We have:
#f(x)=(x+xcos(x))/(x-xcos(x))#
We can simplify this by factoring:
#f(x)=(x(1+cos(x)))/(x(1-cos(x))# We can cross out the #x#'s.
#f(x)=(1+cos(x))/(1-cos(x))#
Now, a function is even if:
#f(x)=f(-x)#
To try this out, we substitute all #x#'s with #-x#
#f(-x)=(1+cos(-x))/(1-cos(-x))#
Now, here is something to remember:
#sin(-x)=-sin(x)# and #cos(-x)=cos(x)#
Using this, we can say that:
#(1+cos(-x))/(1-cos(-x))=(1+cos(x))/(1-cos(x))#
Which tells us that #f(-x)=f(x)#
This means that our function is even.
When you graph this, the graph of the function will be symmetric in respect to the #y#- axis.
If #f(x)!=f(-x)#, you could check whether #f(-x)# equals #-f(x)#. If it does, than the function is odd. If it doesn't, then the function is neither even nor odd.