Out of the following functions, which would be odd?

#g(x)=-1/2(x+5)^3+2#

#g(x)=2(x-3)^2-8#

#g(x)=-4/x#

#g(x)=sqrt(2x+6)-5#

#g(x)=||-x+5||#

1 Answer
Oct 30, 2016

#g(x)=-4/x# is the only odd function given.

Explanation:

I wrote a tutorial for even and odd functions in which all techniques used here are explained. The techniques described may tell you at a glance whether a function is even or odd, but may not be accepted as reasoning by a teacher. As such, alternate reasoning is also provided in some cases.

  • #g(x)=-1/2(x+5)^3+2# - Not odd

If we expanded the cubed binomial, we would have a polynomial with both even and odd exponents. This would give us a sum of terms which are even and odd functions, meaning #g(x)# is neither even nor odd.

Alternately, consider the counterexample #g(-5) != -g(5)#

  • #g(x)=2(x-3)^2-8# - Not odd

Similar to the above, expanding the squared binomial would give us at least an #x^2# term and an #x# term, meaning the whole polynomial is neither even nor odd.

Alternately, consider the counterexample #g(-3) != -g(3)#

  • #g(x)=-4/x# - Odd

This is the product of the constant #-4# (even) and the exponential #x^-1# (odd), and the product of an even and an odd function is odd.

Alternately, using the definition of an odd function:

#g(-x) = -4/(-x) = -(-4/x) = -g(x)#

  • #g(x)=sqrt(2x+6)-5# - Not odd

Consider the counterexample #g(-3) != -g(3)#

  • #g(x)=||-x+5||# - Not odd

Consider the counterexample #g(-5) != -g(5)#