What is the Inverse Function of f^-1 in Terms of x and y?

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Homework Help Overview

The problem involves finding the inverse function of a logarithmic function defined by y = f(x) = 3ln(4x) for the domain 0.01 <= x <= 1. The original poster seeks assistance in solving for x in terms of y and determining the domain of the inverse function f^-1.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to solve for x in terms of y and provides a proposed formula for the inverse function. Other participants question the correctness of the proposed domain for the inverse function and seek confirmation on the values provided.

Discussion Status

The discussion includes attempts to clarify the inverse function and its domain. Some participants express uncertainty about the proposed answers, while others provide feedback on the original poster's attempts. There is no explicit consensus on the correctness of the answers yet.

Contextual Notes

Participants note the importance of understanding the definitions of functions and their inverses, as well as the constraints of the problem, including the specified domain for x.

fazal
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Homework Statement



a)The function f is defined by y=f(x) = 3ln 4x 0.01<=x<=1
solve for x in terms of y and hence find the formula for the inverse function of f^-1

b)Write the domain of f^-1

plse help check my answers below...


Homework Equations


as above


The Attempt at a Solution



solve for x in terms of y
My Ans :x= 1/4*e^(y/3)
and hence find the formula for the inverse function of f^-1
my answer is: f^-1(x)=1/4e^(x/3)

answer to b) plse assist
 
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wikipedia said:
wiki

Let [itex]f[/itex] be a function whose domain is the set [itex]X[/itex], and whose range is the set [itex]Y[/itex]. Then, if it exists, the inverse of [itex]f[/itex] is the function [itex]f^{-1}[/itex] with domain [itex]Y[/itex] and range [itex]X[/itex], defined by the following rule:

[tex]\text{If }f(x) = y\text{, then }f^{-1}(y) = x\text{.}[/tex]

Does this help?:wink:

P.S. Please don't post double threads. Someone was already helping you with this problem in the other thread. If you didn't understand their attempt at helping you, you should have just said so instead of posting the same question in anew thread.
 


iam sorry new to the forum...
 


is the answer for part b) -9.656<=x<=4.159 ??
 


fazal said:
is the answer for part b) -9.656<=x<=4.159 ??

Looks good to me :smile:
 

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