Given f(x) = 2x3 − 5x2 − 4x + 3.

Given $f(x) = 2x^3 - 5x^2 - 4x + 3$.
To find the turning points, we set the first derivative equal to zero.

$$f'(x) = \frac{d}{dx}(2x^3 - 5x^2 - 4x + 3) = 6x^2 - 10x - 4$$

$$6x^2 - 10x - 4 = 0$$

Divide by 2 to simplify:

$$3x^2 - 5x - 2 = 0$$

Factor the quadratic equation:

$$(3x+1)(x-2) = 0$$

This gives two possible x-coordinates for the turning points:

$$3x+1 = 0 \implies x = -\frac{1}{3}$$ $$x-2 = 0 \implies x = 2$$

$$f''(x) = \frac{d}{dx}(6x^2 - 10x - 4) = 12x - 10$$

Evaluate $f''(x)$ at each x-value:
For $x = -\frac{1}{3}$:

$$f''\left(-\frac{1}{3}\right) = 12\left(-\frac{1}{3}\right) - 10 = -4 - 10 = -14$$

Since $f''(-\frac{1}{3}) < 0$, this corresponds to a local maximum (point M).

For $x = 2$:

$$f''(2) = 12(2) - 10 = 24 - 10 = 14$$

Since $f''(2) > 0$, this corresponds to a local minimum (point N).

Substitute $x=2$ into the original function $f(x)$:

$$f(2) = 2(2)^3 - 5(2)^2 - 4(2) + 3$$ $$f(2) = 2(8) - 5(4) - 8 + 3$$ $$f(2) = 16 - 20 - 8 + 3$$ $$f(2) = -4 - 8 + 3$$ $$f(2) = -12 + 3$$ $$f(2) = -9$$

The coordinates of the minimum turning point N are $(2; -9)$.