Question

f(x) = $\frac { ( x - 2 ) ( x - 1 ) } { ( x - 3 ) }$∀ x > 3. The minimum value of f(x) is

equal to

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

Answer 1

Answer: (A)

3 + 2$\sqrt{2}$

Answer 2

Let x – 3 = t ⇒ x – 2 = (t +1), x – 1 = t + 2

⇒ f(x) = $\frac { ( x - 2 ) ( x - 1 ) } { ( x - 3 ) }$ = f(x) = $\frac { ( t + 1 ) ( t + 2 ) } { t }$

= $\frac { \left( t ^ { 2 } + 3 t + 2 \right) } { t }$ = t + + 3 ≥ 3 + $2 \sqrt { 2 }$

(using A.M. ≥ G.M. as t > 0)