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NotTimBest ResponseYou've already chosen the best response.1
dw:1341470686812:dwdw:1341470707694:dw
 one year ago

NotTimBest ResponseYou've already chosen the best response.1
http://www.purplemath.com/modules/polyends.htm
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
sometimes you can just look at the equation itself \[y = x^5 + 2x^3 + x\] all the exponents of x in this equation are 1,3 and 5..all are odd..therefore this is an odd function \[y = 2x^2 + 4\] the exponents are 2 and 0..both are even therefore this is an even function note that this can only be applied sometimes
 one year ago

campbell_stBest ResponseYou've already chosen the best response.0
the simple test is to substitute x for x if f(x) = f(x) the function is even ... e,g, f(x) = x^2 f(x) = (x)^2 = x^2 id f(x) = f(x) the function is odd e.g f(x) = x^3 f(x) = (x)^3 = x^3 the function can also be neither odd nor even as an example f(x) = x^3 + 1 = x^3 + 1 which isn't the negative of the original function.
 one year ago
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