# Out of the following functions, which would be odd?

## $g \left(x\right) = - \frac{1}{2} {\left(x + 5\right)}^{3} + 2$ $g \left(x\right) = 2 {\left(x - 3\right)}^{2} - 8$ $g \left(x\right) = - \frac{4}{x}$ $g \left(x\right) = \sqrt{2 x + 6} - 5$ $g \left(x\right) = | | - x + 5 | |$

Oct 30, 2016

$g \left(x\right) = - \frac{4}{x}$ 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.

• $g \left(x\right) = - \frac{1}{2} {\left(x + 5\right)}^{3} + 2$ - 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 $g \left(x\right)$ is neither even nor odd.

Alternately, consider the counterexample $g \left(- 5\right) \ne - g \left(5\right)$

• $g \left(x\right) = 2 {\left(x - 3\right)}^{2} - 8$ - 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 $g \left(- 3\right) \ne - g \left(3\right)$

• $g \left(x\right) = - \frac{4}{x}$ - 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:

$g \left(- x\right) = - \frac{4}{- x} = - \left(- \frac{4}{x}\right) = - g \left(x\right)$

• $g \left(x\right) = \sqrt{2 x + 6} - 5$ - Not odd

Consider the counterexample $g \left(- 3\right) \ne - g \left(3\right)$

• $g \left(x\right) = | | - x + 5 | |$ - Not odd

Consider the counterexample $g \left(- 5\right) \ne - g \left(5\right)$