Show that the function ` f(x) = |sinx+cosx|` is continuous at `x = pi`.

Explanation

To show that the function \( f(x) = |\sin x + \cos x| \) is continuous at \( x = \pi \), we will follow these steps:

### Step 1: Define the Functions
Let:
- \( g(x) = \sin x + \cos x \)
- \( h(x) = |x| \)

Then, we can express \( f(x) \) as:
\[ f(x) = h(g(x)) = |g(x)| \]

### Step 2: Check Continuity of \( g(x) \)
The function \( g(x) = \sin x + \cos x \) is a sum of two continuous functions (\( \sin x \) and \( \cos x \)). Since the sum of continuous functions is continuous, \( g(x) \) is continuous everywhere.

### Step 3: Check Continuity of \( h(x) \)
The function \( h(x) = |x| \) is also continuous everywhere.

### Step 4: Use the Composite Function Rule
Since both \( g(x) \) and \( h(x) \) are continuous functions, we can use the property that the composition of two continuous functions is also continuous. Therefore, \( f(x) = h(g(x)) \) is continuous everywhere.

### Step 5: Evaluate \( f(\pi) \)
Now, we need to evaluate \( f(\pi) \):
\[ g(\pi) = \sin(\pi) + \cos(\pi) = 0 + (-1) = -1 \]
Thus,
\[ f(\pi) = |g(\pi)| = |-1| = 1 \]

### Step 6: Conclusion
Since \( f(x) \) is continuous everywhere, it is certainly continuous at \( x = \pi \). Therefore, we conclude that:
\[ f(x) = |\sin x + \cos x| \text{ is continuous at } x = \pi. \]