1. **State the problem:** We need to analyze the function $$f(x) = (x - 1)^3 (x + 2)^2$$. 2. **Formula and rules:** This is a polynomial function expressed as a product of powers of linear factors. To understand its behavior, we can find its roots, multiplicities, and analyze the derivative for critical points. 3. **Find roots and multiplicities:** The roots are at $$x=1$$ with multiplicity 3 and $$x=-2$$ with multiplicity 2. 4. **Find the derivative using the product rule:** $$f'(x) = \frac{d}{dx}[(x - 1)^3] \cdot (x + 2)^2 + (x - 1)^3 \cdot \frac{d}{dx}[(x + 2)^2]$$ Calculate each derivative: $$\frac{d}{dx}[(x - 1)^3] = 3(x - 1)^2$$ $$\frac{d}{dx}[(x + 2)^2] = 2(x + 2)$$ So, $$f'(x) = 3(x - 1)^2 (x + 2)^2 + (x - 1)^3 2(x + 2)$$ 5. **Factor the derivative:** $$f'(x) = (x - 1)^2 (x + 2) [3(x + 2) + 2(x - 1)]$$ Simplify inside the bracket: $$3(x + 2) + 2(x - 1) = 3x + 6 + 2x - 2 = 5x + 4$$ Thus, $$f'(x) = (x - 1)^2 (x + 2) (5x + 4)$$ 6. **Find critical points by setting $$f'(x) = 0$$:** $$ (x - 1)^2 = 0 \Rightarrow x = 1$$ $$ (x + 2) = 0 \Rightarrow x = -2$$ $$ (5x + 4) = 0 \Rightarrow x = -\frac{4}{5}$$ 7. **Summary:** - Roots of $$f(x)$$ at $$x=1$$ (multiplicity 3) and $$x=-2$$ (multiplicity 2). - Critical points at $$x=1$$, $$x=-2$$, and $$x=-\frac{4}{5}$$. This analysis helps understand the shape and turning points of the function.