Let `f : [-1, -1/2] rarr [-1, 1]` is defined by `f(x)=4x^(3)-3x`, then `f^(-1) (x)` is

Explanation

To find the inverse function \( f^{-1}(x) \) for the function \( f(x) = 4x^3 - 3x \) defined on the interval \( [-1, -\frac{1}{2}] \) with a range of \( [-1, 1] \), we can follow these steps:

### Step 1: Set up the equation
We start by letting \( y = f(x) \):
\[ y = 4x^3 - 3x \]

### Step 2: Rearrange the equation
We want to express \( x \) in terms of \( y \). Rearranging gives us:
\[ 4x^3 - 3x - y = 0 \]

### Step 3: Substitute \( x \) with \( \cos(\theta) \)
To solve for \( x \), we can use the substitution \( x = \cos(\theta) \):
\[ y = 4(\cos(\theta))^3 - 3(\cos(\theta)) \]

### Step 4: Use the identity for cosine
Using the trigonometric identity for cosine, we know:
\[ 4\cos^3(\theta) - 3\cos(\theta) = \cos(3\theta) \]
Thus, we can rewrite our equation as:
\[ y = \cos(3\theta) \]

### Step 5: Solve for \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ 3\theta = \cos^{-1}(y) \]
So,
\[ \theta = \frac{1}{3} \cos^{-1}(y) \]

### Step 6: Substitute back to find \( x \)
Since \( x = \cos(\theta) \), we substitute \( \theta \):
\[ x = \cos\left(\frac{1}{3} \cos^{-1}(y)\right) \]

### Step 7: Write the inverse function
Thus, we can express the inverse function \( f^{-1}(x) \) as:
\[ f^{-1}(x) = \cos\left(\frac{1}{3} \cos^{-1}(x)\right) \]

### Final Answer
The inverse function is:
\[ f^{-1}(x) = \cos\left(\frac{1}{3} \cos^{-1}(x)\right) \]