1. **State the problem:** Find the critical numbers of the function $$f(x) = 2x^3 - 15x^2 + 36x + 5$$. 2. **Recall the definition:** Critical numbers occur where the derivative $$f'(x)$$ is zero or undefined. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(15x^2) + \frac{d}{dx}(36x) + \frac{d}{dx}(5) = 6x^2 - 30x + 36$$ 4. **Set the derivative equal to zero to find critical points:** $$6x^2 - 30x + 36 = 0$$ 5. **Simplify by dividing both sides by 6:** $$\cancel{6}x^2 - \cancel{6}5x + \cancel{6}6 = 0 \Rightarrow x^2 - 5x + 6 = 0$$ 6. **Factor the quadratic:** $$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$$ 7. **Solve for x:** $$x - 2 = 0 \Rightarrow x = 2$$ $$x - 3 = 0 \Rightarrow x = 3$$ 8. **Conclusion:** The critical numbers of $$f$$ are $$x = 2$$ and $$x = 3$$.