Calculate the derivative of f(x) = 3x − 1 from first principles.

Substitute $x+h$ into the function $f(x)$:

$$f(x+h) = 3(x+h) - 1$$ $$f(x+h) = 3x + 3h - 1$$

Subtract the original function $f(x)$ from $f(x+h)$:

$$f(x+h) - f(x) = (3x + 3h - 1) - (3x - 1)$$ $$f(x+h) - f(x) = 3x + 3h - 1 - 3x + 1$$ $$f(x+h) - f(x) = 3h$$

$$\frac{f(x+h) - f(x)}{h} = \frac{3h}{h}$$ $$\frac{f(x+h) - f(x)}{h} = 3$$

Now, take the limit as $h \to 0$:

$$f'(x) = \lim_{h \to 0} 3$$

Since there is no $h$ in the expression, the limit is simply 3.

$$f'(x) = 3$$

The derivative of $f(x) = 3x - 1$ from first principles is:

$$\text{3}$$