Answer
(a)
Increasing on $(10, \infty)$
Decreasing on $(1, 10)$
(b)
No absolute maximum.
Local maximum: $(1,1) ,\ $
Absolute minimum: $(10,-8)$
No other local minima.
Work Step by Step
$f$ is defined on $[1,\infty).$
$f'(x)=1-\displaystyle \frac{6}{2\sqrt{x-1}}=1-\frac{3}{\sqrt{x-1}} =\displaystyle \frac{\sqrt{x-1}-3}{\sqrt{x-1}}$,
undefined at $ x=1\qquad$ ... critical point.
$f'(x)=0$ for $x=10$ ... critical point.
$f(1)=1,\quad f(10)=-8$
Using testpoints in the intervals between critical points,
$f'(5)=-0.5\lt 0$
$f'(17)=0.25 \gt 0$
Tabular view:
$\begin{array}{l}
K':\\
\\
\\
K:
\end{array} \begin{array}{lllll}
1 & & 10 & & \infty\\
{[} & -- & | & ++ & \\
& & & & \\
1 & \searrow & & \nearrow & \\
& & -8 & & \\
& & & &
\end{array}$
