1. **State the problem:** We need to find the absolute minimum and maximum values of the function $$f(x) = 2.5x^4 + 3x^3 - 2.6x^2 - 5.1x - 5.6$$ on the closed interval $$[-0.5, 1.2]$$. 2. **Recall the method:** To find absolute extrema on a closed interval, evaluate the function at critical points inside the interval and at the endpoints. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(2.5x^4) + \frac{d}{dx}(3x^3) - \frac{d}{dx}(2.6x^2) - \frac{d}{dx}(5.1x) - \frac{d}{dx}(5.6)$$ $$= 10x^3 + 9x^2 - 5.2x - 5.1$$ 4. **Find critical points by solving $$f'(x) = 0$$:** $$10x^3 + 9x^2 - 5.2x - 5.1 = 0$$ 5. **Solve the cubic equation:** This can be done using numerical methods or graphing. Approximate roots inside $$[-0.5, 1.2]$$ are found to be approximately $$x \approx -1.1$$ (outside interval), $$x \approx -0.7$$ (outside interval), and $$x \approx 0.7$$ (inside interval). 6. **Evaluate $$f(x)$$ at critical points inside the interval and at endpoints:** - At $$x = -0.5$$: $$f(-0.5) = 2.5(-0.5)^4 + 3(-0.5)^3 - 2.6(-0.5)^2 - 5.1(-0.5) - 5.6$$ $$= 2.5(0.0625) + 3(-0.125) - 2.6(0.25) + 2.55 - 5.6$$ $$= 0.15625 - 0.375 - 0.65 + 2.55 - 5.6 = -3.91875$$ - At $$x = 0.7$$: $$f(0.7) = 2.5(0.7)^4 + 3(0.7)^3 - 2.6(0.7)^2 - 5.1(0.7) - 5.6$$ $$= 2.5(0.2401) + 3(0.343) - 2.6(0.49) - 3.57 - 5.6$$ $$= 0.60025 + 1.029 - 1.274 - 3.57 - 5.6 = -8.81475$$ - At $$x = 1.2$$: $$f(1.2) = 2.5(1.2)^4 + 3(1.2)^3 - 2.6(1.2)^2 - 5.1(1.2) - 5.6$$ $$= 2.5(2.0736) + 3(1.728) - 2.6(1.44) - 6.12 - 5.6$$ $$= 5.184 + 5.184 - 3.744 - 6.12 - 5.6 = -5.096$$ 7. **Determine absolute extrema:** - Absolute minimum value is approximately $$-8.81475$$ at $$x = 0.7$$. - Absolute maximum value is approximately $$-3.91875$$ at $$x = -0.5$$. **Final answers:** - Absolute minimum: $$f(0.7) \approx -8.815$$ - Absolute maximum: $$f(-0.5) \approx -3.919$$