wrongly asked question about precalculus? Im seeing the following question in a precalc textbook: Suppose $f$ is a function whose domain is $[-5,5]$ and $f(x) = \frac{x}{x+3}$ for every $x$ in $[0,5]$. Suppose $f$ is an odd function. Evaluate $f(-3)$.
Isnt this a bad formulated problem?? I mean, $f(-3)$ is not even defined, but since $f$ is odd, then $f(-3) = - f(3) = - \frac{1}{2}$. 
Or, am I just misinterpreting the question?
 A: You are indeed misinterpreting the question. The definition of $f$ as $f(x) = \frac{x}{x + 3}$ is (as explicitly stated) only defined on the interval $[0, 5],$ Further, it is given that $f$ is odd, which means indeed confirms that the domain is $[-5, 5],$ including at $x = -3.$
A: It is not poorly formulated but may seem a bit tricky. This essentially is a piecewise function as
$f(x)=\frac{x}{x+3}: \quad x\in [0,5]$
$f(x)=-\frac{|x|}{|x|+3}: \quad x\in [-5,0)\quad\!\!$ because the function is odd. 
Thus, $f(-3)=-\frac{1}{2}$ as you have mentioned.
A: To give a direct answer using the properties that you have likely already covered. We are given that $f$ is an odd function, meaning that it satisfies the equation
$f(-x)=-f(x)$
Hence to find $f(-3)$ we let $x=3$ and apply the above formula, yielding
$f(-3)=-f(3)$
I assume you know how to find $f(3)$ using the given formula, which defines $f$ for $x$ in $[0,5]$. (Note the negative sign on $f(3)$ in the above formula, and be sure to include it in your final answer!)
