Question

The maximum value of $(x-1)(x-2)(x-3)$ is

Answer 1

$(x-1)(x-2)(x-3)$

$(x-1)(x^2-5x+6)$

$=x^3-{5x}^2+6x-x^2+5x-6$

$f\left(x\right)=x^3-{6x}^2+11x-6$

Extreme values occurs at $f^1\left(x\right)=0$

$f^1\left(x\right)={3x}^2-12x+11=0$

$x=\frac{12\pm \sqrt{{12}^2-44\left(3\right)}}{6}=\frac{12\pm \sqrt{144-132}}{6}$

$=\frac{12\pm \sqrt{12}}{6}=\frac{12\pm 2\sqrt{3}}{6}=\frac{6\pm \sqrt{3}}{3}$

$x=2+\frac{1}{\sqrt{3}},2-\frac{1}{\sqrt{3}}$

$f(2+\frac{1}{\sqrt{3}})=(2+\frac{1}{\sqrt{3}}-1)(2+\frac{1}{\sqrt{3}}-2)(2+\frac{1}{\sqrt{3}}-3)$

$=(1+\frac{1}{\sqrt{3}})\left(\frac{1}{\sqrt{3}}\right)(-1+\frac{1}{\sqrt{3}})$

$=(\frac{1}{3}-1)\left(\frac{1}{\sqrt{3}}\right)$

$=\frac{-2}{3}\left(\frac{1}{\sqrt{3}}\right)=\frac{-2}{3\sqrt{3}}$

$f(2-\frac{1}{\sqrt{3}})=(2-\frac{1}{\sqrt{3}}-1)(2-\frac{1}{\sqrt{3}}-2)(2-\frac{1}{\sqrt{3}}-3)$

$=(1-\frac{1}{\sqrt{3}})(-1)(1+\frac{1}{\sqrt{3}})\left(\frac{-1}{\sqrt{3}}\right)$

$=\frac{2}{3}\times \frac{1}{\sqrt{3}}=\frac{2}{3\sqrt{3}}$

$\therefore$ , Max value of $f\left(x\right)=\frac{2}{3\sqrt{3}}$