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How do you determine if #f(x) = (x - 2)^2 + 1# is an even or odd function?

1 Answer

Shwetank Mauria

Aug 31, 2016

#f(x)# is neither odd nor even.

Explanation:

If a function #f(x)# is even than #f(-x)=f(x)# and if it is odd than #f(-x)=-f(x)#.

As #f(x)=(x-2)^2+1#,

#f(-x)=(-x-2)^2+1#

= #(x+2)^2+1# and hence

#f(-x)# is neither equal to #f(x)# nor equal to #-f(x)#.

Hence, #f(x)# is neither odd nor even.

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