Question

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

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

Answer 1

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

Given, $f\left(x\right)=\frac{(x-1)(x-2)}{x-3}$ and $x>3$

Taking the derivative and equating to zero. We get

$f^{\prime }\left(x\right)=\frac{x-1}{x-3}+\frac{x-2}{x-3}-\frac{(x-1)(x-2)}{(x-3{)}^2}$

$⟹f^{\prime }\left(x\right)=\frac{1}{x-3}(x-1+x-2-\frac{(x^2-2x-x+2)}{(x-3)})$

$⟹f^{\prime }\left(x\right)=\frac{1}{x-3}(2x-3-\frac{(x^2-3x+2)}{(x-3)})$

$⟹f^{\prime }\left(x\right)=\frac{1}{(x-3)}\left(\frac{(2x-3)(x-3)-(x^2-3x+2)}{(x-3)}\right)$

$⟹f^{\prime }\left(x\right)=\frac{1}{(x-3{)}^2}(2x^2-3x-6x+9-x^2+3x-2)$

$⟹f^{\prime }\left(x\right)=\frac{1}{(x-3{)}^2}(x^2-6x+7)$

$f^{\prime }\left(x\right)=0⟹\frac{1}{(x-3{)}^2}(x^2-6x+7)=0$

therefore $(x^2-6x+7)=0$ and $(x-3{)}^2\ne 0$ since, denominator cannot be equal to zero.

The roots of $(x^2-6x+7)=0$ are

$\frac{6+\sqrt{6^2-7\ast 4}}{2\ast 1}$ and $\frac{6-\sqrt{6^2-7\ast 4}}{2\ast 1}$

$⟹\frac{6+\sqrt{8}}{2}$ and $\frac{6-\sqrt{8}}{2}$

$⟹\frac{6+2\sqrt{2}}{2}$ and $\frac{6-2\sqrt{2}}{2}$

$⟹3+\sqrt{2}$ and $3-\sqrt{2}$

$x\notin 3-\sqrt{2}$ since $x>3$

Therefore $x=3+\sqrt{2}$

substituting $x=3+\sqrt{2}$ in $f\left(x\right)$ we get

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

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

$=\frac{(2+\sqrt{2}+2\sqrt{2}+2)}{\sqrt{2}}$

$=\frac{(4+3\sqrt{2})}{\sqrt{2}}=3+2\sqrt{2}$