Answer is: option1

\( \frac{1}{2} \)

Solution:

Step 1: Recall the inverse function derivative rule

If \( g \) is the inverse of \( f \), then the derivative of the inverse function at any point is given by the formula:

\[ (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} \]

Step 2: Solve for \( f^{-1}(2) \)

We need to find the value of \( x \) such that \( f(x) = 2 \). From the equation \( f(x) = x^3 - x + 2 \), we solve \( x^3 - x + 2 = 2 \), which simplifies to:

\[ x^3 - x = 0 \]

Factoring this gives:

\[ x(x^2 - 1) = 0 \quad \text{or} \quad x(x-1)(x+1) = 0 \]

Thus, \( x = 0 \), \( x = 1 \), and \( x = -1 \) are the solutions. Since we are given \( x \geq 1 \), the valid solution is \( x = 1 \), so \( f^{-1}(2) = 1 \).

Step 3: Find \( f'(x) \)

The derivative of \( f(x) = x^3 - x + 2 \) is:

\[ f'(x) = 3x^2 - 1 \]

Step 4: Evaluate \( f'(1) \)

Substitute \( x = 1 \) into \( f'(x) \):

\[ f'(1) = 3(1)^2 - 1 = 3 - 1 = 2 \]

Step 5: Apply the inverse function derivative formula

Now, apply the inverse function derivative formula:

\[ (f^{-1})'(2) = \frac{1}{f'(1)} = \frac{1}{2} \]

Final Answer:

The value of \( (f^{-1})'(2) \) is \( \frac{1}{2} \), so the correct answer is (A).