Determine algebraically if each function is even, odd, or neither.
1. F(x) = cuberoot{x}
2. F(x) = 2x/|x|
3. f(x) = x + |x|
Like always, seeking the needed steps.
An example:
\begin{align}\text{let} \quad f(x) &= x^2 \\ \text{then} \quad f(-x) &= (-x)^2 \\ &= \big((-1)x\big)^2 \\ &= (-1)^2 x^2 \\ &= (1)x^2 \\ &= x^2 \\ &= f(x) \end{align}
So $f(x)$ Is even.
You can follow similar procedures for your functions.
can we at least assume that you know that |x|= x if $\displaystyle x\ge 0$ and |x|= -x if x< 0.
From that
2) $\displaystyle f(x)= \frac{2x}{|x|}$
if $\displaystyle x\ge 0$ $\displaystyle f(x)= \frac{2x}{x}= 2$ and if $\displaystyle x< 0$, $\displaystyle f(x)= \frac{2x}{-x}= -2$.
3) $\displaystyle f(x)= x+ |x|$
if $\displaystyle x\ge 0$ $\displaystyle f(x)= x+ x= 2x$ and if $\displaystyle x< 0$, $\displaystyle f(x)= x+ (-x)= 0$.
Careful! |-x| is not equal, in general, to x.
$\displaystyle |x| = \begin{cases} -x & x < 0 \\ x & 0 < x \end{cases}$
We don't know if x is negative or positive. So the only thing you can say is
$\displaystyle F(-x) = \dfrac{2(-x)}{|-x|} = \dfrac{-2x}{|x|} = -\dfrac{2x}{|x|} = -F(x)$
As HallsofIvy and Plato have already told you.
You have a similar problem with #3.
-Dan