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

[mathdad]

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?

 

[kaliprasad]

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

 

[mathdad]

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.

 

[Prove It]

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.

 

[mathdad]

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

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

 

[MarkFL]

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

t can be any positive real number, except 1. :D

 

[mathdad]

Didn't I say "...t cannot equal 1 because when t is 1, the fraction becomes 4/0, which is undefined"?