Find the local maxima and local minima, if any, of the followig functions. Find also the local maximum and the local minimum values, as the case may be :
`f(x)=sinx-cosx, 0 lt x lt2pi`

Answer

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

### Step 1: Differentiate the function
We start by differentiating the function \( f(x) \).

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

### Step 2: Set the derivative to zero
Next, we set the derivative equal to zero to find the critical points.

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

Rearranging gives:

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

### Step 3: Use the tangent function
Dividing both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)) gives:

\[ \tan x = -1 \]

### Step 4: Find the general solutions
The general solutions for \( \tan x = -1 \) are:

\[ x = n\pi - \frac{\pi}{4} \quad \text{where } n \text{ is an integer} \]

### Step 5: Find specific solutions in the interval \( 0 < x < 2\pi \)
Now we find the specific values of \( x \) in the interval \( 0 < x < 2\pi \):

1. For \( n = 0 \):
\[ x = -\frac{\pi}{4} \quad (\text{not in the interval}) \]

2. For \( n = 1 \):
\[ x = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \]

3. For \( n = 2 \):
\[ x = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \]

Thus, the critical points are \( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \).

### Step 6: Determine if these points are maxima or minima
To determine whether these critical points are maxima or minima, we will compute the second derivative.

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

Now we evaluate the second derivative at the critical points:

1. For \( x = \frac{3\pi}{4} \):
\[ f''\left(\frac{3\pi}{4}\right) = -\sin\left(\frac{3\pi}{4}\right) + \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} \quad (\text{negative, hence local maximum}) \]

2. For \( x = \frac{7\pi}{4} \):
\[ f''\left(\frac{7\pi}{4}\right) = -\sin\left(\frac{7\pi}{4}\right) + \cos\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \quad (\text{positive, hence local minimum}) \]

### Step 7: Calculate the function values at critical points
Now we find the function values at these critical points:

1. For \( x = \frac{3\pi}{4} \):
\[ f\left(\frac{3\pi}{4}\right) = \sin\left(\frac{3\pi}{4}\right) - \cos\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} - \left(-\frac{\sqrt{2}}{2}\right) = \sqrt{2} \]

2. For \( x = \frac{7\pi}{4} \):
\[ f\left(\frac{7\pi}{4}\right) = \sin\left(\frac{7\pi}{4}\right) - \cos\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \left(\frac{\sqrt{2}}{2}\right) = -\sqrt{2} \]

### Conclusion
- The local maximum occurs at \( x = \frac{3\pi}{4} \) with a maximum value of \( \sqrt{2} \).
- The local minimum occurs at \( x = \frac{7\pi}{4} \) with a minimum value of \( -\sqrt{2} \).