Question

Find the minimum value of a function $y$, represented as $y=x^3-3x^2+6$.

• A. $2$
• B. $0$
• C. $6$
• D. $-2$

Answer 1

The correct option is A $2$

$y=x^3-3x^2+6$
$\frac{dy}{dx}=3x^2-6x$
$\frac{dy}{dx}=0,$ condition for existence of maxima or minima
$\Rightarrow 3x^2-6x=0\Rightarrow x=0,2$
$\frac{d^{2}y}{d{x}^2}=6x-6$
Checking the double derivative at $x=0,2$
${\left[\frac{d^{2}y}{d{x}^2}\right]}_{x=0}=-6,$
Hence $\frac{d^{2}y}{d{x}^2}<0$
$\therefore$ Maxima exists at $x=0$
Similarly, ${\left[\frac{d^{2}y}{d{x}^2}\right]}_{x=2}=6,$
Hence, $\frac{d^{2}y}{d{x}^2}>0$
$\therefore$ Minima exists at $x=2$
$\therefore y_{\text{min}}=2^3-3(2{)}^2+6=2$