Let f(x) = \sin(x)+\cos^2(x). Find all points of absolute minima and maxima on the interval [0,\pi]. As your answer, input the sum of the values of f at all points the absolute minima and maxima occur in the given interval.

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.

Learn how to find the points of absolute minima and maxima for the function f(x) = sin(x) + cos^2(x) on the interval [0,π]. This detailed step-by-step guide covers differentiation using the chain rule, solving for critical points, and evaluating function values at critical points and endpoints. Suitable for calculus students.

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