Math Problem Statement

The function ff is defined by f, of, x, equals, minus, x, cubed, minus, 3, x, minus, 5f(x)=−x3−3x−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 parenthesisg(x)=f−1(x), what is the value of g, prime, of, minus, 9, question markg′(−9)?

Answer

Solution

To find the value of g(9)g'(−9), where g(x)=f1(x)g(x) = f^{-1}(x), we can use the relationship between the derivatives of inverse functions. Specifically, if y=f1(x)y = f^{-1}(x), then:

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

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

f(1)=9f(1) = -9

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

Given f(x)=x33x5f(x) = -x^3 - 3x - 5, the derivative f(x)f'(x) is:

f(x)=ddx(x33x5)=3x23f'(x) = \frac{d}{dx}(-x^3 - 3x - 5) = -3x^2 - 3

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

f(1)=3(1)23=33=6f'(1) = -3(1)^2 - 3 = -3 - 3 = -6

Therefore:

g(9)=1f(1)=16=16g'(-9) = \frac{1}{f'(1)} = \frac{1}{-6} = -\frac{1}{6}

So, the value of g(9)g'(-9) is 16\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)=1f(g(x))g'(x) = \frac{1}{f'(g(x))} and verify that the function is one-to-one.

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Math Problem Analysis

Mathematical Concepts

Inverse Functions
Derivatives

Formulas

Derivative of a cubic function
Derivative of an inverse function

Theorems

Inverse function theorem

Suitable Grade Level

Advanced High School