The minimum value of the function $ f(x) = x^{3/2} + x^{-3/2} - 4\left(x+\frac{1}{x}\right) $ for all permissible real x, is

-10

The correct answer is option (1) : -10

We have,

$f(x) = x^{3/2}+x^{-3/2}-4\left(x+\frac{1}{x}\right)$

$⇒f(x) = \left(\sqrt{x} +\frac{1}{\sqrt{x}}\right)^3-3\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) - 4\begin{Bmatrix}\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 - 2\end{Bmatrix}$

$⇒f(x) = \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^3 - 4\left(\sqrt{x} +\frac{1}{\sqrt{x}}\right)^2 - 3 \left(\sqrt{x} + \frac{1}{\sqrt{x}} \right) + 8 $

Clearly, f(x) is defined $∀ x > 0$.

Let $ \sqrt{x} +\frac{1}{\sqrt{x}}$. Then, $ t≥2 ∀ \, x > 0$.

Also, let $ g (t) = t^3 -4t^2 -3t +8$. Then,

$g'(t) = 3t^2 -8t - 3 $ and $g''(t) = 6t -8$

$⇒g'(t) = (3t +1) (t-3) $ and $ g''(t) = 6t - 8 $

Clearly, $g (t) = 0 $ for $ t = 3 $ and $ g''(t) = 10 > 0 $

Thus, g is minimum when t =3 and the minimum value of g is g (3) = -10.