jmc

Graphing the function, or trying different values of $x,$ we may think that the function is maximized at $x=1,$ which would make the maximum value 2.

To confirm this, we can consider the expression $$2-f(x)=x^3-3x+2.$$We know that this is zero at $x=1,$ so $x-1$ is a factor: $$2-f(x)=(x-1)(x^2+x-2)=(x-1{)}^2(x+2).$$Since $0\le x\le \sqrt{3},$ $x+2$ is always positive. Hence, $f(x)\le 2$ for all $x,$ which confirms that the maximum value is $\boxed{2}.$