How to Find the Derivative of the Inverse Function f(x) for a Given Polynomial?

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SUMMARY

The discussion focuses on finding the derivative of the inverse function \( f^{-1}(x) \) for the polynomial \( f(x) = x^3 - 3x^2 - 1 \) at the point \( x = -1 \), where \( f(3) = -1 \). The key theorem referenced is \( \frac{df^{-1}}{dx}(b) = \frac{1}{\frac{df}{dx}(a)} \) with \( f(a) = b \). The derivative \( \frac{df}{dx} \) must be calculated at \( a = 3 \) to determine \( \frac{df^{-1}}{dx}(-1) \). This theorem is essential for solving inverse function problems in calculus.

PREREQUISITES
  • Understanding of polynomial functions, specifically \( f(x) = x^3 - 3x^2 - 1 \)
  • Knowledge of inverse functions and their properties
  • Familiarity with calculus concepts, particularly derivatives
  • Ability to apply the inverse function theorem
NEXT STEPS
  • Calculate the derivative \( \frac{df}{dx} \) for \( f(x) = x^3 - 3x^2 - 1 \)
  • Study the inverse function theorem in detail
  • Practice finding derivatives of inverse functions with various examples
  • Explore applications of inverse functions in real-world problems
USEFUL FOR

Students in calculus courses, mathematics educators, and anyone interested in understanding the properties and applications of inverse functions in polynomial contexts.

karush
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Let $f(x)={x}^{3}-3{x}^{2}-1, x\ge2$
$\text{find} \ {df}^{-1}/dx$
$ \text{at the point} \, \, x=-1=f(3)$
Not really sure what this is asking for
 
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f^{-1} is the "inverse" of function f. If the function, f, changes "a" to "b" (f(a)= b), then, if f has an inverse, f^{-1} changes "b" to a- it "reverses" the function.
Was that what you didn't understand? It seems peculiar that you would be taking a Calculus course without having seen that before.

Here, f(x)= x^3- 3x^2- 1, for x\ge 2. Notice that f(3)= 3^3- 3(3^2)- 1= 27- 27- 1= -1. That is, f change 3 to -1 so the inverse function changes -1 to 3: f^{-1}(-1)= 3. This problem asks you to find the derivative of f^{-1}(x) at x= -1.

Your Calculus text should have, probably in the same section where you found this problem, a discussion of "inverse" functions as well as a statement (and probably a proof) of the theorem that \frac{df^{-1}}{dx}(b)= \frac{1}{\frac{df}{dx}(a)} where f(a)= b.
 

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