How do you determine if # x / (x^2 + 5)# is an even or odd function?

1 Answer
Apr 8, 2016

#x/(x^2+5)# is odd (but not even)

Explanation:

Definitions

even function

#f(x)# is an even function if (and only if) #f(x)=f(-x)# for all #x# in the Domain of #f(x)#.
Graphically this means that the function is symmetric about the Y-axis (each side of the graph is a mirror image of the other side relative to the Y-axis).

odd function

#f(x)# is an odd function if (and only if) #f(-x)=-f(x)# for all #x# in the Domain of #f(x)#.
Graphically this means that the function is symmetric about the origin (a line from any point on the function through the origin will result in another point in the opposite quadrant at the same distance from the origin).

Application
#color(white)("XXX")f(x)=x/(x^2+5)#

Is it even?

#f(-x)=(-x)/((-x)^2+5) =(-x)/(x^2+5) !=f(x)#
So #f(x)# is not even

Is it odd?

#f(-x)=(-x)/(x^2+5)=-(f(x))#
So #f(x)# is odd

Graphically
graph{x/(x^2+5) [-1.758, 2.57, -1.031, 1.13]}
We can see that the graph is not reflected in the Y-axis and therefore is not even.
Further we can see that the graph is reflected through the origin and therefore is odd