JEE maths#3

Let $u = \sqrt x + \dfrac 1{\sqrt x}$ , then:

$\begin{aligned} \left(\sqrt x + \frac 1{\sqrt x}\right)^2 & = u^2 \\ x + 2 + \frac 1x & = u^2 \\ \implies x + \frac 1x & = u^2 - 2 \end{aligned}$

and

$\begin{aligned} \left(\sqrt x + \frac 1{\sqrt x}\right)^3 & = u^3 \\ x^\frac 32 + 3\sqrt x + \frac 3{\sqrt x} + \frac 1{x^\frac 32} & = u^3 \\ \implies x^\frac 32 + \frac 1{x^\frac 32} & = u^3 - 3u \end{aligned}$

Therefore,

$\begin{aligned} f(x) & = x^\frac 32 + \frac 1{x^\frac 32} - 4\left(x+\frac 1x \right) \\ & = u^3 - 3u - 4u^2 + 8 \\ f(u) & = u^3 - 4u^2 - 3u + 8 \\ f'(u) & = 3u^2 - 8u - 3 \\ & = (3x+1)(x-3) \end{aligned}$

Equating $f'(u)=0$ , we have:

$\begin{cases} u = - \frac 13 & \color{#D61F06} \text{But }u = \sqrt x + \frac 1{\sqrt x} \ge 2 \text{ unacceptable}. \\ u = 3 & f''\left(3\right) = 6\left(3\right)-8 > 0 \color{#3D99F6} \implies f\left(3\right) = \boxed{-10} \text{ is a minimum.} \end{cases}$