what is the domain of the function f(x)=1/square root of (3-x)(7+x)?

Answer 1

3-x≠0 → x≠3
7+x≠0 → x≠-7
(3-x)(7+x) > 0 because it's under square root. → [(3-x > 0) & (7+x > 0)] or [(3-x < 0) & (7+x < 0)]

If [(3-x > 0) & (7+x > 0)], then x<3 & x>-7. It means that xis a number between -7 and 3. -7
If [(3-x < 0) & (7+x < 0)], then x>3 & x<-7. Its impossible.

so the domain is real numbers between -7 and 3.
Domain: -7

Answer 2

X ~ [-inf -7) and (-7 3) and (3 inf]

unless stated otherwise, the function is still defined when the square root is negative resulting in a complex number. That is not undefined. That is complex and is part of the domain of this function.

If however you are asking whats the domain of this function such that it is real-valued, then the below would be correct.

Answer 3

Because it involves a square root, the function is only defined when the denominator is non negative (discounting complex numbers). The 1/ bit further limits you to having a non zero denominator.

The denominator is only positive when both the terms (3-x) and (7+x) are both positive or both negative.

(3-x) is +ve when x < 3 and -ve when x>3

(7+x) is +ve when x>-7 and -ve when x<-7

So both terms are +ve between (-7,3)

They are never both negative, so the domain for your function is

(-7,3)

Answer 4

Not sure I understand your question.

You could restrict the domain to integers, rationals, real numbers or complex numbers since this is just putting a restriction on the value of x.

The co-domain (the value of f(x) ) would be the real numbers or complex numbers depending on the domain you chose.

Answer 5

I think it would be -7 < x < 3.

Answer 6

Ermmmmmmmmmmm 6

Answer 7

Uhhh, don't know, looks like gibberish to me!! LOL