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

Hint: In order to find the maximum and minimum values of any function, it has to be differentiated two times. For the first derivative, critical values are obtained and for the second derivative the sign is to be checked. If the sign is negative, the function has maxima at that value and vice versa.

Complete step by step solution: The given function is
$y = {x^3} - 3{x^2} + 6......(1)$
Differentiate equation (1) concerning to $x$ , to obtain first derivative
$\dfrac{{dy}}{{dx}} = 3{x^2} - 6x......(2)$
Differentiate equation (2) concerning to $x$ , to obtain second derivative
$\dfrac{{{d^2}y}}{{d{x^2}}} = 6x - 6......(3)$
For the critical values, put $\dfrac{{dy}}{{dx}} = 0$
$ 3{x^2} - 6x = 0 \\
  3x\left( {x - 2} \right) = 0 \\
  x = 0,2 \\ $
The critical values are 0 and 2
Substitute $x = 0$ in equation (3)
$ \dfrac{{{d^2}y}}{{d{x^2}}} = 6\left( 0 \right) - 6 \\
  \dfrac{{{d^2}y}}{{d{x^2}}} = - 6 < 0 \\ $
 The sign of the second derivative is negative. $x = 0$ is a point of maxima
Substitute $x = 0$ in equation (1) to obtain maximum value of the function
$ y = 0 - 0 + 6 \\
  y = 6 \\ $
Substitute $x = 2$ in equation (3)
$ \dfrac{{{d^2}y}}{{d{x^2}}} = 6\left( 2 \right) - 6 \\
  \dfrac{{{d^2}y}}{{d{x^2}}} = 6 > 0 \\ $
 The sign of the second derivative is positive. $x = 2$ is a point of minima.
Substitute $x = 2$ in equation (1) to obtain maximum value of the function
$ y = {\left( 2 \right)^3} - 3{\left( 2 \right)^2} + 6 \\
  y = 8 - 12 + 6 \\
  y = 2 \\ $
Thus, maximum and minimum values of the function are $y = 6$ and $y = 2$ which occurs at $x = 0$ and $x = 2$ respectively.

Note: The sign of the second derivative should be correctly observed for obtaining maxima and minima.