Let `f(x) = "min" {1, x^(2), x^(3)}`, then:

Options

A. f(x) is continous `AA x in R`
B. `f'(x) gt 0, AA x gt 1`
C. continous but not differentiable `AA x in R`
D. differentiable everywhere

Answer

To solve the problem, we need to analyze the function \( f(x) = \min \{ 1, x^2, x^3 \} \).

### Step 1: Define the function explicitly

We need to determine the intervals where each of the three functions \( 1 \), \( x^2 \), and \( x^3 \) is the minimum.

1. **For \( x < 0 \)**:
- \( x^2 \) and \( x^3 \) are both positive, while \( 1 \) is constant at \( 1 \).
- Therefore, \( f(x) = 1 \).

2. **For \( 0 \leq x < 1 \)**:
- \( x^2 \) and \( x^3 \) are both less than \( 1 \) and increase as \( x \) increases.
- At \( x = 1 \), \( x^2 = 1 \) and \( x^3 = 1 \).
- Therefore, for \( 0 \leq x < 1 \), \( f(x) = \min \{ 1, x^2, x^3 \} = x^3 \) (since \( x^3 < 1 \) for \( x < 1 \)).

3. **For \( x = 1 \)**:
- \( f(1) = \min \{ 1, 1^2, 1^3 \} = 1 \).

4. **For \( x > 1 \)**:
- Both \( x^2 \) and \( x^3 \) are greater than \( 1 \).
- Therefore, \( f(x) = 1 \).

Combining these intervals, we can define the function as:
\[ f(x) = \begin{cases} 1 & \text{if } x < 0 \\ x^3 & \text{if } 0 \leq x < 1 \\ 1 & \text{if } x \geq 1 \end{cases} \]

### Step 2: Check for continuity

To check if \( f(x) \) is continuous, we need to examine the limits at the points where the definition of the function changes, particularly at \( x = 1 \).

- **Limit from the left**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^3 = 1 \]

- **Limit from the right**:
\[ \lim_{x \to 1^+} f(x) = 1 \]

- **Value at the point**:
\[ f(1) = 1 \]

Since all three values are equal, \( f(x) \) is continuous at \( x = 1 \). It is also continuous everywhere else since it is defined piecewise with continuous functions.

### Step 3: Check for differentiability

Next, we need to check if \( f(x) \) is differentiable.

1. **For \( x < 0 \)**:
- \( f(x) = 1 \) is constant, so \( f'(x) = 0 \).

2. **For \( 0 < x < 1 \)**:
- \( f(x) = x^3 \), so \( f'(x) = 3x^2 \).

3. **At \( x = 1 \)**:
- From the left, \( f'(1^-) = 3(1)^2 = 3 \).
- From the right, \( f'(1^+) = 0 \).

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

### Conclusion

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

### Final Answer

1. **Continuous**: Yes
2. **Differentiable**: No (not differentiable at \( x = 1 \))