Consider the following statement in respect of the function
`f(x)=x^(3)-1 x in [-1,1]`
I f(X) is increasing in [-1,1]
II f(x)has no root in (-1,1]
Which of the statement given above is /are correct

Options

A. only I
B. only II
C. Both I and II
D. Neither I nor II

Explanation

To solve the problem, we need to analyze the function \( f(x) = x^3 - x \) over the interval \([-1, 1]\) and evaluate the two statements provided.

### Step 1: Determine if \( f(x) \) is increasing in \([-1, 1]\)

To check if the function is increasing, we will find the first derivative \( f'(x) \) and analyze its sign.

1. **Find the first derivative:**
\[ f'(x) = \frac{d}{dx}(x^3 - x) = 3x^2 - 1 \]

2. **Set the first derivative to zero to find critical points:**
\[ 3x^2 - 1 = 0 \implies 3x^2 = 1 \implies x^2 = \frac{1}{3} \implies x = \pm \frac{1}{\sqrt{3}} \]

3. **Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points:**
- For \( x < -\frac{1}{\sqrt{3}} \) (e.g., \( x = -1 \)):
\[ f'(-1) = 3(-1)^2 - 1 = 3 - 1 = 2 > 0 \]
- For \( -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \) (e.g., \( x = 0 \)):
\[ f'(0) = 3(0)^2 - 1 = -1 < 0 \]
- For \( x > \frac{1}{\sqrt{3}} \) (e.g., \( x = 1 \)):
\[ f'(1) = 3(1)^2 - 1 = 3 - 1 = 2 > 0 \]

4. **Conclusion on the intervals:**
- \( f(x) \) is increasing on \([-1, -\frac{1}{\sqrt{3}}]\) and \([\frac{1}{\sqrt{3}}, 1]\).
- \( f(x) \) is decreasing on \([- \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)].

### Step 2: Check if \( f(x) \) has roots in \((-1, 1]\)

To find the roots of \( f(x) \), we solve:
\[ f(x) = 0 \implies x^3 - x = 0 \]
Factoring gives:
\[ x(x^2 - 1) = 0 \implies x(x - 1)(x + 1) = 0 \]
Thus, the roots are:
\[ x = 0, \quad x = 1, \quad x = -1 \]

### Conclusion on the statements:

1. **Statement I:** \( f(x) \) is increasing in \([-1, 1]\) - **False** (it is not increasing throughout the interval).
2. **Statement II:** \( f(x) \) has no root in \((-1, 1]\) - **False** (it has roots at \( x = 0 \) and \( x = 1 \)).

### Final Answer:
Both statements are incorrect.

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