How Does the Function Transform from \( x = e^{2t} \) to \( x = e^{-2t} \)?

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Homework Help Overview

The discussion revolves around the transformation of the function from \( x = e^{2t} \) to \( x = e^{-2t} \), exploring the relationships between the variables \( x \), \( y \), and \( t \). Participants are examining the implications of this transformation in terms of graph representation and function behavior.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to express \( y \) in terms of \( x \) and \( t \) for both forms of the function. Questions arise regarding the addition of the \( \pm \) sign in one participant's expression and how to represent the graph for \( t < 0 \).

Discussion Status

The discussion is active, with participants exploring different representations and questioning the transformation between the two forms of the function. Some guidance has been offered regarding the correctness of the expressions, but multiple interpretations are still being explored.

Contextual Notes

There is a mention of specific conditions for \( t \) and the implications for graphing, indicating that assumptions about the variable ranges are under consideration.

nameVoid
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x=e^(2t)

y=t+1

t= ( lnx ) / 2

y= ( lnx ) / 2 + 1

or

blah
 
Last edited:
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nameVoid said:
x=e^(2t)

y=t+1

t= ( lnx ) / 2

y= ( lnx ) / 2 + 1

or

y= +- ( ( lnx ) / 2 + 1 ) ?

Why did you add +- ?
 
what i meant was for t<0 how to represent the graph in terms of y and x
 
x=e^(-2t)
y=t+1

-2t=lnx
y=1-lnx/2
i suppose
 
nameVoid said:
x=e^(-2t)
y=t+1
Is this the same question? How did "[itex]x= e^{2t}[/itex]" become [itex]x= e^{-2t}[/itex]?

-2t=lnx
y=1-lnx/2
i suppose
Assuming you really did mean [itex]x= e^{-2t}[/itex], yes, that is correct.
 

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