Answer
Domain: $(-\infty,0)\cup(1,+\infty)$
Work Step by Step
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: $(-\infty,0)\cup(1,+\infty)$
