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.
Thanks in advance
$x-1$ 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 $(x-1)x(x+1)>0$.