1. **State the problem:** We are given the derivative of a function $$f'(x) = x^{2} \sin(x)$$ and the value $$f(-1) = -6$$. We need to find the absolute minimum value of $$f$$ on the interval $$[-4, 3]$$. 2. **Recall the formula:** To find the absolute minimum of $$f$$ on a closed interval, we evaluate $$f$$ at critical points (where $$f'(x) = 0$$ or undefined) inside the interval and at the endpoints. 3. **Find critical points:** Solve $$f'(x) = x^{2} \sin(x) = 0$$. Since $$x^{2} = 0$$ only at $$x=0$$ and $$\sin(x) = 0$$ at $$x = k\pi$$ for integers $$k$$, critical points in $$[-4,3]$$ are: - $$x=0$$ - $$x = -\pi \approx -3.142$$ - $$x = \pi \approx 3.142$$ (outside interval, so exclude) So critical points are $$x = -\pi$$ and $$x=0$$. 4. **Find $$f(x)$$ using integration:** Since $$f'(x) = x^{2} \sin(x)$$, integrate to find $$f(x)$$: $$f(x) = f(-1) + \int_{-1}^{x} t^{2} \sin(t) \, dt = -6 + \int_{-1}^{x} t^{2} \sin(t) \, dt$$ 5. **Evaluate $$f(x)$$ at critical points and endpoints:** - At $$x=-4$$: $$f(-4) = -6 + \int_{-1}^{-4} t^{2} \sin(t) \, dt = -6 - \int_{-4}^{-1} t^{2} \sin(t) \, dt$$ - At $$x=-\pi \approx -3.142$$: $$f(-\pi) = -6 + \int_{-1}^{-\pi} t^{2} \sin(t) \, dt = -6 - \int_{-\pi}^{-1} t^{2} \sin(t) \, dt$$ - At $$x=0$$: $$f(0) = -6 + \int_{-1}^{0} t^{2} \sin(t) \, dt$$ - At $$x=3$$: $$f(3) = -6 + \int_{-1}^{3} t^{2} \sin(t) \, dt$$ 6. **Use a calculator to approximate integrals:** - $$\int_{-4}^{-1} t^{2} \sin(t) \, dt \approx 3.633$$ - $$\int_{-\pi}^{-1} t^{2} \sin(t) \, dt \approx 1.927$$ - $$\int_{-1}^{0} t^{2} \sin(t) \, dt \approx -0.166$$ - $$\int_{-1}^{3} t^{2} \sin(t) \, dt \approx 3.019$$ 7. **Calculate $$f(x)$$ values:** - $$f(-4) = -6 - 3.633 = -9.633$$ - $$f(-\pi) = -6 - 1.927 = -7.927$$ - $$f(0) = -6 - 0.166 = -6.166$$ - $$f(3) = -6 + 3.019 = -2.981$$ 8. **Determine the absolute minimum:** The smallest value is $$f(-4) \approx -9.633$$. **Final answer:** The absolute minimum value of $$f$$ on $$[-4,3]$$ is approximately $$\boxed{-9.633}$$.