How do you determine if #y=2x^5+x# is an even or odd function?

1 Answer
Alan P.
Apr 5, 2016

#y=2x^5+x# is an odd function

Explanation:

If #y_((-x)) = y_x# then the function is even

If #y_((-x))=-y_x# then the function is odd

Given
#color(white)("XXX")y_x=2x^5+x#

then
#color(white)("XXX")y_((-x)) = 2(-x)^5+(-x)#

#color(white)("XXXXXXX")=-2x^5-x#

#color(white)("XXXXXXX")=-(2x^5+x)#

#color(white)("XXXXXXX")=-y_x#

So the function is odd

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