The maximum value of the function $f(x)=(x-1)(x-2)(x-3)$ is:

Given $f(x)=(x-1)(x-2)(x-3)$.Expanding the expression:$f(x)=x^3-6x^2+11x-6$ Differentiating with respect to $x$:$f^{\prime }(x)=3x^2-12x+11$ To find critical points, set $f^{\prime }(x)=0$:$3x^2-12x+11=0$ Using quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$:$x=\frac{12\pm \sqrt{144-132}}{6}=\frac{12\pm \sqrt{12}}{6}=2\pm \frac{\sqrt{3}}{3}=2\pm \frac{1}{\sqrt{3}}$$x\approx 1.423,2.577$ Finding the second derivative:$f^{\prime \prime }(x)=6x-12$ At $x=2-\frac{1}{\sqrt{3}}\approx 1.423$:$f^{\prime \prime }(1.423)=6(1.423)-12<0$. Thus, a local maximum occurs at $x=2-\frac{1}{\sqrt{3}}$.Value at $x=2-\frac{1}{\sqrt{3}}$:$f(2-\frac{1}{\sqrt{3}})=(2-\frac{1}{\sqrt{3}}-1)(2-\frac{1}{\sqrt{3}}-2)(2-\frac{1}{\sqrt{3}}-3)$$=(1-\frac{1}{\sqrt{3}})(-\frac{1}{\sqrt{3}})(-1-\frac{1}{\sqrt{3}})$$=(-\frac{1}{\sqrt{3}})[(1-\frac{1}{\sqrt{3}})(-1-\frac{1}{\sqrt{3}})]$$=(-\frac{1}{\sqrt{3}})[-(1^2-(\frac{1}{\sqrt{3}}{)}^2)]=(-\frac{1}{\sqrt{3}})[-(1-\frac{1}{3})]=(-\frac{1}{\sqrt{3}})(-\frac{2}{3})=\frac{2}{3\sqrt{3}}$