Find the range of
`f(x)=abs(sinx)+abs(cosx)0lexlepi`

Answer

To find the range of the function \( f(x) = |\sin x| + |\cos x| \) for \( x \) in the interval \( [0, \pi] \), we can analyze the behavior of the function in two parts of the interval: from \( 0 \) to \( \frac{\pi}{2} \) and from \( \frac{\pi}{2} \) to \( \pi \).

### Step 1: Analyze the function in the interval \( [0, \frac{\pi}{2}] \)

In this interval, both \( \sin x \) and \( \cos x \) are non-negative. Therefore, we can write:
\[ f(x) = \sin x + \cos x \]

### Step 2: Find the maximum and minimum values of \( f(x) \) in \( [0, \frac{\pi}{2}] \)

To find the maximum value, we can differentiate \( f(x) \):
\[ f'(x) = \cos x - \sin x \]
Setting the derivative to zero to find critical points:
\[ \cos x - \sin x = 0 \implies \cos x = \sin x \implies x = \frac{\pi}{4} \]

Now we evaluate \( f(x) \) at the endpoints and the critical point:
- At \( x = 0 \):
\[ f(0) = \sin(0) + \cos(0) = 0 + 1 = 1 \]
- At \( x = \frac{\pi}{4} \):
\[ f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \]
- At \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = 1 + 0 = 1 \]

Thus, in the interval \( [0, \frac{\pi}{2}] \), the minimum value is \( 1 \) and the maximum value is \( \sqrt{2} \).

### Step 3: Analyze the function in the interval \( [\frac{\pi}{2}, \pi] \)

In this interval, \( \sin x \) is non-negative and \( \cos x \) is non-positive. Therefore, we can write:
\[ f(x) = \sin x - \cos x \]

### Step 4: Find the maximum and minimum values of \( f(x) \) in \( [\frac{\pi}{2}, \pi] \)

Again, we differentiate:
\[ f'(x) = \cos x + \sin x \]
Setting the derivative to zero:
\[ \cos x + \sin x = 0 \implies \sin x = -\cos x \implies x = \frac{3\pi}{4} \]

Now we evaluate \( f(x) \) at the endpoints and the critical point:
- At \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = 1 \]
- At \( x = \frac{3\pi}{4} \):
\[ f\left(\frac{3\pi}{4}\right) = \sin\left(\frac{3\pi}{4}\right) - \cos\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} - \left(-\frac{\sqrt{2}}{2}\right) = \sqrt{2} \]
- At \( x = \pi \):
\[ f(\pi) = \sin(\pi) - \cos(\pi) = 0 + 1 = 1 \]

Thus, in the interval \( [\frac{\pi}{2}, \pi] \), the minimum value is \( 1 \) and the maximum value is \( \sqrt{2} \).

### Step 5: Combine the results

From both intervals, we find that the minimum value of \( f(x) \) is \( 1 \) and the maximum value is \( \sqrt{2} \).

### Conclusion

The range of the function \( f(x) = |\sin x| + |\cos x| \) for \( x \in [0, \pi] \) is:
\[ \text{Range} = [1, \sqrt{2}] \]