Question

$f\left(x\right)=(x-1)(x-2)(x-3)$ has a local minima when $x$ is equal to

• A. $2+\frac{1}{\sqrt{3}}$
• B. $2-\frac{1}{\sqrt{3}}$
• C. $3-\frac{1}{\sqrt{2}}$
• D. $3+\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $2+\frac{1}{\sqrt{3}}$

$f\left(x\right)=(x-1)(x-2)(x-3)$
$\Rightarrow f\left(x\right)=x^3-6x^2+11x-6$
Differentiate w.r.t. $x$
$f^{\prime }\left(x\right)=3x^2-12x+11$
From $f^{\prime }\left(x\right)=0\Rightarrow x=\frac{12\pm \sqrt{144-132}}{6}$
$\Rightarrow x=2\pm \frac{1}{\sqrt{3}}$
Let $f^{\prime }\left(x\right)=(x-\alpha )(x-\beta ),\alpha <\beta$
and $\alpha =2-\frac{1}{\sqrt{3}},\beta =2+\frac{1}{\sqrt{3}}$

Since, sign of $f^{\prime }\left(x\right)$ changes from negative to positive as $x$ crosses $\beta$ from left to right, therefore $x=\beta$ is a point of local minima.