Let g(x) = [x-1] + [1-x], x ε [-3,3]. Write g(x) as a piecewise function without greatest integer brackets. would someone give me a brief explanation on how to solve such equations? (note: these are greatest integer brackets)
Follow Math Help Forum on Facebook and Google+
Originally Posted by Calc1 Let g(x) = [x-1] + [1-x], x ε [-3,3]. Write g(x) as a piecewise function without greatest integer brackets. would someone give me a brief explanation on how to solve such equations? (note: these are greatest integer brackets) I can't get the latex right but it looks to me like the function is zero for all x 1-x is the additive invers of x-1
Last edited by mr fantastic; Jun 11th 2009 at 03:28 PM. Reason: I only fixed the latex ....
Originally Posted by VonNemo19 I can't get the latex right but it looks to me like the function is zero for all x 1-x is the additive invers of x-1 Counterexample: g(2.5)=[1.5]+[1-2.5] =1+[-1.5] =1+(-2) =-1
i understand the greatest integer theory but i still don't get how it would be solved in this question as you stated g(2.5) = -1 ok i know how you got that but how is it exactly related to the question?
Originally Posted by Calc1 Let g(x) = [x-1] + [1-x], x ε [-3,3]. Write g(x) as a piecewise function without greatest integer brackets. would someone give me a brief explanation on how to solve such equations? (note: these are greatest integer brackets) let f(x)=[x-1] h(x)=[1-x] since we can write f(x) and h(x) like below then find the sum of f(x) and h(x) because g(x)=f(x)+h(x) it is clear or not
you can take 1 from the greatest integer function since the factor of x is one so you can write g(x) like this but the result will be the same as I solve before
Originally Posted by alexmahone Counterexample: g(2.5)=[1.5]+[1-2.5] =1+[-1.5] =1+(-2) =-1 I've made three mistakes within the last 24 hours. I must be losing it.
Thanks Amer, it makes sense to me now
View Tag Cloud