How do you know if #f(x) = (x^2+8)^2# is an even or odd function?

1 Answer
Alan P.
Jan 6, 2016

#f(x)=(x^2+8)^2# is even and not odd

Explanation:

even functions
A function #f(x)# is even #hArr f(x)=f(-x) AAx#

For the specific case of #f(x)=(x^2+8)^2#
#color(white)("XXX")f(-x) = ((-x)^2+8)^2#
#color(white)("XXXXXXX")=(x^2+8)^2#
#color(white)("XXXXXXX")=f(x)#

So the function is even

#bar(color(white)("XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX"))#

odd functions
A function #f(x)# is odd #hArr f(-x)=-f(x) AAx#

For the specific case of #f(x)=(x^2+8)^2#
#color(white)("XXX")f(-x) (=f(x)) > 0 AAx#
Therefore
#color(white)("XXX")-f(x) < 0 AAx#

#f(-x) != -f(x)#

So the function is not odd

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