Can anybody explain to me why this Rational Function is odd? $F(x)= \frac{\sqrt{x^2 + 9}}{-x^3+5x}$
Once that I do the procedure to know if it is even, odd or neither; the numerator remains the same, but the denominator gets opposite signs. I am trying to know if in rational function I always should take the result of the denominator, or if the square root has any influence to know if it is odd or even. Thanks; unfortunately the system doesn't let me post pictures yet.
 A: You have $f(x)=\frac {x^2+9}{-x^3+5x}$ 
$$f(-x)=\frac {x^2+9}{x^3-5x}=-f(x)$$
Thus your function is an odd function. 
We considered the sign of both top and bottom to make a decision. 
A: The quotient of an even function and an odd function is an odd function.  
Your function has an even numerator [because $x^2=(-x)^2$] 
and an odd denominator [because odd power functions are odd, and sum of odd functions is odd]; 
therefore, it is an odd function.
A: First, that's not usually called a rational function since it involves a nontrivial radical. A rational function is a quotient of polynomials.
But on to your question. You have correctly identified that the numerator is even (you simply need to note that it is the square root of an even polynomial), and that the denominator is odd. So what conclusion can you come to about their quotient? In general, if we let $E$ be any even function and $F$ any odd function, then what can we say about the compounds $E\pm F,EF,E/F$? Your case falls into the latter, which is just a special case of the middle one. Thus, what's the product of an odd and an even function? Since an even function remains the same but an odd one changes sign, it follows that their product changes sign too, upon performing $x\mapsto -x.$ Thus, the product $EF$ is also odd. What does this tell you about your function? Can you work out what $E+F$ is? If we let $E'$ be even too, and $F'$ be odd, then what's $EE',FF',E+E',F+F'$?
