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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?

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    The function's domain is [-5,5] so the function exists and is "defined" on [-5,5] even though we might not know what the definition is. However knowing f is odd and knowing the definition on [0,5] is sufficient to know what f(x) is for -5 <= x < 0. – fleablood Jun 21 '16 at 05:24
  • I think "undefined" has multiple contextual ambiguous meaning. To say a function is "undefined" at x=-3 is usually taken to mean f (-3) has no value. Here we are told f has domain [-5,5], so f (-3) does indeed have a value (whether we know what it is or not) so it is "defined". On the other hand it is literally undefined in that we simply haven't been told what f (-3) is. But that doesn't matter as knowing f is odd is enough to deduce it. – fleablood Jun 21 '16 at 05:31

3 Answers3

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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.

Devon
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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 $x = -3.$

K. Jiang
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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!)

Justin Benfield
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  • Why was this voted down? It seems like a correct answer to me. – Aiman Al-Eryani Jun 21 '16 at 11:16
  • @AimanAl-Eryani Furthermore, my approach has the advantage of not having to figure out how to write a formula for the other piece of $f$. You only need to apply the fact that the function is odd along with the given formula to find the answer. – Justin Benfield Jun 23 '16 at 03:38