^2 - x^2 } \)

Here the straight lines drawn parallel to y - axis meets the curve at more than one point.
\( y = \pm \frac{b} {a}\sqrt {a^2 - x^2 } \) is not a function in general, we could say 
    1) an element of 'A' (i.e. domain) should not associate with more than one element in B.
    2) if a graph of function is plotted and any line parallel to y - axis cuts it at more than one point then it does not form a function.
Illustration - III
    Let A = {1,2,3,4}, B = {a,b,c,d,e} be two sets and let f₁, f₂,f₃ and f₄ be rules associating elements from A to B as shown in the following figures.
i)   ii)

iii) iv)

Observations : 
    S.No   Domain    Co-Domain    Range
    i)        A               B                  {b.d.e}
    ii)       A               B                 {a,b,c,e}
    iii)      A               B                  {a,b,e}
    iv)      A               B                 {b,c,d,e}

Note : Co-domain \( \subseteq \) Range 
Illustration IV : 
a) Let A ={1,2,3,4}, B={5,7,9,11} and f={(1,5}, (2.5), (3,11),(4,9)} then the relation 'f' is a function from A to B

b) Let A = {1,2,3,4}, B = {a,b,c,d,e,f} and g = {(1,c), (2,b), (3,b)}, f={(1,a),(2,f),(3,a),(4,f),(4,b)} then 'f' is a relation but not a function from A to B because the element \( 4 \in A \) has two imges and also 'g' is a relation but not a function as the element '4' has no, image in B.