## Precalculus: Mathematics for Calculus, 7th Edition

Domain: $(-\infty,0)\cup(1,+\infty)$
For a logarithmic function to be defined, its argument must be POSITIVE, So, g is defined for $x-x^{2}>0$ We follow the steps outlined in section 3.7 1. All x's are on one side 2. Factor the polynomial $x(x-1)>0$ ... and find the real zeros, (x=0 and x=1) 3. The intervals into which these zeros divide the number line are : $(-\infty,0)$, where we take $-1$ as a test point, $(0,1),$where we take $\displaystyle \frac{1}{2}$ as a test point, and $(1,+\infty)$where we take $2$ as a test point, 4. Make a table or diagram to test the signs of the polynomial $\left[\begin{array}{llll} & (-\infty,0) & (0,1) & (1,+\infty)\\ test.point & -1 & 0.5 & 2\\ sign. of. x(x-1) & + & - & + \end{array}\right]$ so g is defined for $x\in (-\infty,0)\cup(1,+\infty)$ Domain: $(-\infty,0)\cup(1,+\infty)$