Answer
$$\eqalign{
& {\text{There are no inflection points}} \cr
& {\text{Concave downward}}:\left( { - \infty ,9} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = x\sqrt {9 - x} \cr
& {\text{The domain of the function is }}9 - x \geqslant 0 \cr
& x \leqslant 9,{\text{ }}\left( { - \infty ,9} \right] \cr
& {\text{Calculate the first and second derivatives}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {x\sqrt {9 - x} } \right] \cr
& f'\left( x \right) = \sqrt {9 - x} \frac{d}{{dx}}\left[ x \right] + x\frac{d}{{dx}}\left[ {\sqrt {9 - x} } \right] \cr
& f'\left( x \right) = \sqrt {9 - x} + x\left( { - \frac{1}{{2\sqrt {9 - x} }}} \right) \cr
& f'\left( x \right) = \frac{{2\left( {9 - x} \right) - x}}{{2\sqrt {9 - x} }} \cr
& f'\left( x \right) = \frac{{18 - 2x - x}}{{2\sqrt {9 - x} }} \cr
& f'\left( x \right) = \frac{{18 - 3x}}{{2\sqrt {9 - x} }} \cr
& f''\left( x \right) = \underbrace {\frac{d}{{dx}}\left[ {\frac{{18 - 3x}}{{2\sqrt {9 - x} }}} \right]}_{{\text{Use quotient rule}}} \cr
& f''\left( x \right) = \frac{{2\sqrt {9 - x} \left( { - 3} \right) - \left( {18 - 3x} \right)\left( {\frac{{ - 1}}{{\sqrt {9 - x} }}} \right)}}{{{{\left( {2\sqrt {9 - x} } \right)}^2}}} \cr
& f''\left( x \right) = \frac{{ - 6\sqrt {9 - x} + \frac{{18 - 3x}}{{\sqrt {9 - x} }}}}{{4\left( {9 - x} \right)}} \cr
& f''\left( x \right) = \frac{{ - 6\left( {9 - x} \right) + 18 - 3x}}{{4{{\left( {9 - x} \right)}^{3/2}}}} \cr
& f''\left( x \right) = \frac{{ - 54 + 6x + 18 - 3x}}{{4{{\left( {9 - x} \right)}^{3/2}}}} \cr
& f''\left( x \right) = \frac{{3x - 36}}{{4{{\left( {9 - x} \right)}^{3/2}}}} \cr
& {\text{Set the second derivative to }}0 \cr
& 3x - 36 = 0 \cr
& x = 12 \cr
& {\text{This value is not in the domain of the function }}f\left( x \right) = x\sqrt {9 - x} \cr
& {\text{So, there are no inflection points, and the domain is }}x \leqslant 9 \cr
& {\text{Evaluating the second derivative at the interval }}\left( { - \infty ,9} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( { - \infty ,9} \right)} \\
{{\text{Test Value}}}&{x = 0} \\
{{\text{Sign of }}f''\left( x \right)}&{ - \frac{1}{3} < 0} \\
{{\text{Conclusion}}}&{{\text{Concave downward}}}
\end{array}}\]
$$\eqalign{
& {\text{There are no inflection points}} \cr
& {\text{Concave downward}}:\left( { - \infty ,9} \right) \cr} $$
