Question

If $f\left(x\right)=12x^{4/3}-6x^{1/3},x\in [-1,1]$ then the absolute minimum value of the function is

• A. -2.25

Answer 1

The correct option is A -2.25
Given,
$f\left(x\right)=12x^{4/3}-6x^{1/3},x\in [-1,1]$

Differentiating w.r.t. x, we get,

$f^{\prime }\left(x\right)=12.\frac{4}{3}{x}^{1/3}-6\frac{1}{3}{x}^{-2/3}$

$=16x^{1/3}-\frac{2}{x^{2/3}}=\frac{2(8x-1)}{x^{2/3}}$

Now $f^{\prime }\left(x\right)=0$
$2(8x-1)=0$

$x=\frac{1}{8}$

$f\left(\frac{1}{8}\right)=12{\left(\frac{1}{8}\right)}^{4/3}-6{\left(\frac{1}{8}\right)}^{1/3}$

$=12{\left(\frac{1}{2}\right)}^4-6\left(\frac{1}{2}\right)$

$=\frac{12}{16}-3=-\frac{9}{4}$

$f(-1)=12(-1{)}^{4/3}-6(-1{)}^{1/3}=12\left(1\right)-6(-1)=18$

$f\left(1\right)=12(1{)}^{4/3}-6(1{)}^{1/3}=12-6=6$

$\therefore$ Absolute minimum value $=\frac{-9}{4}=-2.25$