Question

Have today off, need some help with AB calc homework.

Answer 1

Ok my question is,
What is the domain of

1/sqrt(x^2 -6x-7) ?

Answer 2

srry f(x)= that

Answer 3

Hmm i think √(1 - x²)
√(1 - x²)² ≥ 0²
1 - x² ≥ 0
-x² ≥ -1
√x² ≤ √1
x ≤ ±1

Domain: [-1, 1] use brackets because both x values are inclusive.

Answer 4

you just wrote a bunch of question marks

Answer 5

It's really confusing to me, i got the answer, x<-1 and x>7, but the answers dont make sense

Answer 6

these are the answers....

Answer 7

when is the sqrt greater or equal to zero?

Answer 8

It asks you which values of x are allowed for f(x) to exist.
\[f(x) = \frac{1}{\sqrt{x^2-6x-7}}\]
Now, if you see. For f(x) to exist, it should always be defined for the domain. So, we essentially have to find out the values of x where f(x) is undefined and remove them from all x.

Are you getting the statement above?

Answer 9

yea exactly

Answer 10

i already got that

Answer 11

its just that my teacher wrote down multiple choice answers and they are confusing

Answer 12

So we have to find when, the denominator becomes zero. [because that's when f(x) is undefined]
Also, \[x^2 - 6x - 7 > 0\]
Solve for this and you'll have your answer!

Answer 13

did you see my post above?
\

Answer 14

isn't that the answer?

Answer 15

the denomontor becomes zero with 7 right..?

Answer 16

That is the perfect answer. Not sure why you got it wrong. Let me see why.

Answer 17

ok do you mind if I write down the answer choices?

Answer 18

A.)x > -1 or x < 7

B.){x<-1} square {x>7}

C.)[-1,7]

D.)(-infinity,-1] U [7, infinity)

Answer 19

those are them

Answer 20

I thought it would be B cause it makes the most sense

Answer 21

What is 'b' again? The 'square' ?

Answer 22

its the square, you know like for when a proof ends

Answer 23

QED

Answer 24

I know QED. I don't know how that 'square' thing is in Maths.
Is only one answer correct?

D is the answer.