Calculus: Solving Inverse Functions for f(x)=2x^3+3x^2+7x+4

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Homework Help Overview

The discussion revolves around finding the inverse of the function f(x) = 2x^3 + 3x^2 + 7x + 4, a topic within calculus related to inverse functions. Participants explore the challenges associated with solving for the inverse of a cubic function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses difficulty in recalling how to find the inverse of a cubic function and seeks guidance on how to proceed. Some participants suggest that using the cubic formula is complicated and question whether an explicit inverse is necessary for the problem at hand. The original poster later clarifies that the goal is to find (f^-1)'(a) for a specific value of a.

Discussion Status

Participants have provided alternative approaches to finding the derivative of the inverse function without explicitly calculating the inverse itself. There is recognition that finding a specific value of x such that f(x) = 4 may be a more manageable task than deriving the general inverse.

Contextual Notes

There is an indication that the original poster is constrained by a lack of familiarity with the cubic formula and is looking for a more straightforward method to solve the problem related to the inverse function.

ace123
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Hi I'm trying to remember inverse functions for calculus but I'm having a few problems. So any help would be appreciated.

f(x)= 2x^3 + 3x^2 + 7x+ 4

So I have no clue how to solve this for the inverse. I know how to do basic ones. But I've forgotten these kind. So can i just get a step in the right direction.
 
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You can get an inverse expression for a general cubic only by using the cubic formula. I DON'T recommend this. It's too complicated. What's the actual problem you need to solve? You probably don't need an explicit formula.
 
I'd agree, it would depend on the problem. You could use the cubic formula, but its pretty nasty.
 
Well I was hoping their was a nice way of doing b/c I knew about the cubic formula but wasn't about to use it for this. The actual question though was to find (f^-1)'(a) and the a= 4. I thought I had to find the inverse of the f(x) to solve the problem. Is there another way?
 
Yes there is. f^(-1)(f(x))=x. Differentiate both sides and use the chain rule. f^(-1)'(f(x))*f'(x)=1. So f^(-1)'(f(x))=1/f'(x). If you want to use this at f(x)=a=4, You still have to find a value of x such that f(x)=4. But that's a lot easier problem than finding the general inverse.
 
Oh, okay I see what your saying I didn't think about using f(x)=a=4. Thank you
 

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