Turn a function into an odd function?

Question

I was asked to let:

$$ f(x) = \begin{cases} 1-x, \text{if $0 \le x < 1$} \\ 0, \text{if $ 1 \le x < 2$} \end{cases}$$

Let fodd be the $4$ periodic odd extension of $f(x)$.

Find the expression of fodd when $-1 < x <0$.

The solution was:

$$f_{\operatorname{odd}}(x) = -f(-x) = -(1-(-x)) = -(1+x) = -1-x$$

My question is: How is $f_{\operatorname{odd}}(x)=-f(-x)$ possible? I thought that $f_{\operatorname{odd}}$ is when $f(-x)=-f(x)$, so I thought that changing $1-x$ to $-1+x$ would be the answer.

edit:

that is what i have fodd(x) is when f(−x)=-f(x) but the solution used fodd(x) = −f(−x) i don't know why this was used.

and for source of why f(-x) = -f(x): http://mathworld.wolfram.com/OddFunction.html https://en.wikipedia.org/wiki/Even_and_odd_functions

Answer 1

$f(-x) = -f(x)$ can be just written (equivalently) as $f(x) = -f(-x)$. That's what's it about.