Find the values of x where function `f(X)m = (sin x + cosx)(e^(x))` in `(0,2pi)` has point of inflection

Explanation

To find the values of \( x \) where the function \( f(x) = (\sin x + \cos x)e^x \) has points of inflection in the interval \( (0, 2\pi) \), we will follow these steps:

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

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

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

Applying the product rule:

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

Calculating the derivatives:

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

Combining the terms:

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

Simplifying this gives:

\[ f'(x) = e^x[2\cos x] \]

### Step 2: Find the second derivative \( f''(x) \)

Now, we differentiate \( f'(x) \):

\[ f''(x) = \frac{d}{dx}[e^x(2\cos x)] \]

Using the product rule again:

\[ f''(x) = e^x \frac{d}{dx}[2\cos x] + 2\cos x \frac{d}{dx}[e^x] \]

Calculating the derivatives:

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

Factoring out \( e^x \):

\[ f''(x) = e^x[2\cos x - 2\sin x] \]

This simplifies to:

\[ f''(x) = 2e^x(\cos x - \sin x) \]

### Step 3: Set the second derivative to zero

To find points of inflection, we set \( f''(x) = 0 \):

\[ 2e^x(\cos x - \sin x) = 0 \]

Since \( e^x \) is never zero, we have:

\[ \cos x - \sin x = 0 \]

This implies:

\[ \cos x = \sin x \]

### Step 4: Solve for \( x \)

Dividing both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)) gives:

\[ 1 = \tan x \]

The solutions for \( \tan x = 1 \) in the interval \( (0, 2\pi) \) are:

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

### Conclusion

Thus, the values of \( x \) where the function has points of inflection in the interval \( (0, 2\pi) \) are:

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

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