Find the domain and graph: $$f(t)=\frac{-t}{|t|}$$

My book says to define it piecewise.

My questions:

$\mathbf{1)}$ Do all rational functions have to be defined piecewise, or just this one because there is an absolute value in the denominator, and the absolute value is always defined piecewise?

$\mathbf{2)}$ Are all rational functions defined piecewise in order to avoid having a denominator be equal to zero, is that the general reason for defining anything piecewise, to avoid having division by zero? (so then this would mean that the domain of a rational function is always defined piecewise, or only when we need to avoid having denominator be $=0$?)

This is how my book defines $f(t)$ piecewise:

If $t>0$, then $|t|$ is $t$ since $t$ is already positive.
For $t>0$, simplify $$f(t)=\frac{-t}{|t|} = \frac{-t}{t}=-1$$

If $t<0$, then $|t|$ is $-t$ since $t$ is negative.
For $t<0$, simplify $$f(t)=\frac{-t}{-|t|} = \frac{-t}{-t}=1$$

(or for the last part, should the negative be inside the absolute value sign for $t<0$, as in $|-t|$ instead of $-|t|$?)

So we define it piecewise to avoid having $0$ in the denominator? Because isn't absolute value defined at $0$, the absolute value is continuous everywhere, and thus defined at $0$?

I'm confused about defining things piecewise, and how to know when to apply a piecewise attempt in order to define a function's domain.

Also, I'm confused about the second part of this, where $t<0$ for $|t|$. How is it that here, $|t|$ is $-t$ if absolute value is always positive?

Maybe I'm not understanding the absolute value concept correctly, because in grade school it has always been drilled into my head that |absolute value| just "turns things positive", so here I don't really understand how it can be negative.

I understand how the domain is $(-\infty,0)\cup(0,\infty)$, because by using open intervals we're not letting it be exactly $=0$, but I'm confused about the whole piecewise thing.

  • $\begingroup$ Let $t=-3$. Then by what you wrote (which is correct), $|t|=-t=-(-3)=3$, it did turn it positive. More generally, $-t$ is not automatically negative. If $t$ itself is negative, then $-t$ is positive. $\endgroup$ – André Nicolas Aug 27 '13 at 22:01
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    $\begingroup$ Your function is not a rational function. A rational function is a ratio of polynomial functions, more briefly polynomials. But $|t|$ is not a polynomial. $\endgroup$ – André Nicolas Aug 27 '13 at 22:21
  • $\begingroup$ @Andre ok, i see. I meant functions that are fractions (ratios), in that case are they always defined piecewise? $\endgroup$ – Emi Matro Aug 27 '13 at 22:22
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    $\begingroup$ Many familiar functions that are ratios are defined by a single formula, like $\frac{x}{1+x^2}$. $\endgroup$ – André Nicolas Aug 27 '13 at 22:29

The absolute value is defined at $0$ with value $0$. It is nicely continuous. The reason your function is not defined at $0$ is because you can't divide by zero. If your function were $g(t)=\frac {-t}t$ with no absolute value you still would not have $0$ in the domain.

  • $\begingroup$ ok, thanks, so then my statements in my questions 1) and 2) are correct? just making sure... $\endgroup$ – Emi Matro Aug 27 '13 at 22:11
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    $\begingroup$ Rational functions need not have a piecewise domain if the denominator does not go to zero. Zero denominators are the problem, not the absolute value. $\endgroup$ – Ross Millikan Aug 27 '13 at 22:25

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