Consider the function $f\left(x\right)=x\sqrt{9-x^2}$ on the interval $[-3,3]$. Find the critical points of $f$ and use the First Derivative Test to classify these points. Then, determine the absolute maximum and minimum values of $f$ on the specified interval (if there are any).

Critical points: $x=\pm \frac{3\sqrt{2}}{2}$

Local maximum at $x=\frac{3\sqrt{2}}{2}$ and local minimum at $x=-\frac{3\sqrt{2}}{2}$

Absolute maximum: $4$ at $x=\frac{3\sqrt{2}}{2}$

Absolute minimum: $-4$ at $x=-\frac{3\sqrt{2}}{2}$

Critical points: $x=\pm \frac{3\sqrt{2}}{2}$, $x=\pm 3$

Local maximum at $x=\frac{3\sqrt{2}}{2}$ and local minimum at $x=-\frac{3\sqrt{2}}{2}$

Absolute maximum: $\frac{9}{2}$ at $x=\frac{3\sqrt{2}}{2}$

Absolute minimum: $-\frac{9}{2}$ at $x=-\frac{3\sqrt{2}}{2}$

Critical points: $x=\pm \frac{2\sqrt{3}}{3}$

Local maximum at $x=\frac{2\sqrt{3}}{3}$ and local minimum at $x=-\frac{2\sqrt{3}}{3}$

Absolute maximum: $4$ at $x=\frac{2\sqrt{3}}{3}$

Absolute minimum: $-4$ at $x=-\frac{2\sqrt{3}}{3}$

Critical points: $x=\pm \frac{2\sqrt{3}}{3}$

Local maximum at $x=\frac{2\sqrt{3}}{3}$ and local minimum at $x=-\frac{2\sqrt{3}}{3}$

No absolute extrema