We are given the function $f(x) = \sin(x) + \cos^2(x)$ and asked to find the sum of the values of $f(x)$ at all points of absolute minima and maxima on the interval $[0, \pi]$.

Step 1: Find the derivative of $f(x)$

To locate the critical points, we first find the derivative of $f(x)$: $f'(x) = \frac{d}{dx} \left( \sin(x) + \cos^2(x) \right)$ Using the chain rule for $\cos^2(x)$: $f'(x) = \cos(x) + 2\cos(x)(-\sin(x)) = \cos(x) - 2\cos(x)\sin(x)$ Thus, the derivative simplifies to: $f'(x) = \cos(x)(1 - 2\sin(x))$

Step 2: Set $f'(x) = 0$

To find the critical points, we solve: $\cos(x)(1 - 2\sin(x)) = 0$ This gives two cases to consider:

1. $\cos(x) = 0$
2. $1 - 2\sin(x) = 0$

Case 1: $\cos(x) = 0$

In the interval $[0, \pi]$, $\cos(x) = 0$ at $x = \frac{\pi}{2}$.

Case 2: $1 - 2\sin(x) = 0$

Solving for $\sin(x)$: $1 - 2\sin(x) = 0 \implies \sin(x) = \frac{1}{2}$ In the interval $[0, \pi]$, $\sin(x) = \frac{1}{2}$ occurs at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.

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

We now evaluate $f(x) = \sin(x) + \cos^2(x)$ at the critical points $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$ and at the endpoints $x = 0$ and $x = \pi$.

• At $x = 0$: $f(0) = \sin(0) + \cos^2(0) = 0 + 1 = 1$

• At $x = \pi$: $f(\pi) = \sin(\pi) + \cos^2(\pi) = 0 + 1 = 1$

• At $x = \frac{\pi}{6}$: $f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) + \cos^2\left(\frac{\pi}{6}\right) = \frac{1}{2} + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} + \frac{3}{4} = \frac{5}{4}$

• At $x = \frac{\pi}{2}$: $f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \cos^2\left(\frac{\pi}{2}\right) = 1 + 0 = 1$

• At $x = \frac{5\pi}{6}$: $f\left(\frac{5\pi}{6}\right) = \sin\left(\frac{5\pi}{6}\right) + \cos^2\left(\frac{5\pi}{6}\right) = \frac{1}{2} + \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} + \frac{3}{4} = \frac{5}{4}$

Step 4: Determine the absolute minima and maxima

• The values of $f(x)$ are: $f(0) = 1$, $f(\pi) = 1$, $f\left(\frac{\pi}{6}\right) = \frac{5}{4}$, $f\left(\frac{\pi}{2}\right) = 1$, and $f\left(\frac{5\pi}{6}\right) = \frac{5}{4}$.
• The absolute maximum occurs at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ with $f(x) = \frac{5}{4}$.
• The absolute minimum occurs at $x = 0, \frac{\pi}{2}, \pi$ with $f(x) = 1$.

Step 5: Calculate the sum of the values of $f(x)$

The sum of the values of $f(x)$ at the absolute minima and maxima is: $1 + 1 + 1 + \frac{5}{4} + \frac{5}{4} = 3 + \frac{10}{4} = 3 + 2.5 = 5.5$

Thus, the sum of the values of $f(x)$ at all points of absolute minima and maxima is $\boxed{5.5}$.

Would you like more details on any step?

Here are 5 related questions:

1. How do you determine the critical points of a function?
2. What is the significance of evaluating endpoints in optimization problems?
3. How do trigonometric identities simplify differentiation in calculus?
4. Why are both $\cos(x) = 0$ and $1 - 2\sin(x) = 0$ considered when solving $f'(x) = 0$?
5. What are some real-world applications of finding absolute minima and maxima in trigonometric functions?

Tip: Always check both the critical points and endpoints when finding absolute extrema on a closed interval.