Finding the domain of a composite function help.

[nukeman]

Homework Statement

Ok, I just worked out a composite function, and it left me with:

√2x^2+5)

Now, how do I find the domain from that? I don't understand that my text says the domain of that is just all Real numbers ?

What makes this different than other square functions that we are able to find the domain like: √2-x ?

Homework Equations

The Attempt at a Solution

 

[Villyer]

This isn't any different from those. How would you go about finding the domain of the example function you posted?

 

[nukeman]

Then why did my book say the domain is all REAL numbers?

I would go:

 

[Mark44]

You did a bunch of work that doesn't get you anywhere.

The square of any real number is ≥ 0.

IOW, x² ≥ 0, for all real x
so 2x² ≥ 0, for all real x
so 2x² + 5 ≥ 5, for all real x

 

[nukeman]

Kinda lost here.

So, 2x^2 + 5

the 2x^2 part is always the same, as 2x^2, x can be any real number. And why is that?

how did you get 2x^2 + 5 <= 5?

What would the domain be?

Thanks Mark!

 

[Saitama]

x² is always positive, do you expect 2x² or 2x²+5 to result a negative number?

Kinda lost here.

So, 2x^2 + 5

the 2x^2 part is always the same, as 2x^2, x can be any real number. And why is that?

x can be any real number because square of every real number is defined.

 

[nukeman]

Oh...ok

So, everytime I get a 2x^2 in a square root, and have to find the domain, its just all real numbers? Then I just worry about what after it, like the +5?

But then, wouldn't 2x^2 + 5 be x >= -5?

 

[Mark44]

Kinda lost here.

So, 2x^2 + 5

This is not a complete statement. So 2x² + 5 is what?

the 2x^2 part is always the same, as 2x^2, x can be any real number. And why is that?

how did you get 2x^2 + 5 <= 5?

I didn't. I got 2x² + 5 ≥ 5

What would the domain be?

Thanks Mark!

Oh...ok

So, everytime I get a 2x^2 in a square root, and have to find the domain, its just all real numbers? Then I just worry about what after it, like the +5?

If you have an expression like x², or 2x² or .1x², it is always guaranteed to be ≥ 0, because the square of any real number can't be negative.

But then, wouldn't 2x^2 + 5 be x >= -5?

NO!

2x² + 5 is an expression that has a value for each value of x.

x ≥ -5,
2x² + 5 = 0,
2x² + 5 < 0,
and
2x² + 5 ≥ 0
are all statements that are either true or false, possibly depending on the value of x, but possibly not.

You can't compare 2x² + 5 (an expression) and x ≥ -5 (a statement).

If you start with this statement:
2x² + 5 ≥ 0,

you can add -5 to both sides to get
2x² ≥ -5

This is legitimate, but unwise. Since x² ≥ 0 for all real x, then 2x² ≥ 0 for all x, so 2x² is automatically ≥ -5.

 

[nukeman]

Oh ok, I see what you mean here. Thanks Mark.

So, for that expression, if they ask for the domain of that expression, what would it be? That is where I am lost.

 

[Mark44]

By that expression, I assume you mean $\sqrt{2x^2 + 5}$, right? I don't see how you can still be asking this. You know the answer, and this thread has explained what's going on in considerable detail.