Carcass
Does having sqaure root of denominator count in rational function? Like x=1

Shouldn't what in the square root be greater than 0, and so the answer should be F only?

yes sir

see our math book for a deep understanding about roots

https://gre.myprepclub.com/forum/gre-math- ... -2609.html

ask if you need more

not quite sure what you meant Sir

I mean for example suppose that B is correct, then plugging in the denominator would be the square root of -9 and that does not make sense because under the root it should be a positive value.

Good question

So

\(\sqrt{-9}\)

\(\sqrt{(9 \times -1)}\)

\(\sqrt{9} \times \sqrt{-1}\)

\(3 \times \sqrt{-1}\) and \(\sqrt{-1}=i\)

so in the end

\(\sqrt{-9}\)

is basically \(3i\)

So we would have \(\frac{0}{3i}\) = correct option because is 0

Good question

So

\(\sqrt{-9}\)

\(\sqrt{(9 \times -1)}\)

\(\sqrt{9} \times \sqrt{-1}\)

\(3 \times \sqrt{-1}\) and \(\sqrt{-1}=i\)

so in the end

\(\sqrt{-9}\)

is basically \(3i\)

So we would have \(\frac{0}{3i}\) = correct option because is 0[/quote]


I agree with what tranhaianh1405 says about the domain of this rational expression; the ETS "Graduate Record Examinations Mathematical Conventions" (google the title...I do not have enough of a post history to include links) state that we are to assume that all numbers used in the test questions are real numbers, and that when the square root symbol is included in an expression, it means the "non negative square root with the domain >= 0:

"1. All numbers used in the test questions are real numbers. In particular, integers and both rational and irrational numbers are to be considered, but imaginary numbers are not. This is the main assumption regarding numbers. Also, all quantities are real numbers, although quantities may involve units of measurement. "

"5. Here are nine examples of other standard symbols with their meanings:
...
Example 5: [sqrt(x)] the nonnegative square root of x, where [x>=0]"

"6. Because all numbers are assumed to be real, some expressions are not defined. Here are three examples:
...
Example 2: If x<0 then [sqrt(x)] is not defined."

"9. Standard function notation is used in the test, as shown in the following three examples.

Example 1: The function g is defined for all [x >= 0] by [g(x) = 2x + sqrt(2)]

Example 2: If the domain of a function f is not given explicitly, it is assumed to be the set of all real numbers x for which f(x) is a real number. "

----------------------------------------

With this in mind, it seems that if we are assuming that the given expression is a "real" function, that the domain would be all real x<-1 and x>3.

This means that the only given answer item that falls into the function's domain is "E".

Theory: Domain of a function f(x) is the set of all possible values of x for which f(x) has a real value

\(f(x)=\frac{x^3-9x}{\sqrt{3x^2-6x-9}}\)
Now, f(x) is a fraction and has a square root.
So, f(x) will not be real if the denominator is 0 or the expression inside the square root evaluates to < 0
=> All values of x for which \(3x^2-6x-9 <=0 \) will not be in Domain.
\(3x^2-6x-9 <=0 \)
Divide both the sides by 3 we get
\(x^2-2x-3 <=0 \)
=> \(x^2 + x -3x -3 <=0 \)
=> x(x+1) -3(x+1) <= 0
=> (x+1) * (x-3) <= 0 [ To learn how to Solve Inequalities watch this video ]
=> -1 <= x <= 3

But for x=0 we will get numerator 0, making the expression 0 anyways.
So, Domain of f(x) = All real values of x except -1 <= x < 0 and 0 < x <= 3

So, Answer will be B and F I think.
Hope it helps!

To learn more about Functions and Inequalities watch the following video

how can option b can give a valid answer.
it gives denominator as square root of -9 which is. undefined

aliceeee : This is because \(\frac{0}{i}\) same as 0*i is equal to zero.