Find the points of local maxima and local minima, if any, of the following functions. Find also the local maximum and local minimum values :
`f(x)=sinx+cosx, 0 lt xlt(pi)/(2)`

Answer

To find the local maxima and minima of the function \( f(x) = \sin x + \cos x \) in the interval \( 0 < x < \frac{\pi}{2} \), we will follow these steps:

### Step 1: Find the derivative of the function
The first step is to find the derivative of the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x \]

### Step 2: Set the derivative to zero to find critical points
Next, we set the derivative equal to zero to find the critical points:
\[ f'(x) = 0 \implies \cos x - \sin x = 0 \]
This simplifies to:
\[ \cos x = \sin x \]
Dividing both sides by \( \cos x \) (valid since \( \cos x \neq 0 \) in the interval \( 0 < x < \frac{\pi}{2} \)):
\[ 1 = \tan x \]
Thus, we find:
\[ \tan x = 1 \implies x = \frac{\pi}{4} \]

### Step 3: Determine the nature of the critical point
To determine whether this critical point is a local maximum or minimum, we will find the second derivative of the function:
\[ f''(x) = \frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x \]
Now, we evaluate the second derivative at the critical point \( x = \frac{\pi}{4} \):
\[ f''\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} \]
Since \( f''\left(\frac{\pi}{4}\right) < 0 \), this indicates that \( x = \frac{\pi}{4} \) is a local maximum.

### Step 4: Find the maximum value
To find the maximum value of the function at this critical point, we substitute \( x = \frac{\pi}{4} \) back into the original function:
\[ f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Conclusion
Thus, the function \( f(x) = \sin x + \cos x \) has a local maximum at \( x = \frac{\pi}{4} \) with a maximum value of \( \sqrt{2} \).

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