state the univers of the following function

[zlatsghost]

 

[Deleted member 4993]

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I suppose you want to state the range and the domain of the given function/s.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[zlatsghost]

Domain

 

[zlatsghost]

Am I at the right track?

 

[Deleted member 4993]

Am I at the right track?

You have circled two problems - whose answer are you posting?

 

[zlatsghost]

f(x)=log2(x2- 2x)

 

[Deleted member 4993]

f(x)=log2(x2- 2x)

According to your textbook/class-notes:

What is the DEFINITION of the universe of a function?​

 

[zlatsghost]

I have no idea just guide me trow me some book or article to follow up thanks

 

[zlatsghost]

I was helping my letile sister it have been 13 + years since I grougua from high school

 

[Steven G]

You have the correct domain. Good job.
Most of us here are not familiar with the definition of the universe of a function? Maybe someone later whose knows the definition will come along and help out. Meanwhile, can you look at your little sister's book for the definition.
We really do want to help!

 

[HallsofIvy]

0Where did you get the word "universe"? Surely the definition should have been given where it was introduced. Is this a translation to English from another language? I suspect it is what I would call the "domain" of a function. That is defined as the set of all number to which the function can be applied. The crucial point is that logarithm to any base has domain "all positive numbers".

So "log₂(x²+ 2x)" has domain x²+ 2x= x(x+ 2)> 0. A product of two numbers is positive if and only if the two numbers have the same sign, both positive of both negative: x> 0 and x+ 2> 0 so x> 0 and x> -2. Both are true for x> 0,

Or x< 0 and x+ 2< 0 so x< 0 and x< -2. Both are true for x< -2.

 

[nasi112]

What if it means: state the inverse?

 

[lex]

 

[pka]

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One wonders about writer of these questions. SEE HERE
It is possible to change the numbers on that site to see the results of all of the above.
It is true that if \(\bf b>0~\&~b\ne 1\) then \(\bf log_b(f(x))\) has domain \(\bf f(x)>0\)
Because \(f(x)=\log_2(3-x)+\log_2(3+x)=\log_2(9-x^2)\) the domain is \(\{x\in\mathbb{R}^+| \;9-x^2>0\}\).