To solve these problems, we will analyze each function within the specified interval to find critical points and determine absolute minima and maxima.

(a) $f(x) = \sin(x) + \cos(x)$, interval: $[0, 2\pi]$

1. Find the derivative: $f'(x) = \cos(x) - \sin(x)$

2. Set the derivative to zero to find critical points: $\cos(x) - \sin(x) = 0$ \tan(x) = 1 \Rightarrow x = \frac{\pi}{4}, \frac{5\pi}{4} \quad \text{(within the interval [0, 2\pi])}

3. Evaluate $f(x)$ at the endpoints and critical points:

   • $f(0) = \sin(0) + \cos(0) = 1$
   • $f(2\pi) = \sin(2\pi) + \cos(2\pi) = 1$
   • $f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$
   • $f\left(\frac{5\pi}{4}\right) = \sin\left(\frac{5\pi}{4}\right) + \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}$

4. Determine absolute extrema:

   • Absolute maximum: $\sqrt{2}$ at $x = \frac{\pi}{4}$
   • Absolute minimum: $-\sqrt{2}$ at $x = \frac{5\pi}{4}$

(b) $f(x) = 3x\sqrt{1 - x^2}$, interval: $[-1, 1]$

1. Find the derivative (using the product rule and chain rule): $f'(x) = 3\sqrt{1 - x^2} - \frac{3x^2}{\sqrt{1 - x^2}}$

2. Set the derivative to zero to find critical points: Simplifying $f'(x) = 0$: $3\sqrt{1 - x^2} = \frac{3x^2}{\sqrt{1 - x^2}}$ $3(1 - x^2) = 3x^2$ $1 = 2x^2 \Rightarrow x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$

3. Evaluate $f(x)$ at the endpoints and critical points:

   • $f(-1) = 3(-1)\sqrt{1 - (-1)^2} = 0$
   • $f(1) = 3(1)\sqrt{1 - 1^2} = 0$
   • $f\left(\frac{\sqrt{2}}{2}\right) = 3 \cdot \frac{\sqrt{2}}{2} \cdot \sqrt{1 - \left(\frac{\sqrt{2}}{2}\right)^2} = \frac{3\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{3}{2}$
   • $f\left(-\frac{\sqrt{2}}{2}\right) = -\frac{3}{2}$

4. Determine absolute extrema:

   • Absolute maximum: $\frac{3}{2}$ at $x = \frac{\sqrt{2}}{2}$
   • Absolute minimum: $-\frac{3}{2}$ at $x = -\frac{\sqrt{2}}{2}$

(c) $g(x) = x^2 - \frac{1}{x}$, interval: $[1, 3]$

1. Find the derivative: $g'(x) = 2x + \frac{1}{x^2}$

2. Set the derivative to zero to find critical points: $2x + \frac{1}{x^2} = 0$ Multiply through by $x^2$: $2x^3 + 1 = 0 \Rightarrow x^3 = -\frac{1}{2}$ This does not have any real solutions within the interval $[1, 3]$.

3. Evaluate $g(x)$ at the endpoints:

   • $g(1) = 1^2 - \frac{1}{1} = 0$
   • $g(3) = 3^2 - \frac{1}{3} = 9 - \frac{1}{3} = \frac{26}{3}$

4. Determine absolute extrema:

   • Absolute maximum: $\frac{26}{3}$ at $x = 3$
   • Absolute minimum: $0$ at $x = 1$

Would you like further clarification on any of these calculations?

Here are five follow-up questions that may help expand your understanding:

1. How do we interpret critical points outside the given interval when analyzing extrema?
2. What role do endpoints play in determining absolute extrema for closed intervals?
3. How would we handle a similar problem if the interval were open instead of closed?
4. Why do some functions lack real critical points within their intervals?
5. How do chain and product rules apply in derivative calculations for complex functions?

Tip: Always remember to evaluate the function at both the critical points and endpoints when seeking absolute extrema on a closed interval.