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 $g\left(t\right)=t^3-3t-4t^2+8$
$g\left(t\right)=t^3-4t^2-3t+8$
$g^{\prime }\left(t\right)=3t^2-8t-3=\left(t-3\right)\left(3t+1\right)$
$g^{\prime }\left(t\right)=0\Rightarrow t=3\left(t\ne -1/3\right)$
$g^{\prime \prime }\left(t\right)=6t-8$
$g^{\prime \prime }\left(3\right)=10>0\Rightarrow g\left(3\right)$ is minimum
$g\left(3\right)=27-9-36+8=-10$