Calculate the average gradient of f(x) = -2x3 between x = -1 and x = 3, and find its derivative from first principles.

Given $f(x) = -2x^3$.
For $x = -1$:

$$f(-1) = -2(-1)^3 = -2(-1) = 2$$

For $x = 3$:

$$f(3) = -2(3)^3 = -2(27) = -54$$

The average gradient is given by $\frac{f(x_2) - f(x_1)}{x_2 - x_1}$.

$$\text{Average gradient} = \frac{f(3) - f(-1)}{3 - (-1)} = \frac{-54 - 2}{3 + 1} = \frac{-56}{4}$$ $$\text{Average gradient} = -14$$

The average gradient between $x = -1$ and $x = 3$ is $\boxed{\mathbf{-14}}$.

Given $f(x) = -2x^3$.

$$f(x+h) = -2(x+h)^3$$

Expand $(x+h)^3$:

$$(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$$

Substitute this back into $f(x+h)$:

$$f(x+h) = -2(x^3 + 3x^2h + 3xh^2 + h^3) = -2x^3 - 6x^2h - 6xh^2 - 2h^3$$

$$f(x+h) - f(x) = (-2x^3 - 6x^2h - 6xh^2 - 2h^3) - (-2x^3)$$ $$f(x+h) - f(x) = -2x^3 - 6x^2h - 6xh^2 - 2h^3 + 2x^3$$ $$f(x+h) - f(x) = -6x^2h - 6xh^2 - 2h^3$$

$$\frac{f(x+h) - f(x)}{h} = \frac{-6x^2h - 6xh^2 - 2h^3}{h}$$

Factor out $h$ from the numerator:

$$\frac{h(-6x^2 - 6xh - 2h^2)}{h} = -6x^2 - 6xh - 2h^2$$

$$f'(x) = \lim_{h \to 0} (-6x^2 - 6xh - 2h^2)$$

Substitute $h=0$:

$$f'(x) = -6x^2 - 6x(0) - 2(0)^2$$ $$f'(x) = -6x^2$$

The derivative $f'(x)$ from first principles is $\boxed{\mathbf{-6x^2}}$.

Substitute $x = -1$ into the derivative function:

$$f'(-1) = -6(-1)^2$$ $$f'(-1) = -6(1)$$ $$f'(-1) = -6$$

The value of $f'(-1)$ is $\boxed{\mathbf{-6}}$.