2(a) If y = 2x3 − 9x2 + 12x, find the stationary points of the function, determine the nature of the stationary point, and find the maximum and minimum value of the function. 2(b) If y = sqrt(x), find dy/dx from first principle.

Given $y = 2x^3 - 9x^2 + 12x$.

$$\frac{dy}{dx} = \frac{d}{dx}(2x^3 - 9x^2 + 12x)$$ $$\frac{dy}{dx} = 6x^2 - 18x + 12$$

$$6x^2 - 18x + 12 = 0$$

Divide the equation by 6:

$$x^2 - 3x + 2 = 0$$

Factor the quadratic equation:

$$(x-1)(x-2) = 0$$

This gives two possible values for $x$:

$$x-1 = 0 \implies x = 1$$ $$x-2 = 0 \implies x = 2$$

For $x=1$:

$$y = 2(1)^3 - 9(1)^2 + 12(1)$$ $$y = 2 - 9 + 12$$ $$y = 5$$

So, the first stationary point is $(1, 5)$.

For $x=2$:

$$y = 2(2)^3 - 9(2)^2 + 12(2)$$ $$y = 2(8) - 9(4) + 24$$ $$y = 16 - 36 + 24$$ $$y = 4$$

So, the second stationary point is $(2, 4)$.

The stationary points are $(1, 5) \text{ and } (2, 4)$.