determine alegebraically whether the function is even, odd or neither
f(x) = x/(x^2+5)

I orginally assumed it was even because of x^2

Question

determine alegebraically whether the function is even, odd or neither
f(x) = x/(x^2+5)

I orginally assumed it was even because of x^2

hartnn · Answer

Find f(-x)
by replacing x in f(x) by -x.
If f(-x) = f(x), then the function is even
If f(-x) = -f(x), then the function is odd
else neither.

anonymous · Answer

how would you write/solve for -f(x)?

hartnn · Answer

-f(x) is just -x/(x^2+5)

anonymous · Answer

so... would it be$$f(x) = - (x \div x^{2} + 5)$$ ??

hartnn · Answer

to get f(-x) you just replace x by -x in the f(x) equation
what u get for f(-x) then ?

anonymous · Answer

He is saying solve f(-x) so it would be $$f(-x)=\frac{ (-x) }{ (-x)^2+5 }$$If f(-x) = f(x), then the function is even
If f(-x) = -f(x), then the function is odd 
else neither.

anonymous · Answer

$$f(-x) = -x \div x^{2}+5$$
because x^2 cancels out the negative

but what I dont understand is why when it's -f(x), shouldn't the 5 be negative too? therefore it would be "neither"?

anonymous · Answer

because then it would be$$f(-x) = -x \div x^{2} + 5   and  -f(x) = -x \div x^{2}-5$$

hartnn · Answer

no, here u get f(-x) exactly equal to -f(x) 
so its odd
and you only replace x by -x
so 65 will remain +5

hartnn · Answer

sorrry, +5 will remain +5

anonymous · Answer

I thought -f(x) made the whole equationo negative

hartnn · Answer

yes, whole means only numerator... so x in numerator became -x

anonymous · Answer

oooo, I had no idea.. I hate overthinking these things -.-

hartnn · Answer

so now its clear, how its odd ?

anonymous · Answer

yep!