If `f(x)=x^(4)(1-x^(2)), x in [0, 1]`, then which of the following is true

Options

A. f(x) is continuous in [0, 1]
B. `f'(x)` exist in (0, 1)
C. `c=2//3 in (0, 1)` such that `f'(c )=0`
D. Rolle's Theorem is not applicable

Answer

To solve the problem, we need to analyze the function \( f(x) = x^4(1 - x^2) \) defined on the interval \( [0, 1] \).

### Step-by-Step Solution

1. **Rewrite the Function**:
We start by expanding the function:
\[ f(x) = x^4(1 - x^2) = x^4 - x^6 \]

2. **Check the Domain**:
The function is defined for \( x \) in the interval \( [0, 1] \). This means we will evaluate the function within this range.

3. **Check Continuity**:
The function \( f(x) = x^4 - x^6 \) is a polynomial function. Polynomial functions are continuous everywhere, including the interval \( [0, 1] \). Therefore, \( f(x) \) is continuous on \( [0, 1] \).

4. **Check Differentiability**:
Since \( f(x) \) is a polynomial, it is also differentiable everywhere on \( [0, 1] \). We can find the derivative:
\[ f'(x) = \frac{d}{dx}(x^4 - x^6) = 4x^3 - 6x^5 \]

5. **Set the Derivative to Zero**:
To find critical points, we set the derivative equal to zero:
\[ 4x^3 - 6x^5 = 0 \]
Factoring out \( 2x^3 \):
\[ 2x^3(2 - 3x^2) = 0 \]
This gives us:
\[ 2x^3 = 0 \quad \text{or} \quad 2 - 3x^2 = 0 \]
From \( 2x^3 = 0 \), we get \( x = 0 \). From \( 2 - 3x^2 = 0 \):
\[ 3x^2 = 2 \quad \Rightarrow \quad x^2 = \frac{2}{3} \quad \Rightarrow \quad x = \sqrt{\frac{2}{3}} \approx 0.816 \]

6. **Apply Rolle's Theorem**:
Since \( f(x) \) is continuous on \( [0, 1] \) and differentiable on \( (0, 1) \), and \( f(0) = f(1) = 0 \), by Rolle's Theorem, there exists at least one \( c \) in \( (0, 1) \) such that \( f'(c) = 0 \). The critical points we found, \( x = 0 \) and \( x = \sqrt{\frac{2}{3}} \), confirm this.

### Conclusion
Based on the analysis:
- The function is continuous on \( [0, 1] \).
- The function is differentiable on \( (0, 1) \).
- There are critical points where the derivative is zero.

Thus, the correct statements regarding the function \( f(x) \) are that it is continuous and differentiable on the given interval, and it has critical points in the interval.