Question

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

• A. $-10$
• B. $-6$
• C. $-7$
• D. $-8$

Answer 1

Answer: (A)

$f \left(x\right)=x^{3/2}+x^{-3/2}-4\left(x+\frac{1}{x}\right)$
$f \left(x\right)=\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{3}-3\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$
$-4\left[\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}-2\right]$
Let $\sqrt{x}+\frac{1}{\sqrt{x}}=t\left(x > 0\right)$
Let $g\left(t\right) = t^{3} - 3t - 4t^{2} + 8$
$g \left(t\right) = t^{3} - 4t^{2} - 3t + 8$
$g'\left(t\right)=3t^{2}-8t-3=\left(t-3\right)\left(3t+1\right)$
$g'\left(t\right)=0 \Rightarrow t=3\left(t\ne -1/3\right)$
$g''\left(t\right)=6t-8$
$g''\left(3\right)=10 > 0 \Rightarrow g\left(3\right)$ is minimum
$g\left(3\right)=27-9-36+8=-10$