Expecto Maxminimum!

The value of a third degree polynomial is equal to the average of its local minimum and a local maximum at its inflection point.

$y(x)=x^3-3x^2+4$

$y'(x)=3x^2-6x$

$y''(x)=6x-x=0$

$x=1$

$A=y(1)=1^3-3\times1^2+4=\boxed{2}$

$\begin{aligned} y & = x^3 - 3x^2 + 4 \\ \frac {dy}{dx} & = 3x^2 - 6x \\ \frac {d^2 y}{dx^2} & = 6x - 6 \end{aligned}$

Extrema of $y$ occurs when $\dfrac {dy}{dx} = 0$ . $\implies 3x^2 - 6x = 0$ $\implies x(x-2) = 0$ .

$\implies \begin{cases} x = 0 & \implies \dfrac {d^2y}{dx^2} = - 6 < 0 & \implies y(0) = 4 & \text{is maximum.} \\ x = 2 & \implies \dfrac {d^2y}{dx^2} = 6 > 0 & \implies y(2) = 0 & \text{is minimum.} \end{cases}$

Therefore, the average of maximum and minimum $A = \dfrac {4+0}2 = \boxed{2}$ .

The function $y = x^3 - 3x^2 + 4$ has no maximum or minimum, at least over all real numbers.