# Domain of definition of the function

I was going through some questions of Relations and Functions and now I am stuck to one. Question says

Question: Domain of definition of the function $$f(x)=\frac{9}{9-x^2}+\log_{10}(x^3-x)$$ is?

Till yet I know that the denominator of first term (i.e. $\frac{9}{9-x^2}$) will be non-zero. So I figured,

$$9-x^2\ne0$$ $$\implies x\ne\pm3$$

And from second term (i.e. $\log_{10}x^3-x$), we know that the term inside the logarithm can never be less than or equal to zero. So,

$$\log_{10}(x^3-x)$$ $$\implies x^3-x>0$$ $$\implies (x-1)x(x+1)>0$$

From here I know that if $x−1$ is negative when $x<1$ and positive when $x>1$. $x+1$ is negative when $x<−1$ and positive when $x>−1$. But I am still not getting the domain in a specific interval.

I seriously need help in this.

$x-1$ is negative when $x<1$ and positive when $x>1$. $x+1$ is negative when $x<-1$ and positive when $x>-1$. Using this and the fact that the product of two negative numbers is positive and a negative times a positive is negative, you can find all $x$ such that $(x-1)x(x+1)>0$.
• Then what can be the domain of the function. I want it to be in kind of intervals like $(i,j)\cup(k,l)$. Like this. – Saharsh Jun 3 '14 at 5:37
• The intervals you want to look at are $(-\infty,-1)$, $(-1,0)$, $(0,1)$, $(1,\infty)$. Different combinations of the factors will be positive or negative depending on which interval you are in, because of what I wrote in my "Answer". – mathematician Jun 3 '14 at 21:39