Find the range of function, f: [0,1] -> R, `f(x)= x^3-x^2+4x+2sin^-1x .`

Options

A. (a)`[-pi -2, 0]`
B. (b)`[2,3]`
C. (c)`[0,4+pi]`
D. (d)`(0,2+pi)`

Explanation

To find the range of the function \( f: [0,1] \to \mathbb{R} \) defined by

\[ f(x) = x^3 - x^2 + 4x + 2\sin^{-1}(x), \]

we will follow these steps:

### Step 1: Analyze the function
The function is defined on the interval \([0, 1]\). We will first find the derivative of \( f(x) \) to determine its behavior (increasing or decreasing) on this interval.

### Step 2: Find the derivative
The derivative \( f'(x) \) is given by:

\[ f'(x) = \frac{d}{dx}(x^3 - x^2 + 4x + 2\sin^{-1}(x)). \]

Calculating this, we have:

\[ f'(x) = 3x^2 - 2x + 4 + \frac{2}{\sqrt{1-x^2}}. \]

### Step 3: Analyze the derivative
We need to check if \( f'(x) \) is greater than 0 for \( x \in [0, 1] \).

1. The term \( 3x^2 - 2x + 4 \) is a quadratic function that opens upwards. The discriminant of this quadratic is:

\[ D = (-2)^2 - 4 \cdot 3 \cdot 4 = 4 - 48 = -44, \]

which is negative, indicating that \( 3x^2 - 2x + 4 \) is always positive.

2. The term \( \frac{2}{\sqrt{1-x^2}} \) is also positive for \( x \in [0, 1) \).

Thus, \( f'(x) > 0 \) for all \( x \in [0, 1) \), meaning that \( f(x) \) is an increasing function on this interval.

### Step 4: Find the function values at the endpoints
Now we will evaluate \( f(x) \) at the endpoints of the interval:

1. At \( x = 0 \):

\[ f(0) = 0^3 - 0^2 + 4 \cdot 0 + 2\sin^{-1}(0) = 0 + 0 + 0 + 0 = 0. \]

2. At \( x = 1 \):

\[ f(1) = 1^3 - 1^2 + 4 \cdot 1 + 2\sin^{-1}(1) = 1 - 1 + 4 + 2 \cdot \frac{\pi}{2} = 4 + \pi. \]

### Step 5: Determine the range
Since \( f(x) \) is an increasing function on \([0, 1]\), the minimum value occurs at \( x = 0 \) and the maximum value occurs at \( x = 1 \). Therefore, the range of \( f(x) \) is:

\[ [0, 4 + \pi]. \]

### Conclusion
The range of the function \( f(x) \) is

\[ \boxed{[0, 4 + \pi]}. \]