Consider the following. (If an answer does not exist, enter DNE.)

*f*(*x*) = 7 sin(*x*) + 7 cos(*x*),    0 ≤ *x* ≤ 2𝜋

(a)

Find the interval(s) on which *f* is increasing. (Enter your answer using interval notation.)
 
(b)

Find the interval(s) on which *f* is decreasing. (Enter your answer using interval notation.)
 
(c)

Find the local minimum and maximum values of *f*.

To analyze the function \( f(x) = 7\sin(x) + 7\cos(x) \) over the interval \( 0 \leq x \leq 2\pi \), let's proceed step by step.

---

### Step 1: Derivative of \( f(x) \)
To determine intervals of increase and decrease, find the derivative:
\[
f'(x) = 7\cos(x) - 7\sin(x).
\]

---

### Step 2: Critical Points
Set \( f'(x) = 0 \) to find critical points:
\[
7\cos(x) - 7\sin(x) = 0
\]
\[
\cos(x) = \sin(x).
\]
This occurs when:
\[
x = \frac{\pi}{4}, \frac{5\pi}{4}.
\]

---

### Step 3: Sign of \( f'(x) \)
Determine the sign of \( f'(x) \) in each interval created by the critical points \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \):
1. **Interval \( (0, \frac{\pi}{4}) \):**
   - Choose \( x = \frac{\pi}{8} \):
   \[
   f'\left(\frac{\pi}{8}\right) = 7\cos\left(\frac{\pi}{8}\right) - 7\sin\left(\frac{\pi}{8}\right).
   \]
   Since \( \cos(x) > \sin(x) \) in this range, \( f'(x) > 0 \).

2. **Interval \( (\frac{\pi}{4}, \frac{5\pi}{4}) \):**
   - Choose \( x = \pi \):
   \[
   f'(\pi) = 7\cos(\pi) - 7\sin(\pi) = -7 - 0 = -7.
   \]
   \( f'(x) < 0 \) in this interval.

3. **Interval \( (\frac{5\pi}{4}, 2\pi) \):**
   - Choose \( x = \frac{7\pi}{4} \):
   \[
   f'\left(\frac{7\pi}{4}\right) = 7\cos\left(\frac{7\pi}{4}\right) - 7\sin\left(\frac{7\pi}{4}\right).
   \]
   Since \( \cos(x) > \sin(x) \) in this range, \( f'(x) > 0 \).

Thus, \( f'(x) > 0 \) in \( (0, \frac{\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi) \) and \( f'(x) < 0 \) in \( (\frac{\pi}{4}, \frac{5\pi}{4}) \).

---

### Step 4: Local Maxima and Minima
- At \( x = \frac{\pi}{4} \), \( f'(x) \) changes from positive to negative, so this is a **local maximum**.
- At \( x = \frac{5\pi}{4} \), \( f'(x) \) changes from negative to positive, so this is a **local minimum**.

To find these values, evaluate \( f(x) \):
\[
f\left(\frac{\pi}{4}\right) = 7\sin\left(\frac{\pi}{4}\right) + 7\cos\left(\frac{\pi}{4}\right) = 7\sqrt{2}/2 + 7\sqrt{2}/2 = 7\sqrt{2}.
\]
\[
f\left(\frac{5\pi}{4}\right) = 7\sin\left(\frac{5\pi}{4}\right) + 7\cos\left(\frac{5\pi}{4}\right) = -7\sqrt{2}/2 - 7\sqrt{2}/2 = -7\sqrt{2}.
\]

---

### Final Results

**(a)** \( f(x) \) is increasing on:
\[
(0, \frac{\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi).
\]

**(b)** \( f(x) \) is decreasing on:
\[
(\frac{\pi}{4}, \frac{5\pi}{4}).
\]

**(c)** Local extrema:
- Local maximum: \( 7\sqrt{2} \) at \( x = \frac{\pi}{4} \).
- Local minimum: \( -7\sqrt{2} \) at \( x = \frac{5\pi}{4} \).

---

Let me know if you'd like further details or clarifications!

**Follow-up questions:**
1. How do you find the critical points of a function?
2. Why is the derivative useful for identifying increasing and decreasing intervals?
3. What is the significance of the trigonometric identity \( \cos(x) = \sin(x) \)?
4. How can we evaluate trigonometric functions at key angles like \( \frac{\pi}{4} \)?
5. Can a function have more than one local maximum or minimum?

**Tip:** Always sketch the graph of a function when analyzing intervals—it can provide valuable insight!

Consider the following. (If an answer does not exist, enter DNE.)

f(x) = 7 sin(x) + 7 cos(x), 0 ≤ x ≤ 2π

(a) Find the interval(s) on which f is increasing. (Enter your answer using interval notation.)
(b) Find the interval(s) on which f is decreasing. (Enter your answer using interval notation.)
(c) Find the local minimum and maximum values of f.

Explore the detailed solution to finding intervals of increase and decrease and determining local extrema for f(x) = 7sin(x) + 7cos(x) on [0, 2π]. Learn about critical points, derivative analysis, and trigonometric applications, suitable for advanced high school or college-level calculus.

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