Answer
$ a.\quad$ shown below
$ b.\quad$ shown below
Work Step by Step
$ a.\quad$
$f(x)=x-\ln x$
$f'(x)=1-\displaystyle \frac{1}{x}$
When $ x\gt1$, the second term$, \displaystyle \frac{1}{x}$ is less than 1,
so $f'(x)\gt 0 \quad $on $(1,\infty)$,
which by definition, means that $f$ is increasing on $(1,\infty)$.
$ b.\quad$
We find that $f(1)=1-0=1$
(the function value of x=1 is positive)
By part (a), for any x greater than 1, f(x) is greater than f(1)
(which is 1, which is positive),
because f is increasing on $(1,\infty)$.
So, on $(1,\infty)$
f(x) is positive $\quad x-\ln x\gt 0$,
or
$x \gt \ln x$,
which we needed to show.
