Answer is: option1
\( \frac{1}{23} \)Solution:
We are asked to find \( f'(-10) \), where \( g^{-1}(x) = f(x) \). From the inverse derivative rule:
\[ f'(y) = \frac{1}{g'(x)} \quad \text{where} \quad y = g(x) \]
Step 1: Find \( g'(x) \)
For \( g(x) = 2x^3 - x^2 - 5x \), we differentiate:
\[ g'(x) = 6x^2 - 2x - 5 \]
Step 2: Evaluate \( g'(x) \) at \( x = -2 \)
At \( x = -2 \):
\[ g'(-2) = 6(-2)^2 - 2(-2) - 5 = 6(4) + 4 - 5 = 24 + 4 - 5 = 23 \]
Step 3: Find \( f'(-10) \)
Since \( g(-2) = -10 \), we have:
\[ f'(-10) = \frac{1}{g'(-2)} = \frac{1}{23} \]
Final Answer:
The correct answer is: a) \( \frac{1}{23} \)