Question #0dd78

1 Answer
Jan 8, 2018

Answer:

#g(x)# is an even function.

Explanation:

If #g(-x)=g(x)#, the function is even, and if #g(-x)=-g(x)#, it is odd.

To find out which one it is, we plug in #-x#:
#g(-x)=(-x)^2+cos^2(-x)#

Cosine is an even function, and a negative squared is positive, so the minus signs just disappear:
#g(-x)=x^2+cos^2(x)=g(x)#

Since #g(-x)=g(x)#, we can conclude that #g(x)# is an even function.