Question

Find the intervals on which \( f(x) \) is increasing or decreasing for \( x \in [0, \pi] \).

Answer 1

To determine increasing or decreasing intervals, we first find the first derivative \( f'(x) \). 

Step 1: Compute \( f'(x) \) using the product rule. 
Given: \[ f(x) = e^x \sin x. \] Using the product rule: \[ f'(x) = \frac{d}{dx} (e^x \sin x) = e^x \frac{d}{dx} (\sin x) + \sin x \frac{d}{dx} (e^x). \] \[ f'(x) = e^x \cos x + e^x \sin x. \] \[ f'(x) = e^x (\cos x + \sin x). \] 

Step 2: Find the critical points. 
To find critical points, set \( f'(x) = 0 \): \[ e^x (\cos x + \sin x) = 0. \] Since \( e^x>0 \) for all \( x \), we set: \[ \cos x + \sin x = 0. \] Dividing both sides by \( \cos x \): \[ 1 + \tan x = 0. \] \[ \tan x = -1. \] Solving in \( [0, \pi] \): \[ x = \frac{3\pi}{4}. \] 

Step 3: Determine sign changes in \( f'(x) \). 
For \( x \in [0, \pi] \), check the sign of \( f'(x) \) in the intervals: 
1. \( (0, \frac{3\pi}{4}) \): Choose \( x = \frac{\pi}{2} \). \[ \cos \frac{\pi}{2} + \sin \frac{\pi}{2} = 0 + 1 = 1>0. \] So, \( f'(x)>0 \), meaning \( f(x) \) is increasing. 
2. \( (\frac{3\pi}{4}, \pi) \): Choose \( x = \pi \). \[ \cos \pi + \sin \pi = -1 + 0 = -1<0. \] So, \( f'(x)<0 \), meaning \( f(x) \) is decreasing. 

Final Answer: 
- \( f(x) \) is increasing for \( x \in (0, \frac{3\pi}{4}) \).
- \( f(x) \) is decreasing for \( x \in (\frac{3\pi}{4}, \pi) \).

Answer 2

Step 1: Compute the second derivative \( f''(x) \).
Using the derivative \( f'(x) = e^x (\cos x + \sin x) \), apply the product rule again: \[ f''(x) = e^x (\cos x + \sin x) + e^x (\cos x - \sin x). \] \[ = e^x [(\cos x + \sin x) + (\cos x - \sin x)]. \] \[ = e^x (2\cos x). \] Step 2: Evaluate \( f''(x) \) at the critical point \( x = \frac{3\pi}{4} \).
\[ f''\left(\frac{3\pi}{4}\right) = e^{3\pi/4} (2\cos (3\pi/4)). \] Since: \[ \cos(3\pi/4) = -\frac{1}{\sqrt{2}}, \] \[ f''(3\pi/4) = e^{3\pi/4} \left(2 \times -\frac{1}{\sqrt{2}}\right). \] \[ = -e^{3\pi/4} \sqrt{2}. \] Since \( f''(3\pi/4)<0 \), the function has a local maximum at \( x = \frac{3\pi}{4} \). Final Answer:
The critical point \( x = \frac{3\pi}{4} \) is a local maximum.