Finding the domain of this function

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Homework Help Overview

The discussion revolves around finding the domain of the function 1/(sqrt(2x)-1)², specifically addressing the conditions under which the denominator is defined and non-zero.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of the denominator being zero and explore the conditions for the square root to be defined. Questions arise about the reasoning behind the constraints on x, particularly regarding the values that make the denominator greater than zero.

Discussion Status

Participants are actively questioning the assumptions related to the domain, particularly the conditions under which the square root is valid and the denominator is non-zero. Some guidance has been offered regarding the implications of the square root's argument needing to be non-negative.

Contextual Notes

There is a noted concern about the values of x that lead to the denominator being zero or negative, and the discussion reflects on the implications of these conditions for determining the domain.

nothingsus
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Homework Statement


Find the domain of 1/(sqrt(2x)-1)2
Note: the minus one is not included under the square root


Homework Equations


Title


The Attempt at a Solution



So the denominator cannot be zero

so that means that sqrt(2x) cannot equal 1
x cannot be 1/2

letting the denominator be greater than 0 gets me x is greater than 1/2

my calculator says x is equal to or greater than 0 but cannot be 1/2

Where did the equal to greater than 0 part come from?
 
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nothingsus said:

Homework Statement


Find the domain of 1/(sqrt(2x)-1)2
Note: the minus one is not included under the square root


Homework Equations


Title


The Attempt at a Solution



So the denominator cannot be zero

so that means that sqrt(2x) cannot equal 1
x cannot be 1/2

letting the denominator be greater than 0 gets me x is greater than 1/2

my calculator says x is equal to or greater than 0 but cannot be 1/2

Where did the equal to greater than 0 part come from?

From the square root. The argument of a square root can't be negative, so you need 2x ≥ 0.
 
nothingsus said:
...

letting the denominator be greater than 0 gets me x is greater than 1/2
...
Why are you concerned about the denominator being greater than zero?
 
The simplest way is, with the answer provided, plug in x=-1 and x=1/2 and you should know why.
 

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