Question

Let $f\left(x\right)=\frac{4}{3}{x}^3-4x,0\le x\le 2.$ Then the global minimum value of the function is

• A. $0$
• B. $-8/3$
• C. $-4$
• D. none of these

Answer 1

The correct option is B $-8/3$

$f\left(x\right)=\frac{4}{3}{x}^3-4x$
$f^{\prime }\left(x\right)=4(x^2-1)$
Hence $f\left(x\right)$ will be increasing in $(-\infty ,-1)\cup (1,\infty )$ and decreasing in $(-1,1)$
So in $xϵ[0,2]$ $f\left(x\right)$ will attain its minimum at $x=1$
The global minimum $=\frac{4}{3}-4=\frac{-8}{3}$