The function `f(x)=e^(sinx+cosx)AA x in [0, 2pi]` attains local extrema at `x=alpha and x= beta,` then `alpha+beta` is equal to

Options

A. `pi`
B. `2pi`
C. `(3pi)/(2)`
D. `(pi)/(2)`

Answer

To solve the problem, we need to find the local extrema of the function \( f(x) = e^{\sin x + \cos x} \) on the interval \( [0, 2\pi] \) and then determine the sum \( \alpha + \beta \) where \( \alpha \) and \( \beta \) are the points of local extrema.

### Step 1: Find the derivative of \( f(x) \)

Using the chain rule, we differentiate \( f(x) \):

\[ f'(x) = e^{\sin x + \cos x} \cdot (\cos x - \sin x) \]

### Step 2: Set the derivative equal to zero

To find the critical points, we set the derivative equal to zero:

\[ e^{\sin x + \cos x} \cdot (\cos x - \sin x) = 0 \]

Since \( e^{\sin x + \cos x} \) is never zero, we focus on the term \( \cos x - \sin x = 0 \):

\[ \cos x = \sin x \]

### Step 3: Solve for \( x \)

The equation \( \cos x = \sin x \) implies:

\[ \tan x = 1 \]

The solutions for \( x \) in the interval \( [0, 2\pi] \) are:

\[ x = \frac{\pi}{4} \quad \text{and} \quad x = \frac{5\pi}{4} \]

### Step 4: Identify the local extrema

We have found the critical points \( \alpha = \frac{\pi}{4} \) and \( \beta = \frac{5\pi}{4} \). To determine which is a maximum and which is a minimum, we can use the second derivative test.

### Step 5: Find the second derivative \( f''(x) \)

To find \( f''(x) \), we differentiate \( f'(x) \):

Using the product rule:

\[ f''(x) = \frac{d}{dx} \left( e^{\sin x + \cos x} \cdot (\cos x - \sin x) \right) \]

This gives us:

\[ f''(x) = e^{\sin x + \cos x} \cdot (\cos x - \sin x)^2 + e^{\sin x + \cos x} \cdot (-\sin x - \cos x) \]

### Step 6: Evaluate the second derivative at critical points

1. For \( x = \frac{\pi}{4} \):

\[ f''\left(\frac{\pi}{4}\right) < 0 \quad \text{(indicating a local maximum)} \]

2. For \( x = \frac{5\pi}{4} \):

\[ f''\left(\frac{5\pi}{4}\right) > 0 \quad \text{(indicating a local minimum)} \]

### Step 7: Calculate \( \alpha + \beta \)

Now we can find \( \alpha + \beta \):

\[ \alpha + \beta = \frac{\pi}{4} + \frac{5\pi}{4} = \frac{6\pi}{4} = \frac{3\pi}{2} \]

### Final Answer

Thus, the value of \( \alpha + \beta \) is:

\[ \boxed{\frac{3\pi}{2}} \]