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

1 Answer
Nov 17, 2016

I would say neither odd or even.

Explanation:

If it is odd then: #f(-x)=-f(x)# for example the sine function;
if it is even then #f(-x)=f(x)# as for the cossine function.
In our case let us choose #x=2#
#f(x)=f(2)=(2)^2+(2)=4+2=6#
So, let us check:
#f(-x)=f(-2)=(-2)^2+(-2)=4-2=2#
#-f(x)=-f(2)=-[(2)^2+2]=-6#

#f(-x)!=f(x)!=-f(x)#
So no condition is satisfied!