Answer
$(0,1)$
Work Step by Step
We have to determine the values of $x$ for which $f(x) > g(x)$.
Given $f(x) = \frac {1}{x}$ and $g(x) = \frac {1}{x-1}$, we have
$\frac{1}{x} > \frac{1}{x-1}$
$\frac{1}{x} - \frac {1}{x-1} > 0 $
$\frac {-1}{(x)(x-1)} > 0$
$(x)(x-1) < 0$
Find the zeros of the expressions in the numerator AND the denominator
$x = 0, x=1$
Test numbers in the intervals $(-\infty,0)$, $(0,1)$ and $(1,\infty)$to determine if the function is negative or positive:
$(-\infty, 0)$: $\frac {(+)}{(-)(-)} = (+)$
$(0, 1)$: $\frac {(+)}{(+)(-)} = (-)$
$(1,\infty)$: $\frac {(+)}{(+)(+)} = (+)$
Thus the solution is $(0,1)$.