Consider the function on the interval \( (0,2 \pi) \). Use a graphing utility to confirm your answers for parts (a) and (b). \[ f(x)=\sin (x)+\cos (x) \] (a) Find the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation.) increasing \( \square \) decreasing \( \square \) (b) Apply the First Derivative Test to identify the relative extrema. (If an answer does not exist, enter DNE.) relative maximum \( \quad(x, y)=(\square) \) \( \square \) relative minimum \( \quad(x, y)=(\square) \)

**Step 1. Compute the derivative** The function is \[ f(x)=\sin x+\cos x. \] Differentiate with respect to \(x\): \[ f'(x)=\cos x-\sin x. \] **Step 2. Find the critical points** Set the derivative equal to zero: \[ \cos x-\sin x=0 \quad \Longrightarrow \quad \cos x=\sin x. \] Dividing both sides by \(\cos x\) (assuming \(\cos x\neq 0\)) gives: \[ 1=\tan x \quad \Longrightarrow \quad \tan x=1. \] The solutions for \(\tan x=1\) are given by \[ x=\frac{\pi}{4}+n\pi, \quad n\in \mathbb{Z}. \] On the interval \((0,2\pi)\), the solutions are: \[ x=\frac{\pi}{4} \quad \text{and} \quad x=\frac{5\pi}{4}. \] **Step 3. Determine the intervals on which \(f(x)\) is increasing or decreasing** The critical points divide the interval \((0,2\pi)\) into three subintervals: - Interval I: \((0,\frac{\pi}{4})\) - Interval II: \(\left(\frac{\pi}{4},\frac{5\pi}{4}\right)\) - Interval III: \(\left(\frac{5\pi}{4},2\pi\right)\) Test the sign of \(f'(x)=\cos x-\sin x\) in each interval. 1. **Interval I: \((0,\frac{\pi}{4})\)** Choose \(x=\frac{\pi}{8}\). \[ \cos\left(\frac{\pi}{8}\right) > \sin\left(\frac{\pi}{8}\right), \quad \text{so} \quad f'\left(\frac{\pi}{8}\right)>0. \] Thus, \(f(x)\) is increasing on \(\left(0,\frac{\pi}{4}\right)\). 2. **Interval II: \(\left(\frac{\pi}{4},\frac{5\pi}{4}\right)\)** Choose \(x=\pi\). \[ \cos \pi=-1 \quad \text{and} \quad \sin \pi=0, \quad \text{thus} \quad f'(\pi)=-1-0=-1<0. \] Thus, \(f(x)\) is decreasing on \(\left(\frac{\pi}{4},\frac{5\pi}{4}\right)\). 3. **Interval III: \(\left(\frac{5\pi}{4},2\pi\right)\)** Choose \(x=\frac{3\pi}{2}\). \[ \cos\left(\frac{3\pi}{2}\right)=0 \quad \text{and} \quad \sin\left(\frac{3\pi}{2}\right)=-1, \quad \text{so} \quad f'\left(\frac{3\pi}{2}\right)=0-(-1)=1>0. \] Thus, \(f(x)\) is increasing on \(\left(\frac{5\pi}{4},2\pi\right)\). **Answer for Part (a):** - Increasing on: \(\left(0,\frac{\pi}{4}\right)\cup\left(\frac{5\pi}{4},2\pi\right)\) - Decreasing on: \(\left(\frac{\pi}{4},\frac{5\pi}{4}\right)\) **Step 4. Apply the First Derivative Test to identify the relative extrema** - At \(x=\frac{\pi}{4}\): \(f'(x)\) changes from positive (increasing) to negative (decreasing), indicating a **relative maximum**. - At \(x=\frac{5\pi}{4}\): \(f'(x)\) changes from negative (decreasing) to positive (increasing), indicating a **relative minimum**. Now, compute the function values at the critical points. 1. **At \(x=\frac{\pi}{4}\):** \[ f\left(\frac{\pi}{4}\right)=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}. \] So, the relative maximum is at \(\left(\frac{\pi}{4},\sqrt{2}\right)\). 2. **At \(x=\frac{5\pi}{4}\):** \[ f\left(\frac{5\pi}{4}\right)=\sin\frac{5\pi}{4}+\cos\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=-\sqrt{2}. \] So, the relative minimum is at \(\left(\frac{5\pi}{4},-\sqrt{2}\right)\). **Answer for Part (b):** - Relative maximum: \(\left(\frac{\pi}{4}, \sqrt{2}\right)\) - Relative minimum: \(\left(\frac{5\pi}{4}, -\sqrt{2}\right)\)