If f(-x) = f(x) then the function is even.
If f(-x) = -f(x) then the function is odd.
We have: [tex]f(x)=\dfrac{-1}{x^2+x^4}[/tex]
calculate f(-x):
[tex] f(-x)=\dfrac{-1}{(-x)^2+(-x)^4}=\dfrac{-1}{(-1x)^2+(-1x)^4}\\\\=\dfrac{-1}{(-1)^2x^2+(-1)^4x^4}=\dfrac{-1}{1x^2+1x^4}=\dfrac{-1}{x^2+x^4}=f(x) [/tex]
f(-x) = f(x), therefore your answer: The function f(x) is even.
If we have: [tex]f(x)=\dfrac{-1}{x^2}+x^4[/tex]
[tex]f(-x)=\dfrac{-1}{(-x)^2}+(-x)^4=\dfrac{-1}{x^2}+x^4[/tex]
f(-x) = f(x)
f(x) is even
