1. **State the problem:** Given the function $f(x) = \frac{1}{3}x^3 - \frac{1}{2}x^2 + 6x$, find the values of $a$ such that $f'(a) = 18$. 2. **Find the derivative:** Use the power rule for derivatives: $\frac{d}{dx} x^n = nx^{n-1}$. $$f'(x) = \frac{d}{dx}\left(\frac{1}{3}x^3 - \frac{1}{2}x^2 + 6x\right) = \frac{1}{3} \cdot 3x^{2} - \frac{1}{2} \cdot 2x + 6 = x^2 - x + 6$$ 3. **Set the derivative equal to 18:** $$x^2 - x + 6 = 18$$ 4. **Solve for $x$:** $$x^2 - x + 6 - 18 = 0 \implies x^2 - x - 12 = 0$$ 5. **Factor the quadratic:** $$x^2 - x - 12 = (x - 4)(x + 3) = 0$$ 6. **Find the roots:** $$x - 4 = 0 \implies x = 4$$ $$x + 3 = 0 \implies x = -3$$ 7. **Answer:** The values of $a$ for which $f'(a) = 18$ are $-3$ and $4$. **Final answer:** $-3, 4$