Is #f(x)=x^2+sin x# an even or odd function?

2 Answers
Sep 26, 2015

Answer:

It is neither.

Explanation:

A function #f# is even if and only if #f(-x)=f(x)# for every #x# in the domain of #f#.
A function #f# is odd if and only if #f(-x)=-f(x)# for every #x# in the domain of #f#.

For #f(x)=x^2+sin x#, we check #f(-x)#.

If #f(-x)# simplifies to #f(x)#, then #f# is even,
If #f(-x)# simplifies to #-f(x) =-(x^2+sin x)#, the #f# is odd.
If #f(-x)# does not simplify to one of the above, then #f# is neither even nor odd.

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

# = x^2-sinx# which is neither #f(x)# nor #-f(x)#

We cannot show that #f(-x)=f(x)# for every #x# by showing that it is true for only one #x#,
but we can show that is fails to be true for all #x# by showing that it fails for one value.

#f(pi/2)=pi^2/4+1#
#f(-pi/2) =pi^2/4-1#

It is, I think, clear that these numbers are neither equal nor negatives of each other.

Sep 26, 2015

Answer:

Neither.

#f(-pi/6) = pi^2/36 - 1/2 ~~ 10/36 - 1/2 = -2/9#

#f(pi/6) = pi^2/36 + 1/2 ~~ 10/36 + 1/2 = 7/9#

So #f(-pi/6) != +-f(pi/6)#

Explanation:

An even function requires #f(-x) = f(x)#

An odd function requires #f(-x) = -f(x)#

Our example satisfies neither of these conditions.