Answer
(a) The domain of $f(x)$ is $(1, \infty)$
The range of $f(x)$ is $(-\infty, \infty)$
(b) The x-intercept is $~~e+1$
(c) We can see a sketch of the graph of $f$ below.
Work Step by Step
(a) The domain of $ln(x)$ is $x \gt 0$
We can find the domain of $f(x) = ln(x-1)-1$:
$x-1 \gt 0$
$x \gt 1$
The domain of $f(x)$ is $(1, \infty)$
The range of $ln(x)$ is $(-\infty, \infty)$
Therefore the range of $f(x)$ is $(-\infty, \infty)$
(b) We can find the x-intercept:
$f(x) = ln(x-1)-1 = 0$
$ln(x-1)=1$
$x-1=e^1$
$x=e+1$
The x-intercept is $~~e+1$
(c) We can see a sketch of the graph of $f$ below.
