What is the domain of 15x^2 + 3x - sqrt2 (x)?

[trip7]

Find the domain?

How would I go about finding the domain of: 15x^2 + 3x - sqrt2 (x)

trip7

 

[maverick280857]

The domain is the set of all x greater than or equal to zero, if you mean

$$15x^{2} + 3x - \sqrt x$$

How did you do this? And where did you go wrong?

Cheers
Vivek

 

[AKG]

How would I go about finding the domain of: 15x^2 + 3x - sqrt2 (x)

trip7

For what values of x can you evaluate this expression? Note, as a general rule:
if the domain of f(x) is Da
if the domain of g(x) is Db
if the domain of h(x) is Dc
The domain of f(x) + g(x) + h(x) is the intersection of Da, Db, and Dc. So, what is the domain of 15x^2? I.e. what values can you plug in for x and get real value for 15x^2? Now, it depends if you're restricting yourself to real numbers, or are you including complex numbers, or are we talking about vectors, or anything else? But in most situations, I'm guessing you're talking about reals? What reals can you plug into 3x? What reals can you plug into -sqrt{x}?

I'll give you a similar example, this is really too easy to give away though. Take the expression:

3x^3 - 1/x + x

For all Reals, 3x^3 is defined, so R (the entire set of Reals) is the domain for this part. -1/x is defined for all x except 0. That's the set R\{0}. The last part, x, is also defined for all Reals (obviously). So, the interaction of these sets are:
R INTERSECT R\{0} INTERSECT R = R\{0}, or all the reals except zero.

 

[trip7]

The domain is the set of all x greater than or equal to zero, if you mean

$$15x^{2} + 3x - \sqrt x$$

How did you do this? And where did you go wrong?

Cheers
Vivek

I mean $$f(x)=15x^{2} + 3x - \sqrt 2 x$$

Its the $$- \sqrt 2 x$$ that I don't know what to do with.
Any help would be appreciated.

 

[AKG]

Okay, well is it $\sqrt{(2x)}$ or $(\sqrt{2})(x)$?

 

[matt grime]

See, another question that demonstrates most maths courses aren't taught well.
The domain could br Q, R, C, F_2, the p-adics, the p-locals, Z, N...

 

[trip7]

Okay, well is it $\sqrt{(2x)}$ or $(\sqrt{2})(x)$?

$(-\sqrt{2})(x)$ is the rational.
My guess is that the domain of (f) is the set of all real numbers because you can plug any number into x and get an answer. I haven't done this in a long time and remembering that you can't get the square root of a negative number, I thought I would ask the forum how this is done. Since the x is not under the radical, anything can be substituted for x in this equation. Is this correct?

trip7

 

[matt grime]

can i ask you to complain to your teacher if that's exactly how the question was stated? the domain and codomain are part of the definition of function.

even if there were are square root of x in there it would still be a function from R, just the domain would be C.

that is why these questions should all be excised from courses.

 

[trip7]

can i ask you to complain to your teacher if that's exactly how the question was stated? the domain and codomain are part of the definition of function.

even if there were are square root of x in there it would still be a function from R, just the domain would be C.

that is why these questions should all be excised from courses.

Its not from a teacher. Its from an Algebre II Prentice Hall book. I may not have gotten far enough in the book to see what your meaning is :-)

trip7

 

[Gokul43201]

$(-\sqrt{2})(x)$ is the rational.
My guess is that the domain of (f) is the set of all real numbers because you can plug any number into x and get an answer. I haven't done this in a long time and remembering that you can't get the square root of a negative number, I thought I would ask the forum how this is done. Since the x is not under the radical, anything can be substituted for x in this equation. Is this correct?

trip7

Setting aside matt's objection, yes. That is what the book wants you to say.

However, as matt said, it is meaningless to ask for a domain without also specifying the co-domain.

EDIT : That's not what matt said, it's what AKG said.

 

[trip7]

Thanks all for your input. I can now sleep comfortably hehe.

trip7