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

Question

Madelyn Anderson · Accepted Answer

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

Definitions even operate #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). strange operation #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 this strange? #f(-x)=(-x)/(x^2+5)=-(f(x))# So #f(x)# is odd From the graphic representation, graph{x/(x^2+5) [-1.758, 2.57, -1.031, 1.13]}, we can observe that the graph is odd because it is reflected through the origin and not in the Y-axis.

Lily Anderson · Answer

undefined To determine if $ \frac{x}{x^2 + 5} $ is an even or odd function, we evaluate the function at $ -x $ and compare it to the original function.

For an even function, $ f(-x) = f(x) $.
For an odd function, $ f(-x) = -f(x) $.

If $ \frac{x}{x^2 + 5} $ is even, then $ \frac{-x}{(-x)^2 + 5} $ should equal $ \frac{x}{x^2 + 5} $.
If $ \frac{x}{x^2 + 5} $ is odd, then $ \frac{-x}{(-x)^2 + 5} $ should equal $ -\frac{x}{x^2 + 5} $.