If `f(x)=cos^(-1)(4x^(3)-3x), x in [-1,1]`, then

Options

A. `f'(x)=(-3)/(sqrt(1-x^(2)))"for all "x in [-1,1]`
B. `f'(x)=(-3)/(sqrt(1-x^(2)))"for all "x in [-1,1]`
C. `f'(x)={{:(,(-3)/(sqrt(1-x^(2))),"if "|x|lt (1)/(2)),(,(3)/(sqrt(1-x^(2))),"if "(1)/(2) lt |x| lt (1)/(2)):}`
D. `f'(x)={{:(,(-3)/(sqrt(1-x^(2))),"if "|x|lt (1)/(2)),(,(-3)/(sqrt(1-x^(2))),"if "lt x lt(-1)/(2)","(1)/(2) lt x lt 1):}`

Explanation

To solve the problem, we need to analyze the function given by \( f(x) = \cos^{-1}(4x^3 - 3x) \) for \( x \) in the interval \([-1, 1]\).

### Step 1: Understand the function
The function \( f(x) = \cos^{-1}(4x^3 - 3x) \) is defined in terms of the inverse cosine function. The expression \( 4x^3 - 3x \) is a polynomial that we need to analyze to ensure it stays within the range of the cosine function, which is \([-1, 1]\).

### Step 2: Determine the range of \( 4x^3 - 3x \)
To find the range of \( 4x^3 - 3x \) for \( x \in [-1, 1] \), we can find the critical points by differentiating the function:

\[ g(x) = 4x^3 - 3x \]

Calculating the derivative:

\[ g'(x) = 12x^2 - 3 \]

Setting the derivative to zero to find critical points:

\[ 12x^2 - 3 = 0 \implies 12x^2 = 3 \implies x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2} \]

### Step 3: Evaluate \( g(x) \) at critical points and endpoints
Now we evaluate \( g(x) \) at the critical points and the endpoints of the interval:

1. At \( x = -1 \):
\[ g(-1) = 4(-1)^3 - 3(-1) = -4 + 3 = -1 \]

2. At \( x = -\frac{1}{2} \):
\[ g\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{2}\right)^3 - 3\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{8}\right) + \frac{3}{2} = -\frac{1}{2} + \frac{3}{2} = 1 \]

3. At \( x = \frac{1}{2} \):
\[ g\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) = 4\left(\frac{1}{8}\right) - \frac{3}{2} = \frac{1}{2} - \frac{3}{2} = -1 \]

4. At \( x = 1 \):
\[ g(1) = 4(1)^3 - 3(1) = 4 - 3 = 1 \]

### Step 4: Determine the range of \( g(x) \)
From the evaluations, we find:
- \( g(-1) = -1 \)
- \( g\left(-\frac{1}{2}\right) = 1 \)
- \( g\left(\frac{1}{2}\right) = -1 \)
- \( g(1) = 1 \)

Thus, the range of \( g(x) \) for \( x \in [-1, 1] \) is \([-1, 1]\).

### Step 5: Analyze \( f(x) \)
Since \( g(x) \) takes values in \([-1, 1]\), \( f(x) = \cos^{-1}(g(x)) \) is well-defined for all \( x \in [-1, 1] \).

### Step 6: Find the derivative \( f'(x) \)
Using the chain rule, we find the derivative of \( f(x) \):

\[ f'(x) = -\frac{1}{\sqrt{1 - (4x^3 - 3x)^2}} \cdot (12x^2 - 3) \]

### Step 7: Determine the intervals for \( f'(x) \)
We analyze the sign of \( f'(x) \) based on the critical points and the intervals determined by \( g'(x) = 0 \).

### Final Answer
The function \( f(x) \) is continuous and differentiable on the interval \([-1, 1]\).