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We are given a piecewise function \( f(x) \) and need to discuss its differentiability at \( x = 0 \) and \( x = \frac{\pi}{2} \).

The function is defined as follows:

\[
f(x) = \begin{cases}
1 & \text{when } x < 0, \\
1 + \sin(x) & \text{when } 0 \leq x < \frac{\pi}{2}, \\
2 + \left( x - \frac{\pi}{2} \right)^2 & \text{when } x \geq \frac{\pi}{2}.
\end{cases}
\]

### Step 1: Differentiability at \( x = 0 \)

For differentiability at \( x = 0 \), we need to check two things:
1. **Continuity at \( x = 0 \)**: The left-hand limit, right-hand limit, and function value at \( x = 0 \) must be equal.
   \[
   \lim_{x \to 0^-} f(x) = 1, \quad \lim_{x \to 0^+} f(x) = f(0) = 1 + \sin(0) = 1.
   \]
   Therefore, \( f(x) \) is continuous at \( x = 0 \).

2. **Derivative from both sides**:
   - For \( x < 0 \), \( f(x) = 1 \), so \( f'(x) = 0 \) for \( x < 0 \).
   - For \( 0 \leq x < \frac{\pi}{2} \), \( f(x) = 1 + \sin(x) \), so \( f'(x) = \cos(x) \). Therefore,
     \[
     f'(0^+) = \cos(0) = 1.
     \]
   
Since the left-hand derivative is \( 0 \) and the right-hand derivative is \( 1 \), the derivative is not continuous at \( x = 0 \). Thus, **the function is not differentiable at \( x = 0 \)**.

### Step 2: Differentiability at \( x = \frac{\pi}{2} \)

Now, we analyze differentiability at \( x = \frac{\pi}{2} \).

1. **Continuity at \( x = \frac{\pi}{2} \)**:
   \[
   \lim_{x \to \frac{\pi}{2}^-} f(x) = 1 + \sin\left(\frac{\pi}{2}\right) = 2, \quad f\left(\frac{\pi}{2}\right) = 2, \quad \lim_{x \to \frac{\pi}{2}^+} f(x) = 2.
   \]
   Therefore, \( f(x) \) is continuous at \( x = \frac{\pi}{2} \).

2. **Derivative from both sides**:
   - For \( 0 \leq x < \frac{\pi}{2} \), \( f'(x) = \cos(x) \), so
     \[
     f'\left(\frac{\pi}{2}^-\right) = \cos\left(\frac{\pi}{2}\right) = 0.
     \]
   - For \( x \geq \frac{\pi}{2} \), \( f(x) = 2 + \left(x - \frac{\pi}{2}\right)^2 \), so
     \[
     f'(x) = 2\left(x - \frac{\pi}{2}\right), \quad f'\left(\frac{\pi}{2}^+\right) = 0.
     \]

Since the left-hand and right-hand derivatives are both \( 0 \), the function is differentiable at \( x = \frac{\pi}{2} \).

### Conclusion:
- The function is **not differentiable** at \( x = 0 \).
- The function **is differentiable** at \( x = \frac{\pi}{2} \).

Would you like further details or have any questions?

Here are some related questions for further exploration:
1. What are the conditions for a function to be differentiable at a point?
2. Can a function be continuous but not differentiable? Provide examples.
3. How can we generalize differentiability for piecewise functions?
4. What is the geometrical interpretation of differentiability?
5. How does the differentiability of a function relate to the behavior of its graph?

**Tip**: Always check both continuity and the derivatives from both sides when determining differentiability at a point for piecewise functions.

A function f(x) is defined as follows:

f(x) =
1 when x < 0
1 + sin(x) when 0 ≤ x < π/2
2 + (x - π/2)^2 when x ≥ π/2

Discuss the differentiability at x = 0 and x = π/2.

Explore the differentiability of a piecewise function defined by f(x) = 1, 1 + sin(x), and 2 + (x - π/2)^2 at x = 0 and x = π/2. This problem covers key calculus concepts such as limits, continuity, and piecewise differentiability. Suitable for students in Grades 11-12.

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