1. **Problem statement:** Given a cubic function $f(x) = ax^3 + bx^2 + cx + d$, find the stationary points, determine their nature, find the point of inflexion, and sketch the graph. 2. **Choose a cubic function:** Let's take $f(x) = 2x^3 - 3x^2 - 12x + 5$. 3. **Find the stationary points:** Stationary points occur where the first derivative $f'(x)$ is zero. $$f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 5) = 6x^2 - 6x - 12$$ Set $f'(x) = 0$: $$6x^2 - 6x - 12 = 0$$ Divide both sides by 6: $$\cancel{6}x^2 - \cancel{6}x - \cancel{6}2 = 0 \Rightarrow x^2 - x - 2 = 0$$ 4. **Solve the quadratic:** $$x^2 - x - 2 = 0$$ Factor: $$(x - 2)(x + 1) = 0$$ So stationary points at $x = 2$ and $x = -1$. 5. **Find the nature of stationary points:** Use the second derivative test. $$f''(x) = \frac{d}{dx}(6x^2 - 6x - 12) = 12x - 6$$ Evaluate at $x=2$: $$f''(2) = 12(2) - 6 = 24 - 6 = 18 > 0$$ So $x=2$ is a local minimum. Evaluate at $x=-1$: $$f''(-1) = 12(-1) - 6 = -12 - 6 = -18 < 0$$ So $x=-1$ is a local maximum. 6. **Find the point of inflexion:** Point of inflexion occurs where $f''(x) = 0$. $$12x - 6 = 0$$ Divide both sides by 6: $$\cancel{6}2x - \cancel{6}1 = 0 \Rightarrow 2x - 1 = 0$$ Solve: $$2x = 1 \Rightarrow x = \frac{1}{2}$$ Find $f\left(\frac{1}{2}\right)$: $$f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right)^2 - 12\left(\frac{1}{2}\right) + 5 = 2\left(\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) - 6 + 5 = \frac{1}{4} - \frac{3}{4} - 6 + 5 = -\frac{1}{2} - 1 = -1.5$$ So point of inflexion at $\left(\frac{1}{2}, -1.5\right)$. 7. **Summary:** - Stationary points: $(-1, f(-1))$ and $(2, f(2))$ Calculate $f(-1)$: $$f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12$$ Calculate $f(2)$: $$f(2) = 2(8) - 3(4) - 24 + 5 = 16 - 12 - 24 + 5 = -15$$ - Nature: $(-1, 12)$ is a local maximum, $(2, -15)$ is a local minimum. - Point of inflexion: $\left(\frac{1}{2}, -1.5\right)$.