Use derivatives to sketch the graph of a function

Let

$f(x) = \frac{1}{(x-1)(x-3)}.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f'(x) = \frac{-2x + 4}{(x-1)^2 (x-3)^2}.$

Thus,

$f'(x) = 0 \quad \implies \quad \frac{-2x+4}{(x-1)^2(x-3)^2} = 0 \quad \implies \quad x = 2.$
$f$ is increasing if $x < 1$ or $1 < x < 2$ and $f$ is decreasing if $2 < x < 3$ or $x > 3$ . (We have to take some care here to leave out the points $x = 1$ and $x = 3$ since the function is not defined at these points.)
Taking the second derivative,
$f''(x) = \frac{1}{(x-3)^3} - \frac{1}{(x-1)^3}.$

Thus, $f'$ is increasing for $x < 1$ and $x > 3$ and $f'$ is decreasing for $1 < x < 3$ .
We sketch the curve,