What is the domain for the expression 4/[(t - 1)•sqrt{t}]?

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Discussion Overview

The discussion revolves around determining the domain of the expression 4/[(t - 1)•sqrt{t}]. Participants explore the implications of the square root and the conditions under which the denominator becomes undefined.

Discussion Character

  • Homework-related, Debate/contested, Mathematical reasoning

Main Points Raised

  • Some participants propose that the domain is t > 0, excluding t = 1, due to the undefined nature of the expression at these points.
  • Others argue that the domain can be expressed as 0 < t < 1 U t > 1, indicating that t can take any positive value except for 1.
  • A participant highlights that t = 0 also leads to an undefined expression, reinforcing the need to exclude it from the domain.
  • There is a repeated emphasis on the undefined nature of the expression when t = 1, as it results in division by zero.

Areas of Agreement / Disagreement

Participants express disagreement regarding the exact formulation of the domain, with some suggesting t > 0 while others propose a more specific range excluding 1. No consensus is reached on the final domain description.

Contextual Notes

Participants reference the conditions for the square root and the implications of division by zero, but there are unresolved aspects regarding the precise boundaries of the domain.

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Precalculus by David Cohen, 3rd Edition
Chapter 1, Section 1.3.
Question 5b

Specify the domain.

4/[(t - 1)•sqrt{t}]

Solution:

Set (t - 1) = 0 and solve for t.

t - 1 = 0

t = 1

For sqrt{t}, the radicand cannot be negative.

Domain: t can be any number except 1; t > or = 0.

Yes?
 
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RTCNTC said:
Precalculus by David Cohen, 3rd Edition
Chapter 1, Section 1.3.
Question 5b

Specify the domain.

4/[(t - 1)•sqrt{t}]

Solution:

Set (t - 1) = 0 and solve for t.

t - 1 = 0

t = 1

For sqrt{t}, the radicand cannot be negative.

Domain: t can be any number except 1; t > or = 0.

Yes?

I think you mean $\frac{4}{(t - 1)•\sqrt{t}}$
t =0 makes denominator zero so t >0 and not 1
 
So, t cannot equal 1 because when t is 1, the fraction becomes 4/0, which is undefined.

There is also a square root in the denominator.
I know that the radicand cannot be negative for square roots.

Also, t cannot be 0 for the sqrt{t} because the fraction also becomes 4/0, which is undefined.

Ok. I got it. This is why the domain is t > 0.
 
RTCNTC said:
So, t cannot equal 1 because when t is 1, the fraction becomes 4/0, which is undefined.

There is also a square root in the denominator.
I know that the radicand cannot be negative for square roots.

Also, t cannot be 0 for the sqrt{t} because the fraction also becomes 4/0, which is undefined.

Ok. I got it. This is why the domain is t > 0.

It's not, it's 0 < t < 1 U t > 1.
 
Prove It said:
It's not, it's 0 < t < 1 U t > 1.

Can you explain 0 < t < 1 U t > 1 in words?
 
RTCNTC said:
Can you explain 0 < t < 1 U t > 1 in words?

t can be any positive real number, except 1. :D
 
Didn't I say "...t cannot equal 1 because when t is 1, the fraction becomes 4/0, which is undefined"?
 

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