Question #7801b

1 Answer
Apr 4, 2017

#f# is even.

Explanation:

We need to define even and odd functions:

  • #f# is an even function if #f(-x)=f(x)#.
  • #f# is an odd function if #f(-x)=-f(x)#.

It's also useful to know that #cos(x)# is an even function—that is, #cos(-x)=cos(x)#.

We have:

#f(x)=x^2cos(x)#

Test this function by finding #f(-x)#:

#f(-x)=(-x)^2cos(-x)#

We already know that #cos(-x)=cos(x)#, Furthermore, #(-x)^2=x^2#. You can think about this as #(-x)^2=(-1)^2x^2=x^2#.

#f(-x)=x^2cos(x)=f(x)#

Since #f(-x)=f(x)#, #f# is an even function.