1. The problem asks whether the Extreme Value Theorem (EVT) guarantees an absolute maximum and minimum for each function on the closed interval $[0,6]$. 2. The EVT states: If a function is continuous on a closed interval $[a,b]$, then it must have both an absolute maximum and an absolute minimum on that interval. 3. For each function, we check continuity on $[0,6]$: a. $f(x) = 8x^5 - 7x^2 + 3x - 4$ is a polynomial, which is continuous everywhere, so it is continuous on $[0,6]$. b. $g(x) = \frac{2x + 1}{x - 5}$ has a denominator zero at $x=5$, which lies in $[0,6]$, so $g$ is not continuous on $[0,6]$. c. $h(x) = \begin{cases} 2x + 2, & 0 \leq x < 3 \\ x^2, & 3 < x \leq 6 \end{cases}$ is piecewise defined with a jump at $x=3$ (since $\lim_{x \to 3^-} h(x) = 2(3)+2=8$ and $\lim_{x \to 3^+} h(x) = 3^2=9$), so $h$ is not continuous on $[0,6]$. 4. Conclusion: a. $f$ satisfies EVT conditions, so it has absolute max and min on $[0,6]$. b. $g$ is not continuous on $[0,6]$, so EVT does not guarantee absolute extrema. c. $h$ is not continuous on $[0,6]$, so EVT does not guarantee absolute extrema. --- 5. Next problem: Find critical points of $f(x) = x^3 + 10x^2 + 12x - 4$. 6. Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined. 7. Compute derivative: $$f'(x) = 3x^2 + 20x + 12$$ 8. Set derivative equal to zero: $$3x^2 + 20x + 12 = 0$$ 9. Divide entire equation by 3: $$\cancel{3}x^2 + \cancel{20}x + \cancel{12} = 0 \Rightarrow x^2 + \frac{20}{3}x + 4 = 0$$ 10. Use quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=\frac{20}{3}$, $c=4$: $$x = \frac{-\frac{20}{3} \pm \sqrt{\left(\frac{20}{3}\right)^2 - 4 \cdot 1 \cdot 4}}{2}$$ 11. Calculate discriminant: $$\left(\frac{20}{3}\right)^2 - 16 = \frac{400}{9} - 16 = \frac{400}{9} - \frac{144}{9} = \frac{256}{9}$$ 12. Square root of discriminant: $$\sqrt{\frac{256}{9}} = \frac{16}{3}$$ 13. Substitute back: $$x = \frac{-\frac{20}{3} \pm \frac{16}{3}}{2} = \frac{\frac{-20 \pm 16}{3}}{2} = \frac{-20 \pm 16}{6}$$ 14. Calculate both solutions: $$x_1 = \frac{-20 + 16}{6} = \frac{-4}{6} = -\frac{2}{3}$$ $$x_2 = \frac{-20 - 16}{6} = \frac{-36}{6} = -6$$ 15. Therefore, the critical points are at $x = -\frac{2}{3}$ and $x = -6$. --- Final answers: - EVT guarantees absolute extrema only for $f(x)$ in part 1. - Critical points of $f(x) = x^3 + 10x^2 + 12x - 4$ are $x = -\frac{2}{3}$ and $x = -6$.