The function f*f* is defined by f, of, x, equals, minus, x, cubed, minus, 3, x, minus, 5*f*(*x*)=−*x*3−3*x*−5 and the point left parenthesis, 1, comma, minus, 9, right parenthesis(1,−9) is on the graph of f, .*f*. If g, of, x, equals, f, to the power minus 1 , left parenthesis, x, right parenthesis*g*(*x*)=*f*−1(*x*), what is the value of g, prime, of, minus, 9, question mark*g*′(−9)?

Answer

To find the value of \( g'(−9) \), where \( g(x) = f^{-1}(x) \), we can use the relationship between the derivatives of inverse functions. Specifically, if \( y = f^{-1}(x) \), then:

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

In this case, we're interested in \( g'(-9) \). Notice that since the point \( (1, -9) \) is on the graph of \( f(x) \), it means that:

\[
f(1) = -9
\]

So, \( g(-9) = 1 \). To find \( g'(-9) \), we need to calculate \( f'(1) \) first.

Given \( f(x) = -x^3 - 3x - 5 \), the derivative \( f'(x) \) is:

\[
f'(x) = \frac{d}{dx}(-x^3 - 3x - 5) = -3x^2 - 3
\]

Now, substitute \( x = 1 \) into \( f'(x) \):

\[
f'(1) = -3(1)^2 - 3 = -3 - 3 = -6
\]

Therefore:

\[
g'(-9) = \frac{1}{f'(1)} = \frac{1}{-6} = -\frac{1}{6}
\]

So, the value of \( g'(-9) \) is \( \boxed{-\frac{1}{6}} \).

Would you like any further details or have any other questions?

Here are 5 related questions you might find interesting:
1. How do you find the inverse of a cubic function algebraically?
2. What are the conditions for a function to have an inverse?
3. Can a function with a negative derivative have an increasing inverse?
4. How would you find the derivative of an inverse function without knowing the original function explicitly?
5. How does the graph of a function compare to the graph of its inverse?

**Tip:** When dealing with derivatives of inverse functions, always remember the key relationship \( g'(x) = \frac{1}{f'(g(x))} \) and verify that the function is one-to-one.

Learn how to find g'(-9) using the inverse function theorem and derivatives. Solve for the derivative of the inverse function g(x) = f^{-1}(x) where f(x) = -x^3 - 3x - 5, and determine the value of g'(-9). Suitable for advanced high school students studying calculus and inverse functions.

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