lnl lH Order. Justify how all work clearly and in order. Justi when you do use c
ID: 3148856 • Letter: L
Question
lnl lH Order. Justify how all work clearly and in order. Justi when you do use calculator your answers algebraically whenever possible; ur calculator, sketch all relevant graphs and write down all relevant hematics. You have 2 hours and 45 minutes. 1. (20 points) Give examples, along with a brief explanation, of functions f : AB satisfying each set of conditions: (a) f is both one-to-one and onto. (b) f is one-to-one, but it is not onto. (c) f is onto, but not one-to-one. (d) f is neither one-to-one nor ontoExplanation / Answer
A function f: A -> B is one-to-one if all the elements in the co-domain have only a single pre-image (or mapping from) the domain;
i.e. any line drawn parallel to the x-axis should cut the graph at only 1 point;
A function f:A->B is onto if for all elements in B there is a pre-image (or mapping) in the domain A;
a) f(x) = x for all x is a one to one and onto fuction since any line drawn parallel to x-axis will cut the graph only one and all elments in the co-domain will have a pre-image in the domain
b) A fuction f(x) = 2x +2 for f:A-> B such that A = set of positive natural numbers and B= set of even numbers
Here it is one-to-one as any line drawn parallel to the x-axis will cut the graph at only one point;
But it is not onto as the element {2} in its co-domain does not have any pre-image in the domain of set A;
c) f(x) = x2 such that f:A -> B for x belonging to [-1,1] and B = [0,1] ;
The function is not one-one as a line drawn parallel to x-axis at y=1/4 will cut the graph at two points of x=1/4 and x=-1/4;
The function is onto as all the elements in its co-domain [0,1] will have a pre-image in its domain of set A = [-1,1]
d) A fuction f(x) = sin x such that f: A -> B for A = [0,pi] and B = [-1,1]
This function is not one-to-one as a line drawn parallel to x-axis at y=1/2 will cut the graph at two points of x=pi/6 and x= 5pi/6
The function is not onto as elements in its co-domain of B that are negative will not have a pre-image in domain A since sin x takes only non-negative values for the set A = [0,pi]
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