1. **State the problem:** Find the second derivative of the function $$f(x) = \frac{2x^2 + 3x + 4}{x}$$. 2. **Rewrite the function:** Simplify the expression by dividing each term by $x$: $$f(x) = \frac{2x^2}{x} + \frac{3x}{x} + \frac{4}{x} = 2x + 3 + 4x^{-1}$$ 3. **Find the first derivative $f'(x)$:** Use the power rule $\frac{d}{dx} x^n = nx^{n-1}$: $$f'(x) = \frac{d}{dx}(2x) + \frac{d}{dx}(3) + \frac{d}{dx}(4x^{-1}) = 2 + 0 + 4(-1)x^{-2} = 2 - 4x^{-2}$$ 4. **Find the second derivative $f''(x)$:** Differentiate $f'(x)$ again: $$f''(x) = \frac{d}{dx}(2) - \frac{d}{dx}(4x^{-2}) = 0 - 4(-2)x^{-3} = 8x^{-3}$$ 5. **Conclusion:** The second derivative is $$f''(x) = 8x^{-3}$$ which corresponds to option (c). This means the correct answer is (c) $f''(x) = 8x^{-3}$. This process uses the power rule for derivatives and careful simplification to find the second derivative step-by-step.