Mathematics 35 Online
OpenStudy (anonymous):

Find the domain of the function.

OpenStudy (anonymous):

OpenStudy (anonymous):

So, you need to know where the function is defined. Where are radicals with even roots not defined? When they are less than 0. So, you need to set u > 0 and 9-u>0. So, the domain for the sqrt u would be x>0. The domain for sqrt 9-u would be to solve when 9-u>0

OpenStudy (anonymous):

When u is less than 9, the radicand stays positive. So x>9. You need to follow the restrictions for both the radicals, so choose the one with more restrictions.

OpenStudy (anonymous):

I really dont understand this. It is sort of confusing

OpenStudy (anonymous):

The domain is the set of numbers you are allowed to input into the function, does that make sense?

OpenStudy (anonymous):

For instance...1/x...what number are you NOT allowed to input?

OpenStudy (anonymous):

For 1/x, you are not allowed to have x=0. So, the domain would be all number except 0.

OpenStudy (anonymous):

So, for a square root, let's just say sqrt x...we know we are not allowed to take the square root of a negative number.

OpenStudy (anonymous):

So, the domain would be all positive numbers.

OpenStudy (anonymous):

For your problem, you are given 2 functions. You need to figure out what numbers are you allowed to plug in so that those square roots are legit.

OpenStudy (anonymous):

Oh! I am going to have to try to get a better understanding of these problems.. THANKS

OpenStudy (anonymous):

The answer wasnt all positive numbers....it was [0,9]

OpenStudy (anonymous):

yes, sweetie, scroll up..was giving you simpler examples to help you understand

OpenStudy (anonymous):

Do you understand why that's the answer?

OpenStudy (anonymous):

yes I do

OpenStudy (anonymous):

it's because one of the functions is defined for x greater than or equal to 0 and the other is only defined for x less than or equal to 9

OpenStudy (anonymous):

Right on. Glad you got it.

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