Question

Find the greatest and the least values of the following functions.
$y=\frac{4}{3}{x}^3-4x$ on the interval $[0,2]$.

Answer 1

$y=\frac{4}{3}{x}^3-4x$

for greatest and least value in interval $[0,2]$

$\frac{dy}{dx}=\frac{4}{3}\times 3x^2-4=0$

$\Rightarrow 4x^2-4=0$

$\Rightarrow x^2=1$

$\Rightarrow x=\pm 1$

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(4x^2-4)$

$=8x-0$

at $x=1$

$\frac{d^{2}y}{d{x}^2}=8>0$ (point of local minima)

at $x=-1$

$\frac{d^{2}y}{d{x}^2}=-8<0$ (point of local maxima)

so at $x=1$

$y=\frac{4}{3}(1{)}^3-4(1)$

$=\frac{4}{3}-4$

$=\frac{-8}{3}$

$x=-1$

$y=\frac{-4}{3}+4$

$=\frac{8}{3}$

at $x=0$

$y=0$

and at $x=2$

$y=\frac{4}{3}\times 8-8$

$=\frac{32}{3}-8=\frac{8}{3}$

Greatest value is $=\frac{8}{3}$

Minimum value is $=0$.