Question

Find the greatest and the least values of the following functions.
$y=x^2(2x-3)-12(3x-2)$ on the interval [-3, 6].

Answer 1

$y=x^2(2x-3)-12(3x-2)$

$\frac{dy}{dx}=6x^2-6x-36=0$

$x=3,-2$

$\left(\frac{d^{2}y}{d{x}^2}\right)=12x-6$

${\left(\frac{d^{2}y}{d{x}^2}\right)}_{y=3}=30>0$, Hence minimum will be $y\left(3\right)$

${\left(\frac{d^{2}y}{d{x}^2}\right)}_{y=-2}=-30<0$, Hence maximum will be $y(-2)$

$y\left(3\right)=-57$ is the minimum of the function

$y(-2)=82$ is the maximum of the function