Author 
Topic 

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 onetoone. [Hint: Use a piecewisedefined 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 onetoone; and for x < 0, y=x is a decreasing function and therefore a onetoone function.
Am I correct? Thank you for any help you can give me.



skeeter
Advanced Member
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 piecewise defined function (I'm using the hint) ... f(x) = 1/x for all x not equal to 0 f(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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