1. Problem: Determine if the function $y = h(x)$ has any absolute extreme values on $[a,b]$ from the given graph. Step 1: Theorem 1 states that a continuous function on a closed interval $[a,b]$ must have an absolute maximum and minimum. Step 2: From the graph, $h(x)$ has two local maxima at $c_1$ and $c_2$ and endpoints $a$ and $b$. Step 3: Check values at $a$, $b$, $c_1$, and $c_2$ to find absolute extrema. Step 4: Since the graph shows the highest point at $c_1$ and lowest at $a$, absolute max at $c_1$, absolute min at $a$. Answer: Absolute maximum at $x=c_1$, absolute minimum at $x=a$. 2. Problem: Determine absolute extrema of $y=f(x)$ on $[a,b]$. Step 1: The graph shows a single peak at $c$. Step 2: Check values at $a$, $b$, and $c$. Step 3: The highest value is at $c$, lowest at $a$ or $b$ depending on graph. Answer: Absolute maximum at $x=c$, absolute minimum at endpoint with lowest value. 3. Problem: Determine absolute extrema of $y=f(x)$ on $[a,b]$. Step 1: Graph shows a sharp peak at $c$. Step 2: Check values at $a$, $b$, and $c$. Step 3: Absolute max at $c$, absolute min at either $a$ or $b$. Answer: Absolute maximum at $x=c$, absolute minimum at endpoint with lowest value. 4. Problem: Determine absolute extrema of $y=h(x)$ on $[a,b]$. Step 1: Graph shows endpoints $a$ and $b$ as highest points. Step 2: Check values at $a$, $b$, and $c$. Step 3: Absolute maxima at $a$ and $b$, absolute minimum at $c$. Answer: Absolute maxima at $x=a$ and $x=b$, absolute minimum at $x=c$. 5. Problem: Determine absolute extrema of $y=g(x)$ on $[a,b]$. Step 1: Graph shows a peak at $a$ and decreasing to $b$. Step 2: Check values at $a$, $b$, and $c$. Step 3: Absolute maximum at $a$, absolute minimum at $b$. Answer: Absolute maximum at $x=a$, absolute minimum at $x=b$. 6. Problem: Determine absolute extrema of $y=g(x)$ on $[a,b]$. Step 1: Graph shows maximum at $a$ and decreasing to $b$. Step 2: Check values at $a$, $b$, and $c$. Step 3: Absolute maximum at $a$, absolute minimum at $b$. Answer: Absolute maximum at $x=a$, absolute minimum at $x=b$. 7. Problem: Find absolute max and min of $f(x) = \frac{2}{3}x - 5$ on $[-2,3]$. Step 1: Evaluate at endpoints: $f(-2) = \frac{2}{3}(-2) - 5 = -\frac{4}{3} - 5 = -\frac{19}{3} \approx -6.33$ $f(3) = \frac{2}{3}(3) - 5 = 2 - 5 = -3$ Step 2: Since $f$ is linear and increasing, min at $x=-2$, max at $x=3$. Answer: Absolute minimum $-6.33$ at $x=-2$, absolute maximum $-3$ at $x=3$. 8. Problem: $f(x) = -x - 4$ on $[-4,1]$. Step 1: Evaluate at endpoints: $f(-4) = -(-4) - 4 = 4 - 4 = 0$ $f(1) = -1 - 4 = -5$ Step 2: $f$ is decreasing, so max at $x=-4$, min at $x=1$. Answer: Absolute max $0$ at $x=-4$, absolute min $-5$ at $x=1$. 9. Problem: $f(x) = x^2 - 1$ on $[-1,2]$. Step 1: Evaluate at endpoints and critical points. Derivative $f'(x) = 2x$; critical point at $x=0$. $f(-1) = 1 - 1 = 0$ $f(0) = 0 - 1 = -1$ $f(2) = 4 - 1 = 3$ Step 2: Absolute min at $x=0$, absolute max at $x=2$. Answer: Absolute minimum $-1$ at $x=0$, absolute maximum $3$ at $x=2$. 10. Problem: $f(x) = 4 - x^2$ on $[-3,1]$. Step 1: Derivative $f'(x) = -2x$, critical point at $x=0$. $f(-3) = 4 - 9 = -5$ $f(0) = 4 - 0 = 4$ $f(1) = 4 - 1 = 3$ Step 2: Absolute max at $x=0$, absolute min at $x=-3$. Answer: Absolute maximum $4$ at $x=0$, absolute minimum $-5$ at $x=-3$. 11. Problem: $F(x) = -\frac{1}{x^2}$ on $[0.5,2]$. Step 1: $F$ is negative and increasing on $(0,\infty)$. Evaluate at endpoints: $F(0.5) = -\frac{1}{0.25} = -4$ $F(2) = -\frac{1}{4} = -0.25$ Step 2: Absolute min at $x=0.5$, absolute max at $x=2$. Answer: Absolute minimum $-4$ at $x=0.5$, absolute maximum $-0.25$ at $x=2$. 12. Problem: $F(x) = -\frac{1}{x}$ on $[-2,-1]$. Step 1: Evaluate at endpoints: $F(-2) = -\frac{1}{-2} = 0.5$ $F(-1) = -\frac{1}{-1} = 1$ Step 2: $F$ is increasing on negative domain. Answer: Absolute minimum $0.5$ at $x=-2$, absolute maximum $1$ at $x=-1$. 13. Problem: $h(x) = \sqrt{x}$ on $[-1,8]$. Step 1: Domain of $h$ is $[0,\infty)$, so $h$ undefined for $x<0$. Step 2: On $[0,8]$, evaluate: $h(0) = 0$ $h(8) = \sqrt{8} = 2\sqrt{2} \approx 2.83$ Step 3: Absolute min at $0$, max at $8$. Answer: Absolute minimum $0$ at $x=0$, absolute maximum $2.83$ at $x=8$. 14. Problem: $h(x) = -3x^{2/3}$ on $[-1,1]$. Step 1: Evaluate at endpoints and critical points. Derivative $h'(x) = -3 \cdot \frac{2}{3} x^{-1/3} = -2 x^{-1/3}$ undefined at $x=0$. Evaluate: $h(-1) = -3(-1)^{2/3} = -3(1) = -3$ $h(0) = 0$ $h(1) = -3(1) = -3$ Step 2: Absolute max at $x=0$, absolute min at $x=\pm 1$. Answer: Absolute maximum $0$ at $x=0$, absolute minimum $-3$ at $x=\pm 1$. 15. Problem: $g(x) = \sqrt{4 - x^2}$ on $[-2,1]$. Step 1: Evaluate at endpoints and critical points. Domain: $|x| \leq 2$. Derivative $g'(x) = \frac{-x}{\sqrt{4 - x^2}}$. Critical point at $x=0$. Evaluate: $g(-2) = 0$ $g(0) = 2$ $g(1) = \sqrt{3} \approx 1.732$ Step 2: Absolute max at $x=0$, absolute min at $x=-2$. Answer: Absolute maximum $2$ at $x=0$, absolute minimum $0$ at $x=-2$. 16. Problem: $g(x) = -\sqrt{5 - x^2}$ on $[-\sqrt{5},0]$. Step 1: Evaluate at endpoints and critical points. Derivative $g'(x) = - \frac{-x}{\sqrt{5 - x^2}} = \frac{x}{\sqrt{5 - x^2}}$. Critical point at $x=0$. Evaluate: $g(-\sqrt{5}) = -0 = 0$ $g(0) = -\sqrt{5} \approx -2.236$ Step 2: Absolute max at $x=-\sqrt{5}$, absolute min at $x=0$. Answer: Absolute maximum $0$ at $x=-\sqrt{5}$, absolute minimum $-2.236$ at $x=0$. 17. Problem: $f(\theta) = \sin \theta$ on $[-\frac{\pi}{2}, \frac{5\pi}{6}]$. Step 1: Evaluate at endpoints and critical points. Derivative $f'(\theta) = \cos \theta$. Critical points where $\cos \theta = 0$: $\theta = \frac{\pi}{2}$ in interval. Evaluate: $f(-\frac{\pi}{2}) = -1$ $f(\frac{\pi}{2}) = 1$ $f(\frac{5\pi}{6}) = \sin(150^\circ) = \frac{1}{2}$ Step 2: Absolute max at $\frac{\pi}{2}$, absolute min at $-\frac{\pi}{2}$. Answer: Absolute maximum $1$ at $\theta=\frac{\pi}{2}$, absolute minimum $-1$ at $\theta=-\frac{\pi}{2}$. 18. Problem: $f(\theta) = \tan \theta$ on $[-\frac{\pi}{3}, \frac{\pi}{4}]$. Step 1: Evaluate at endpoints: $f(-\frac{\pi}{3}) = -\sqrt{3} \approx -1.732$ $f(\frac{\pi}{4}) = 1$ Step 2: $\tan \theta$ is increasing on this interval. Answer: Absolute minimum $-1.732$ at $\theta=-\frac{\pi}{3}$, absolute maximum $1$ at $\theta=\frac{\pi}{4}$. 19. Problem: $g(x) = \csc x$ on $[\frac{\pi}{3}, \frac{2\pi}{3}]$. Step 1: $\csc x = \frac{1}{\sin x}$. Evaluate at endpoints: $g(\frac{\pi}{3}) = \frac{1}{\sin(60^\circ)} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} \approx 1.155$ $g(\frac{2\pi}{3}) = \frac{1}{\sin(120^\circ)} = \frac{1}{\sqrt{3}/2} = 1.155$ Step 2: Minimum at $\frac{\pi}{2}$ where $\csc x = 1$. Answer: Absolute minimum $1$ at $x=\frac{\pi}{2}$, absolute maxima $1.155$ at endpoints. 20. Problem: $g(x) = \sec x$ on $[-\frac{\pi}{3}, \frac{\pi}{6}]$. Step 1: $\sec x = \frac{1}{\cos x}$. Evaluate at endpoints: $g(-\frac{\pi}{3}) = \frac{1}{\cos(-60^\circ)} = \frac{1}{0.5} = 2$ $g(\frac{\pi}{6}) = \frac{1}{\cos(30^\circ)} = \frac{1}{\sqrt{3}/2} = 1.155$ Step 2: Minimum at $x=0$ where $\sec 0 = 1$. Answer: Absolute minimum $1$ at $x=0$, absolute maximum $2$ at $x=-\frac{\pi}{3}$. 21. Problem: $f(t) = 2 - |t|$ on $[-1,3]$. Step 1: Evaluate at endpoints and critical point $t=0$. $f(-1) = 2 - 1 = 1$ $f(0) = 2 - 0 = 2$ $f(3) = 2 - 3 = -1$ Step 2: Absolute max at $t=0$, absolute min at $t=3$. Answer: Absolute maximum $2$ at $t=0$, absolute minimum $-1$ at $t=3$. 22. Problem: $f(t) = |t - 5|$ on $[4,7]$. Step 1: Evaluate at endpoints and critical point $t=5$. $f(4) = |4 - 5| = 1$ $f(5) = 0$ $f(7) = 2$ Step 2: Absolute min at $t=5$, absolute max at $t=7$. Answer: Absolute minimum $0$ at $t=5$, absolute maximum $2$ at $t=7$. 23. Problem: $f(x) = x^{4/3}$ on $[-1,8]$. Step 1: Derivative $f'(x) = \frac{4}{3} x^{1/3}$. Critical point at $x=0$. Evaluate: $f(-1) = (-1)^{4/3} = 1$ $f(0) = 0$ $f(8) = 8^{4/3} = (8^{1/3})^4 = 2^4 = 16$ Step 2: Absolute min at $x=0$, absolute max at $x=8$. Answer: Absolute minimum $0$ at $x=0$, absolute maximum $16$ at $x=8$. 24. Problem: $f(x) = x^{5/3}$ on $[-1,8]$. Step 1: Derivative $f'(x) = \frac{5}{3} x^{2/3}$. Critical point at $x=0$. Evaluate: $f(-1) = (-1)^{5/3} = -1$ $f(0) = 0$ $f(8) = 8^{5/3} = (8^{1/3})^5 = 2^5 = 32$ Step 2: Absolute min at $x=-1$, absolute max at $x=8$. Answer: Absolute minimum $-1$ at $x=-1$, absolute maximum $32$ at $x=8$. 25. Problem: $g(\theta) = \theta^{3/5}$ on $[-32,1]$. Step 1: Derivative $g'(\theta) = \frac{3}{5} \theta^{-2/5}$ undefined at $\theta=0$. Evaluate: $g(-32) = (-32)^{3/5} = -8$ $g(0) = 0$ $g(1) = 1$ Step 2: Absolute min at $-32$, absolute max at $1$. Answer: Absolute minimum $-8$ at $\theta=-32$, absolute maximum $1$ at $\theta=1$. 26. Problem: $h(\theta) = 3 \theta^{2/3}$ on $[-27,8]$. Step 1: Derivative $h'(\theta) = 3 \cdot \frac{2}{3} \theta^{-1/3} = 2 \theta^{-1/3}$ undefined at $\theta=0$. Evaluate: $h(-27) = 3 (-27)^{2/3} = 3 (9) = 27$ $h(0) = 0$ $h(8) = 3 (8^{2/3}) = 3 (4) = 12$ Step 2: Absolute max at $x=-27$, absolute min at $x=0$. Answer: Absolute maximum $27$ at $\theta=-27$, absolute minimum $0$ at $\theta=0$.