If `f(x)=min(1,x^2,x^3),` then

Options

A. f(x) is everywhere continuous
B. f(x) is continuous and differentiable everywhere
C. f(x) is not differentiable at two points
D. f(x) is not differentiable at one points

Explanation

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

### Step 1: Identify the components of the function
The function consists of three parts:
1. \( y = 1 \)
2. \( y = x^2 \)
3. \( y = x^3 \)

### Step 2: Find the points of intersection
We need to find where these functions intersect to determine the intervals where each function is the minimum.

1. **Intersection of \( y = 1 \) and \( y = x^2 \)**:
\[ x^2 = 1 \implies x = \pm 1 \]

2. **Intersection of \( y = 1 \) and \( y = x^3 \)**:
\[ x^3 = 1 \implies x = 1 \]

3. **Intersection of \( y = x^2 \) and \( y = x^3 \)**:
\[ x^2 = x^3 \implies x^2(1 - x) = 0 \implies x = 0 \text{ or } x = 1 \]

### Step 3: Analyze the intervals
Now we will analyze the intervals determined by the points of intersection: \( x = -1, 0, 1 \).

1. **For \( x < -1 \)**:
- \( x^2 \) and \( x^3 \) are both positive, but \( 1 \) is the minimum.
- Thus, \( f(x) = 1 \).

2. **For \( -1 < x < 0 \)**:
- Here, \( x^2 \) is positive and \( x^3 \) is negative.
- Thus, \( f(x) = x^3 \) (since it is the minimum).

3. **For \( 0 < x < 1 \)**:
- Here, both \( x^2 \) and \( x^3 \) are positive, but \( x^2 \) is less than \( 1 \).
- Thus, \( f(x) = x^2 \).

4. **For \( x \geq 1 \)**:
- Here, \( x^2 \) and \( x^3 \) are both greater than or equal to \( 1 \).
- Thus, \( f(x) = 1 \).

### Step 4: Summary of the function
We can summarize the function as follows:
\[ f(x) = \begin{cases} 1 & \text{if } x < -1 \\ x^3 & \text{if } -1 < x < 0 \\ x^2 & \text{if } 0 < x < 1 \\ 1 & \text{if } x \geq 1 \end{cases} \]

### Step 5: Check continuity and differentiability
1. **Continuity**:
- At \( x = -1 \): \( f(-1) = 1 \) from both sides.
- At \( x = 0 \): \( f(0) = 0 \) from both sides.
- At \( x = 1 \): \( f(1) = 1 \) from both sides.
- Thus, \( f(x) \) is continuous everywhere.

2. **Differentiability**:
- At \( x = -1 \): The derivative from the left is \( 0 \) (since \( f(x) = 1 \)) and from the right is \( 3(-1)^2 = 3 \). Not differentiable.
- At \( x = 0 \): The derivative from the left is \( 0 \) and from the right is \( 0 \). Differentiable.
- At \( x = 1 \): The derivative from the left is \( 2(1) = 2 \) and from the right is \( 0 \). Not differentiable.

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

### Final Answer
- The function \( f(x) \) is continuous everywhere but not differentiable at \( x = -1 \) and \( x = 1 \).