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jfg4707
Average Member

USA
19 Posts

 Posted - 10/13/2007 :  01:54:28 Hi,I am not sure of my solution to the following problem.Problem:Give an example of a function whose domain is the set of real numbers and that is neither increasing nor decreasing on its domain, but is one-to-one. [Hint: Use a piecewise-defined function.]Solution:I am assuming that the author means that the function is neither increasing nor decreasing on the function's entire domain. So, I chose the absolute value function, y = x, if x0, or y = -x, if x < 0, because y = |x| is neither increasing over its entire domain nor decreasing over its entiere domain and, therefore, is neither an increasing function nor a decreasing function. But, for x0, y=|x| is an increasing function and therefore one-to-one; and for x < 0, y=|x| is a decreasing function and therefore a one-to-one function.Am I correct? Thank you for any help you can give me.

skeeter

USA
5634 Posts

 Posted - 10/13/2007 :  07:18:10 y = |x| is not one to one.go back and look at the definition of a one to one function.consider this piece-wise defined function (I'm using the hint) ...f(x) = 1/x for all x not equal to 0f(x) = 0 for x = 0

jfg4707
Average Member

USA
19 Posts

 Posted - 10/13/2007 :  13:55:16 Thank you, Skeeter.I have two question. (1) When they say a function is decreasing (or increasing) on its domain, do they mean that the function is decreasing (or increasing) over its entire domain?(2) How do you describe f(x) = 1/x? It is decreasing in Quadrant IV and it is decreasing in Quadrant I, but the Quadrant I values > Quadrant IV values. Is this an example of a function that is neither increasing nor decreasing?Thank you.

jfg4707
Average Member

USA
19 Posts

 Posted - 10/13/2007 :  15:13:26 I have been able to answer my two questions.(1) If a function is increasing or decreasing on its domain, it means its entire domain. (2) A function that is neither increasing nor decreasing on its domain is a function that is not doing one or the other over its entire domain.Now I understand Skeeter's response much better. Thanks to Skeeter and to everyone for the help I have received from this forum.
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