Question

Find the minimum and maximum values of the function $y=x^3-3x^2+6$. Also find the values of $x$ at which these occur.

Answer 1

Given $y=x^3-3x^2+6$
Differentiating $y$ w.r.t. $x,\frac{dy}{dx}=3x^2-6x$
Putting $dy/dx=0$, we get the values at which the function is maximum or minimum. So
$3x^2-6x=0$
$\Rightarrow x(3x-6)=0\Rightarrow x=0,+2$
To distinguish the values of $x$ as the point of maximum or minimum, we need second derivative of the function.
$\therefore \frac{d^{2}y}{d{x}^2}=6x-6$; Now ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=0}=-6<0$.
So $x=0$ is a point of maximum.
Similarly, ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=+2}=6>0$
So $x=+2$ is a point of minimum.
Hence, the maximum value of $y$ is $0^3-3\times 0+6=6$ and the minimum value of $y$ is $(2{)}^3-3(2{)}^2+6=2$.