1. The problem asks for the equation of a function with two stationary points and a point of inflection. 2. A stationary point occurs where the first derivative of the function equals zero: $$f'(x) = 0$$. 3. A point of inflection occurs where the second derivative changes sign, often where $$f''(x) = 0$$. 4. To have two stationary points and one inflection point, a cubic polynomial is a good candidate: $$f(x) = ax^3 + bx^2 + cx + d$$. 5. The first derivative is $$f'(x) = 3ax^2 + 2bx + c$$, which is a quadratic and can have two distinct roots (stationary points). 6. The second derivative is $$f''(x) = 6ax + 2b$$, which is linear and has exactly one root (inflection point). 7. Example: Let $$f(x) = x^3 - 3x^2 + 2$$. 8. Compute first derivative: $$f'(x) = 3x^2 - 6x$$. 9. Set $$f'(x) = 0$$: $$3x^2 - 6x = 0 \Rightarrow 3x(x - 2) = 0$$, so stationary points at $$x=0$$ and $$x=2$$. 10. Compute second derivative: $$f''(x) = 6x - 6$$. 11. Set $$f''(x) = 0$$: $$6x - 6 = 0 \Rightarrow x=1$$, the inflection point. 12. Thus, the function $$f(x) = x^3 - 3x^2 + 2$$ has two stationary points at $$x=0$$ and $$x=2$$ and one inflection point at $$x=1$$. 13. Final answer: $$f(x) = x^3 - 3x^2 + 2$$.