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Let's solve exercise 265, where we have \( f(x) = 2x^3 - 3 \) and \( x = 1 \). We need to find the derivative of the inverse function \((f^{-1})'(a)\).

### Solution
For the derivative of the inverse function, we use the formula:
\[
(f^{-1})'(a) = \frac{1}{f'(x)} \quad \text{where } a = f(x).
\]

1. **Step 1: Find \( f(1) \):**

   Given \( f(x) = 2x^3 - 3 \), plug in \( x = 1 \):
   \[
   f(1) = 2(1)^3 - 3 = 2 - 3 = -1.
   \]

   So, \( a = -1 \).

2. **Step 2: Find the derivative \( f'(x) \):**

   Differentiate \( f(x) \):
   \[
   f'(x) = \frac{d}{dx}(2x^3 - 3) = 6x^2.
   \]

3. **Step 3: Evaluate \( f'(1) \):**

   Plug in \( x = 1 \):
   \[
   f'(1) = 6(1)^2 = 6.
   \]

4. **Step 4: Find \((f^{-1})'(-1)\):**

   Using the formula:
   \[
   (f^{-1})'(-1) = \frac{1}{f'(1)} = \frac{1}{6}.
   \]

### Final Answer:
\[
(f^{-1})'(-1) = \frac{1}{6}.
\]

Would you like a deeper explanation or have any questions about this problem? Here are some related problems:

1. How do you derive the formula \((f^{-1})'(a) = \frac{1}{f'(x)}\)?
2. What is the meaning of the derivative of an inverse function in a practical context?
3. If \( f(x) = 3x^2 \), how would you find \((f^{-1})'(a)\)?
4. What are the conditions under which a function has an inverse?
5. How do we determine the domain and range of an inverse function?

**Tip:** Remember, for a function to have an inverse, it must be one-to-one (bijective), meaning it should pass both the horizontal line test (injective) and cover the range (surjective).

Exercise 265: Find the derivative of the inverse function for f(x) = 2x^3 - 3 at x = 1.

Find the derivative of the inverse function for f(x) = 2x^3 - 3 at x = 1. This exercise uses the inverse function derivative formula, covering key calculus concepts. Suitable for students in Grades 11-12.

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