`f(x) = sin^(-1)x+x^(2)-3x + (x^(3))/(3),x in[0,1]`

Options

A. f(x) has a point of maxima
B. f(x) has a point of minima
C. f(x) is increasing
D. f(X) is decreasing

Explanation

To solve the problem, we need to analyze the function \( f(x) = \sin^{-1}(x) + x^2 - 3x + \frac{x^3}{3} \) over the interval \( [0, 1] \). We will find the first and second derivatives to determine the monotonicity and the nature of the critical points.

### Step 1: Find the first derivative \( f'(x) \)

The first derivative of the function \( f(x) \) is calculated as follows:

\[ f'(x) = \frac{d}{dx}(\sin^{-1}(x)) + \frac{d}{dx}(x^2) - \frac{d}{dx}(3x) + \frac{d}{dx}\left(\frac{x^3}{3}\right) \]

Using the derivative formulas:
- The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1 - x^2}} \)
- The derivative of \( x^2 \) is \( 2x \)
- The derivative of \( 3x \) is \( 3 \)
- The derivative of \( \frac{x^3}{3} \) is \( x^2 \)

Thus, we have:

\[ f'(x) = \frac{1}{\sqrt{1 - x^2}} + 2x - 3 + x^2 \]

### Step 2: Set the first derivative to zero to find critical points

To find the critical points, we set \( f'(x) = 0 \):

\[ \frac{1}{\sqrt{1 - x^2}} + 2x - 3 + x^2 = 0 \]

This equation may be complex to solve analytically, so we will analyze it numerically or graphically later.

### Step 3: Find the second derivative \( f''(x) \)

Now, we compute the second derivative:

\[ f''(x) = \frac{d}{dx}\left(\frac{1}{\sqrt{1 - x^2}}\right) + \frac{d}{dx}(2x) + \frac{d}{dx}(-3) + \frac{d}{dx}(x^2) \]

Using the chain rule for the first term:

\[ f''(x) = -\frac{x}{(1 - x^2)^{3/2}} + 2 + 0 + 2x \]

Thus, we have:

\[ f''(x) = -\frac{x}{(1 - x^2)^{3/2}} + 2 + 2x \]

### Step 4: Analyze the second derivative

To determine the concavity of \( f(x) \), we need to evaluate \( f''(x) \) over the interval \( [0, 1] \).

1. At \( x = 0 \):
\[ f''(0) = 2 > 0 \quad (\text{concave up}) \]

2. At \( x = 1 \):
\[ f''(1) \text{ is undefined, but we can check values approaching 1.} \]

Since \( f''(x) \) is positive for \( x \) in \( (0, 1) \), the function is concave upward in this interval.

### Step 5: Conclusion about maxima and minima

Since \( f'(x) \) changes from negative to positive at the critical point (which we can find numerically), and since \( f''(x) > 0 \) in the interval, we conclude that:

- There is a point of minima in the interval \( [0, 1] \).
- There is no point of maxima in the interval.

### Final Answer

The correct option is that \( f(x) \) has a point of minima.