Is the function #f(s) = 4s^(3/2) # even, odd or neither?

1 Answer
Oct 21, 2015

Answer:

Neither.

Explanation:

As a Real valued function #f(s)# is only defined for #s >= 0#, so it's neither odd nor even.

As a Complex valued function we find:

If #s >= 0# then #f(s) >= 0#

If #s < 0# then:

#f(s) = 4 sqrt(s)^3 = 4(i sqrt(-s))^3 = -4i sqrt(-s) = -i f(-s)#

So neither #f(-s) = -f(s)# nor #f(-s) = f(s)#