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

Options

A. f(x) is everywhere differentiable
B. `f(x) gt 0 "for "x gt 1`
C. f(x) is not differentiable at three points but continuous for all `x in R`
D. f(x) is not differerntiable for two values of x

Explanation

To solve the problem, we need to analyze the function \( f(x) = \min(x, x^2, x^3) \) and determine its properties, including differentiability and continuity.

### Step 1: Identify Intervals
First, we need to find the intervals where each of the functions \( x \), \( x^2 \), and \( x^3 \) dominate. We can do this by comparing the functions pairwise.

1. **Compare \( x \) and \( x^2 \)**:
- Solve \( x = x^2 \) → \( x^2 - x = 0 \) → \( x(x - 1) = 0 \)
- Roots are \( x = 0 \) and \( x = 1 \).

2. **Compare \( x \) and \( x^3 \)**:
- Solve \( x = x^3 \) → \( x^3 - x = 0 \) → \( x(x^2 - 1) = 0 \)
- Roots are \( x = 0 \), \( x = 1 \), and \( x = -1 \).

3. **Compare \( x^2 \) and \( x^3 \)**:
- Solve \( x^2 = x^3 \) → \( x^3 - x^2 = 0 \) → \( x^2(x - 1) = 0 \)
- Roots are \( x = 0 \) and \( x = 1 \).

### Step 2: Determine the Dominating Function in Each Interval
From the comparisons, we can summarize the intervals:

- For \( x < -1 \): \( x^3 < x^2 < x \) → \( f(x) = x^3 \)
- For \( -1 < x < 0 \): \( x^2 < x < x^3 \) → \( f(x) = x^2 \)
- For \( 0 < x < 1 \): \( x < x^2 < x^3 \) → \( f(x) = x \)
- For \( x > 1 \): \( x < x^2 < x^3 \) → \( f(x) = x \)

### Step 3: Write the Piecewise Function
Now we can write the piecewise function:

\[ f(x) = \begin{cases} x^3 & \text{if } x < -1 \\ x^2 & \text{if } -1 \leq x < 0 \\ x & \text{if } 0 \leq x < 1 \\ x & \text{if } x \geq 1 \end{cases} \]

### Step 4: Check Continuity
To check continuity at the points \( x = -1 \), \( x = 0 \), and \( x = 1 \):

1. **At \( x = -1 \)**:
- \( \lim_{x \to -1^-} f(x) = (-1)^3 = -1 \)
- \( \lim_{x \to -1^+} f(x) = (-1)^2 = 1 \)
- Not continuous at \( x = -1 \).

2. **At \( x = 0 \)**:
- \( \lim_{x \to 0^-} f(x) = (0)^2 = 0 \)
- \( \lim_{x \to 0^+} f(x) = (0) = 0 \)
- Continuous at \( x = 0 \).

3. **At \( x = 1 \)**:
- \( \lim_{x \to 1^-} f(x) = (1) = 1 \)
- \( \lim_{x \to 1^+} f(x) = (1) = 1 \)
- Continuous at \( x = 1 \).

### Step 5: Check Differentiability
Now we check differentiability at the points \( x = -1 \), \( x = 0 \), and \( x = 1 \):

1. **At \( x = -1 \)**:
- \( f'(x) \) does not exist because the left-hand and right-hand derivatives are not equal.

2. **At \( x = 0 \)**:
- \( f'(0) \) exists and is equal to 1 (since \( f(x) = x^2 \) for \( x < 0 \) and \( f(x) = x \) for \( x > 0 \)).

3. **At \( x = 1 \)**:
- \( f'(1) \) exists and is equal to 1 (since \( f(x) = x \) for both sides).

### Conclusion
- The function \( f(x) \) is not differentiable at \( x = -1 \) and \( x = 0 \).
- The function is continuous everywhere except at \( x = -1 \).

### Final Answer
The correct option is:
- **Option C**: \( f(x) \) is not differentiable at two points but continuous for all \( x \in \mathbb{R} \).