1. **Stating the problem:** We are given the function $f(x) = x^2 \cdot (x - 5)^4$ and asked to analyze its local extrema. 2. **Formula and rules:** To find local extrema, we first find the derivative $f'(x)$ and then solve $f'(x) = 0$ to find critical points. Local extrema occur at critical points where the derivative changes sign. 3. **Find the derivative:** Using the product rule and chain rule, $$f(x) = x^2 (x-5)^4$$ $$f'(x) = 2x (x-5)^4 + x^2 \cdot 4 (x-5)^3$$ 4. **Factor the derivative:** $$f'(x) = 2x (x-5)^3 \left((x-5) + 2x\right) = 2x (x-5)^3 (3x - 5)$$ 5. **Find critical points by setting $f'(x) = 0$:** $$2x (x-5)^3 (3x - 5) = 0$$ Critical points are at: - $x=0$ - $x=5$ - $x=\frac{5}{3}$ 6. **Determine the nature of each critical point:** - At $x=0$, the factor $x^2$ in $f(x)$ means the function touches the x-axis and does not cross it, indicating a local minimum. - At $x=5$, the factor $(x-5)^4$ is even-powered, so the function also touches the x-axis here, indicating a local minimum. - At $x=\frac{5}{3}$, the derivative changes sign, indicating a local maximum. 7. **Summary:** - Local minima at $x=0$ and $x=5$ - Local maximum at $x=\frac{5}{3}$ This matches the polynomial degree 6 behavior with roots at 0 and 5 and the shape of the graph.