1. The problem asks to find the value of $f(0)$ given the derivative $f'(x) = -10x^4 + 8x^3$ and the value $f(1) = 9$. 2. To find $f(0)$, we first need to find the original function $f(x)$ by integrating $f'(x)$. 3. Integrate $f'(x)$: $$f(x) = \int (-10x^4 + 8x^3) \, dx = \int -10x^4 \, dx + \int 8x^3 \, dx$$ 4. Calculate each integral: $$\int -10x^4 \, dx = -10 \cdot \frac{x^5}{5} = -2x^5$$ $$\int 8x^3 \, dx = 8 \cdot \frac{x^4}{4} = 2x^4$$ 5. So, $$f(x) = -2x^5 + 2x^4 + C$$ where $C$ is the constant of integration. 6. Use the condition $f(1) = 9$ to find $C$: $$f(1) = -2(1)^5 + 2(1)^4 + C = -2 + 2 + C = C = 9$$ 7. Therefore, $C = 9$ and the function is: $$f(x) = -2x^5 + 2x^4 + 9$$ 8. Finally, find $f(0)$: $$f(0) = -2(0)^5 + 2(0)^4 + 9 = 0 + 0 + 9 = 9$$ Answer: $f(0) = 9$