1. The problem is to find the derivative of the function $$f(x) = \frac{3x^2 + 2x - 1}{x - 4}$$. 2. We use the quotient rule for derivatives, which states: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ where $u = 3x^2 + 2x - 1$ and $v = x - 4$. 3. Compute the derivatives of $u$ and $v$: $$u' = \frac{d}{dx}(3x^2 + 2x - 1) = 6x + 2$$ $$v' = \frac{d}{dx}(x - 4) = 1$$ 4. Apply the quotient rule: $$f'(x) = \frac{(6x + 2)(x - 4) - (3x^2 + 2x - 1)(1)}{(x - 4)^2}$$ 5. Expand the numerator: $$(6x + 2)(x - 4) = 6x^2 - 24x + 2x - 8 = 6x^2 - 22x - 8$$ 6. Substitute back: $$f'(x) = \frac{6x^2 - 22x - 8 - (3x^2 + 2x - 1)}{(x - 4)^2}$$ 7. Simplify the numerator: $$6x^2 - 22x - 8 - 3x^2 - 2x + 1 = (6x^2 - 3x^2) + (-22x - 2x) + (-8 + 1) = 3x^2 - 24x - 7$$ 8. Final derivative: $$f'(x) = \frac{3x^2 - 24x - 7}{(x - 4)^2}$$ This is the derivative of the given function using the quotient rule.