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f: R--> R f(x) = sin^2 (x)  +  cos^2 (x) to show: f is not a surjection  

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rocky.goyal | eNoter

Posted June 11, 2013 at 4:32 PM via web

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f: R--> R

f(x) = sin^2 (x)  +  cos^2 (x)

to show: f is not a surjection

 

Tagged with functions, math

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted June 11, 2013 at 4:43 PM (Answer #1)

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Show that `f: RR->RR,f(x)=sin^2(x)+cos^2(x)` is not surjective:

Let X be the domain and Y the codomain (both of which are the set of reals in this case.) A function mapping from the domain to the codomain is surjective if for every `y in Y` there exists an `x in X` such that f(x)=y.

We need only show a counterexample to show that f is not surjective. Let y=2. There is no `x in RR` such that f(x)=2. (f(x) is identically 1 -- `sin^2(x)+cos^2(x)=1` for all `x in RR` )

Therefore the function is not surjective.

** Change to `f:RR ->{1}` and the function is surjective (onto).**

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rocky.goyal | eNoter

Posted June 11, 2013 at 4:48 PM (Reply #1)

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thank you sir...

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