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Mathematics 7 Online
OpenStudy (anonymous):

What is the domain and range of y?

OpenStudy (anonymous):

OpenStudy (amorfide):

The domain is all of the possible x values you can put into the equation Range is all the possible y values you can obtain

OpenStudy (anonymous):

See, all of that confuses me.

OpenStudy (mertsj):

Do you know that the denominator of a fraction cannot be 0?

OpenStudy (anonymous):

Yes, the denominator cannot be a fraction.

OpenStudy (mertsj):

The denominator can be a fraction. It CANNOT be 0

OpenStudy (anonymous):

D: x ≠ 0; R: y > 2 ? @Mertsj

OpenStudy (mertsj):

What denominator do you have in your problem?

OpenStudy (anonymous):

x-2

OpenStudy (mertsj):

What value of x would cause that denominator to be 0?

OpenStudy (anonymous):

2

OpenStudy (mertsj):

So x cannot be 2 and we will exclude 2 from the domain.

OpenStudy (mertsj):

Do you also know that the number under the radical cannot be negative?

OpenStudy (anonymous):

No, i do not:/

OpenStudy (mertsj):

Well, that is why you came here...to learn something. The radicand must be positive since we are operating within the set of real numbers.

OpenStudy (mertsj):

Now I see that the radicand is a fraction and that its numerator is positive. Do you see that?

OpenStudy (anonymous):

Yes

OpenStudy (mertsj):

If the numerator of a fraction is positive and its denominator is negative, then the fraction will be negative. Do you agree?

OpenStudy (anonymous):

Yes

OpenStudy (mertsj):

That means that x-2 cannot be negative which means it must be positive or 0. But we have already said earlier that the denominator cannot be 0. So now we know that it must be positive. Do you understand?

OpenStudy (anonymous):

Yes!

OpenStudy (mertsj):

So write x-2>0 and solve

OpenStudy (anonymous):

so √1/(x-2>0) ?

OpenStudy (mertsj):

x>2

OpenStudy (mertsj):

The solution of x-2>0 is x>2 and that is the domain of the function.

OpenStudy (anonymous):

and y cannot be 0 ?

OpenStudy (mertsj):

Yes. y cannot be 0 because it says that y is the positive root.

OpenStudy (mertsj):

Now we know the domain and range of the original function. The problem asks for the domain and range of the inverse function so remember: In the inverse function, the domain and range are reversed: The domain of the function is the range of the inverse and the range of the function is the domain of the inverse.

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