even or odd or neither functions

[kpx001]

1. f(x) = x^-3
so
f(x) = 1/x^3

f(-x) = 1/(-x)^3
f(-x) = -1/x
so is this neither?

2. y = -2|x| - 1

f(-x) = -2|-x| - 1
f(-x) = -2x - 1

so even?

3. y = (1/x) + x

y = (1/-x) - x

odd

im a bit confused how i can compare to original when the exponents disapear

 

[pka]

\(\displaystyle \L \begin{array}{l}
\left( {x^{ - 3} } \right) = - \left[ {\left( { - x} \right)^{ - 3} } \right]\quad \Rightarrow \quad \mbox{odd} \\
- 2\left| x \right| - 1 = - 2\left| { - x} \right| - 1\quad \Rightarrow \quad \mbox{even} \\
\end{array}\)

 

[galactus]

Odd functions are symmetric about the origin and evens about the y-axis.

\(\displaystyle \L\\\frac{1}{x^{3}}=\frac{1}{(-x)^{3}}=\frac{-1}{x^{3}}\)

Since \(\displaystyle \L\\ODD=f(-x)=-f(x)\)

 

[pka]

Here is one way to use the graph to see oddness or evenness.
An even function has a graph that is symmetric about the y-axis.

An odd function has a graph that is symmetric about the origin.
Look at the graph above. Start at any point on the graph.
Draw a line through that point and the origin.
It will intersect the graph at a point equally distant on the other side of (0,0).

 

[kpx001]

3. y = (1/x) + x

y = (1/-x) - x

odd right?

i graphed it and it looks both symmetrical to originand y, but more origin?

 

[galactus]

It looks symmetrical to the origin, doesn't it?.

 

[pka]

Note that it could not possibly be symmetric about the y-axis.
To be symmetric about the y-axis means that the y-axis acts like a mirror.
To be symmetric about the origin means that the orgin acts like “pin-hole” camera.

It is an odd function.