Question

$f\left(x\right)=\frac{(x-2)(x-1)}{(x-3)}\forall x>3$. The minimum value of f(x) is equal to

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

Answer 1

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

Let x-3 = t $\Rightarrow$ x-2 = (t+1), x-1 = t+2
$\Rightarrow$ f(x) = $\frac{(x-2)(x-1)}{(x-3)}$ i.e f(x) = $\frac{(t+1)(t+2)}{t}$
$f\left(t\right)=\frac{t^2+3t+2}{t}\Rightarrow f^{\prime }\left(t\right)=\frac{t^2-2}{t^2}{f}^{\prime }\left(t\right)=0\Rightarrow t=\sqrt{2}[\because t\ne -\sqrt{2}]\therefore Minimumvalue=f\left(\sqrt{2}\right)=3+2\sqrt{2}$