Question

Find the greatest and the least values of the following functions.
$y=x^3+4x^2+4x+3$ on the interval [-1, 3].

Answer 1

$y=x^3+4x^2+4x+3$

$\frac{dy}{dx}=3x^2+8x+4=(3x+2)(x+2)=0$

$x=-\frac{2}{3},-2$

$\frac{d^{2}y}{d{x}^2}=6x+8$

${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=-2}=-4<0$,Hence $f(-2)=3$ is the maximum

${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=-\frac{2}{3}}=4>0$,Hence $f(-\frac{2}{3})=\frac{49}{27}$ is the minimum in the interval $[-1,3]$