1. **State the problem:** We have a function $$y = -x^2 + 2(\log(2)a)x + \log(2)b$$ defined for $$0 \leq x \leq 4$$. The maximum value of the function is 5 and the minimum value is -4. We need to find the values of $$a$$ and $$b$$. 2. **Identify the type of function:** This is a quadratic function in the form $$y = ax^2 + bx + c$$ where $$a = -1$$, $$b = 2(\log(2)a)$$, and $$c = \log(2)b$$. 3. **Find the vertex:** The vertex of a quadratic $$y = ax^2 + bx + c$$ is at $$x = -\frac{b}{2a}$$. Here, $$x_v = -\frac{2(\log(2)a)}{2(-1)} = \log(2)a$$. 4. **Calculate the vertex value:** Substitute $$x_v$$ into the function: $$$y_v = - (\log(2)a)^2 + 2(\log(2)a)(\log(2)a) + \log(2)b = - (\log(2)a)^2 + 2(\log(2)a)^2 + \log(2)b = (\log(2)a)^2 + \log(2)b$$$ 5. **Determine max and min values:** Since $$a = -1 < 0$$, the parabola opens downward, so the vertex is a maximum point. Given max value $$y_v = 5$$, so: $$$(\log(2)a)^2 + \log(2)b = 5$$$ 6. **Check values at endpoints:** The minimum value is -4, which must occur at one of the endpoints $$x=0$$ or $$x=4$$. At $$x=0$$: $$$y = -0 + 0 + \log(2)b = \log(2)b$$$ At $$x=4$$: $$$y = -16 + 8(\log(2)a) + \log(2)b$$$ 7. **Set minimum value:** The minimum is -4, so either: - $$\log(2)b = -4$$ or - $$-16 + 8(\log(2)a) + \log(2)b = -4$$ 8. **Solve system:** From step 5: $$$(\log(2)a)^2 + \log(2)b = 5$$ If minimum at $$x=0$$: $$\log(2)b = -4$$ Substitute into max equation: $$(\log(2)a)^2 - 4 = 5 \Rightarrow (\log(2)a)^2 = 9 \Rightarrow \log(2)a = \pm 3$$ If minimum at $$x=4$$: $$-16 + 8(\log(2)a) + \log(2)b = -4 \Rightarrow 8(\log(2)a) + \log(2)b = 12$$ Using max equation: $$(\log(2)a)^2 + \log(2)b = 5$$ Subtracting these two: $$8(\log(2)a) + \log(2)b - ((\log(2)a)^2 + \log(2)b) = 12 - 5$$ $$8(\log(2)a) - (\log(2)a)^2 = 7$$ Rewrite: $$(\log(2)a)^2 - 8(\log(2)a) + 7 = 0$$ Solve quadratic: $$\log(2)a = \frac{8 \pm \sqrt{64 - 28}}{2} = \frac{8 \pm \sqrt{36}}{2} = \frac{8 \pm 6}{2}$$ So: $$\log(2)a = 7 \text{ or } 1$$ Use max equation to find $$\log(2)b$$: - For $$\log(2)a = 7$$: $$(7)^2 + \log(2)b = 5 \Rightarrow 49 + \log(2)b = 5 \Rightarrow \log(2)b = -44$$ - For $$\log(2)a = 1$$: $$(1)^2 + \log(2)b = 5 \Rightarrow 1 + \log(2)b = 5 \Rightarrow \log(2)b = 4$$ 9. **Summary of solutions:** - Case 1 (min at $$x=0$$): $$\log(2)a = \pm 3$$, $$\log(2)b = -4$$ - Case 2 (min at $$x=4$$): $$\log(2)a = 7$$, $$\log(2)b = -44$$ or $$\log(2)a = 1$$, $$\log(2)b = 4$$ 10. **Find $$a$$ and $$b$$:** Recall $$\log(2)a = \log_2 a$$, so: $$a = 2^{\log(2)a}$$ and $$b = 2^{\log(2)b}$$. For example, for $$\log(2)a = 3$$, $$a = 2^3 = 8$$. **Final answers:** - If minimum at $$x=0$$: $$a = 2^{\pm 3} = 8 \text{ or } \frac{1}{8}$$, $$b = 2^{-4} = \frac{1}{16}$$ - If minimum at $$x=4$$: - $$a = 2^7 = 128$$, $$b = 2^{-44}$$ (very small) - $$a = 2^1 = 2$$, $$b = 2^4 = 16$$