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Let's break down and solve the problems.

### Problem 1: Differentiability at \( x = 1 \)
The function is defined as:

\[
f(x) = 
\begin{cases} 
x^3 & \text{if } x \leq 1 \\
3x - 2 & \text{if } x > 1 
\end{cases}
\]

**Step 1: Check if \( f(x) \) is continuous at \( x = 1 \)**

For \( f(x) \) to be differentiable at \( x = 1 \), it must first be continuous at \( x = 1 \). We need to check if the left-hand limit and right-hand limit at \( x = 1 \) are equal, and if they match the function value at \( x = 1 \).

- Left-hand limit: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^3 = 1^3 = 1 \)
- Right-hand limit: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3x - 2) = 3(1) - 2 = 1 \)
- Function value: \( f(1) = 1^3 = 1 \)

Since the left-hand limit, right-hand limit, and the function value are all equal to 1, \( f(x) \) is continuous at \( x = 1 \).

**Step 2: Check if \( f(x) \) is differentiable at \( x = 1 \)**

Next, we check the derivative from both sides at \( x = 1 \).

- Left-hand derivative: \( f'(x) = \frac{d}{dx} (x^3) = 3x^2 \)
  \[
  f'(1^-) = 3(1)^2 = 3
  \]

- Right-hand derivative: \( f'(x) = \frac{d}{dx} (3x - 2) = 3 \)
  \[
  f'(1^+) = 3
  \]

Since the left-hand derivative and the right-hand derivative are both equal to 3, \( f(x) \) is differentiable at \( x = 1 \).

**Conclusion for Problem 1:**

- \( f(x) \) is differentiable at \( x = 1 \).
- The derivative at \( x = 1 \) is \( f'(1) = 3 \).

---

### Problem 2: Continuity and Non-differentiability at \( x = -1 \)
The function is defined as:

\[
f(x) = 
\begin{cases} 
x^2 & \text{if } x \leq -1 \\
1 & \text{if } x > -1 
\end{cases}
\]

**Step 1: Check if \( f(x) \) is continuous at \( x = -1 \)**

For continuity at \( x = -1 \), we need the left-hand limit and right-hand limit to be equal, and to match the function value.

- Left-hand limit: \( \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} x^2 = (-1)^2 = 1 \)
- Right-hand limit: \( \lim_{x \to -1^+} f(x) = 1 \)
- Function value: \( f(-1) = (-1)^2 = 1 \)

Since the left-hand limit, right-hand limit, and function value are all equal to 1, \( f(x) \) is continuous at \( x = -1 \).

**Step 2: Check if \( f(x) \) is differentiable at \( x = -1 \)**

Now, let's check the derivative from both sides.

- Left-hand derivative: \( f'(x) = \frac{d}{dx} (x^2) = 2x \)
  \[
  f'(-1^-) = 2(-1) = -2
  \]

- Right-hand derivative: For \( x > -1 \), \( f(x) = 1 \), which is constant, so the derivative is 0.
  \[
  f'(-1^+) = 0
  \]

Since the left-hand derivative and right-hand derivative are not equal (one is -2, the other is 0), \( f(x) \) is **not differentiable** at \( x = -1 \).

**Conclusion for Problem 2:**

- \( f(x) \) is continuous at \( x = -1 \).
- \( f(x) \) is not differentiable at \( x = -1 \).

---

Do you have any questions or would you like more details on any part? Here are 5 related questions for further exploration:

1. How do the conditions for continuity and differentiability differ?
2. Why does a function being continuous not necessarily imply it is differentiable?
3. Can a function be differentiable at a point but not continuous?
4. How can we use limits to check for differentiability?
5. What are some typical cases where a function is continuous but not differentiable?

**Tip**: Remember, differentiability implies continuity, but continuity alone does not imply differentiability!

Problem 1: Show that f(x) is differentiable at x = 1 and compute f'(1).
Problem 2: Show that f(x) is continuous at x = -1 and that f(x) is not differentiable at x = -1.

Explore the differentiability and continuity of piecewise functions. Problem 1 examines differentiability at x = 1 and finds the derivative. Problem 2 checks continuity at x = -1 and demonstrates non-differentiability. Learn essential calculus concepts of limits, continuity, and piecewise functions.

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