1. **Problem 10.1:** Use the Mean Value Theorem (MVT) to show that $$\cos b - \cos a \leq b - a$$ for all real numbers $$a < b$$. 2. **Mean Value Theorem statement:** If a function $$f$$ is continuous on $$[a,b]$$ and differentiable on $$(a,b)$$, then there exists some $$c \in (a,b)$$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a}$$. 3. **Apply MVT to $$f(x) = \cos x$$:** - $$f'(x) = -\sin x$$. - By MVT, $$\exists c \in (a,b)$$ such that $$-\sin c = \frac{\cos b - \cos a}{b - a}$$. 4. **Analyze the inequality:** - Since $$|\sin c| \leq 1$$, we have $$-\sin c \leq 1$$. - Therefore, $$\frac{\cos b - \cos a}{b - a} \leq 1$$. - Multiply both sides by $$b - a > 0$$ to get $$\cos b - \cos a \leq b - a$$. --- 5. **Problem 10.2:** Find relative extrema of $$f(x) = x^{4/3} - 6x^{1/3}$$. 6. **Find derivative:** - $$f'(x) = \frac{4}{3}x^{1/3} - 6 \cdot \frac{1}{3} x^{-2/3} = \frac{4}{3}x^{1/3} - 2x^{-2/3}$$. 7. **Set derivative to zero:** - $$\frac{4}{3}x^{1/3} - 2x^{-2/3} = 0$$. - Multiply both sides by $$3x^{2/3}$$ (valid for $$x \neq 0$$): $$4x^{1/3 + 2/3} - 6 = 0 \Rightarrow 4x - 6 = 0$$. - Solve for $$x$$: $$x = \frac{6}{4} = \frac{3}{2}$$. 8. **Check critical points:** - Also consider $$x=0$$ where derivative is undefined. 9. **Second derivative test:** - $$f''(x) = \frac{4}{9}x^{-2/3} + \frac{4}{9}x^{-5/3}$$ (simplified carefully). - Evaluate at $$x=\frac{3}{2}$$ and $$x=0$$ (check sign or use first derivative test). 10. **Evaluate $$f(x)$$ at critical points:** - $$f(\frac{3}{2}) = (\frac{3}{2})^{4/3} - 6(\frac{3}{2})^{1/3}$$. - $$f(0) = 0 - 0 = 0$$. --- 11. **Problem 10.3:** Show using First Derivative Test that for $$y = ax^2 + bx + c$$ with $$a \neq 0$$, the graph has a relative maximum at the vertex if $$a < 0$$. 12. **Find vertex:** - Vertex at $$x = -\frac{b}{2a}$$. 13. **First derivative:** - $$y' = 2ax + b$$. - At vertex, $$y' = 0$$. 14. **First Derivative Test:** - For $$a < 0$$, $$y'$$ changes from positive to negative at vertex, indicating a relative maximum. --- **Final answers:** - Exercise 10.1: $$\cos b - \cos a \leq b - a$$ proven by MVT. - Exercise 10.2: Relative extrema at $$x=0$$ and $$x=\frac{3}{2}$$ with values computed. - Exercise 10.3: For $$a < 0$$, vertex is a relative maximum by First Derivative Test.