Let f(x)=`min(x^(3),x^(4))"for all "x in R`. Then,

Options

A. f(x) is continuous for all x
B. f(x) is indifferentiable for all x
C. `f'(x)=3x^(2)"for all "x gt 1`
D. f(x) is not differentiable at two points

Explanation

To solve the problem, we need to analyze the function \( f(x) = \min(x^3, x^4) \) for all \( x \in \mathbb{R} \). We will break down the function into different intervals and check for continuity and differentiability.

### Step 1: Determine the intervals for \( f(x) \)

We will consider three intervals based on the behavior of \( x^3 \) and \( x^4 \):
1. \( x < 0 \)
2. \( 0 \leq x < 1 \)
3. \( x \geq 1 \)

### Step 2: Evaluate \( f(x) \) in each interval

**Interval 1: \( x < 0 \)**
For negative values of \( x \):
- \( x^3 < 0 \) (since the cube of a negative number is negative)
- \( x^4 > 0 \) (since the fourth power of a negative number is positive)

Thus, in this interval:
\[ f(x) = x^3 \]

**Interval 2: \( 0 \leq x < 1 \)**
For values of \( x \) between 0 and 1:
- \( x^3 \) is less than \( x^4 \) (since \( x^3 < x^4 \) for \( 0 < x < 1 \))

Thus, in this interval:
\[ f(x) = x^4 \]

**Interval 3: \( x \geq 1 \)**
For values of \( x \) greater than or equal to 1:
- \( x^3 < x^4 \) (since \( x^3 < x^4 \) for \( x \geq 1 \))

Thus, in this interval:
\[ f(x) = x^3 \]

### Step 3: Define the piecewise function

Combining the results from the intervals, we can define \( f(x) \) as:
\[ f(x) = \begin{cases} x^3 & \text{if } x < 0 \\ x^4 & \text{if } 0 \leq x < 1 \\ x^3 & \text{if } x \geq 1 \end{cases} \]

### Step 4: Check for continuity

To check for continuity at the points where the function changes (i.e., at \( x = 0 \) and \( x = 1 \)), we need to compute the left-hand limit (LHL), right-hand limit (RHL), and the function value at these points.

**At \( x = 0 \)**:
- LHL: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^3 = 0 \)
- RHL: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^4 = 0 \)
- \( f(0) = 0^4 = 0 \)

Since LHL = RHL = \( f(0) \), \( f(x) \) is continuous at \( x = 0 \).

**At \( x = 1 \)**:
- LHL: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^4 = 1 \)
- RHL: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^3 = 1 \)
- \( f(1) = 1^3 = 1 \)

Since LHL = RHL = \( f(1) \), \( f(x) \) is continuous at \( x = 1 \).

Thus, \( f(x) \) is continuous for all \( x \in \mathbb{R} \).

### Step 5: Check for differentiability

Next, we need to check differentiability at the points \( x = 0 \) and \( x = 1 \).

**At \( x = 0 \)**:
- LHD: \( \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} 3x^2 = 0 \)
- RHD: \( \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} 4x^3 = 0 \)

Since LHD = RHD, \( f(x) \) is differentiable at \( x = 0 \).

**At \( x = 1 \)**:
- LHD: \( \lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} 4x^3 = 4 \)
- RHD: \( \lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} 3x^2 = 3 \)

Since LHD ≠ RHD, \( f(x) \) is not differentiable at \( x = 1 \).

### Conclusion

1. \( f(x) \) is continuous for all \( x \in \mathbb{R} \).
2. \( f(x) \) is differentiable at \( x = 0 \) but not at \( x = 1 \).

### Final Answer

- \( f(x) \) is continuous for all \( x \).
- \( f(x) \) is differentiable at \( x = 0 \) but not at \( x = 1 \).