Find domain of 1/squareroot(x^2-7x-8)?

Answer 1

sqrt(x^2-7x-8) is only real if (x^2-7x-8) >= 0
sqrt(x^2-7x-8) <> 0 because the denominator <> 0

So wherever (x^2-7x-8)>0 then x will be in the domain..

(x-8)(x+1)>0

Two posibilities
1] (x-8) > 0 and (x+1)>0
or x> 8 and x > -1
meaning x > 8
or
2] (x-8) < 0 and (x+1)<0
or x < 8 and x < - 1
meaning x < -1

So domain = {xεR: x <-1 or x >8}

Answer 2

Ok, let me add a few details:

1. As mentioned before, the denominator cannot be zero. Finding the roots of the expression x^2 -7x-8

= (x-8)(x+1)

so x=8, x=-1 are the roots of the function f(x). So, the values of x cannot be equal to 8, -1.

2. There is on additional constraint...the value of the expression cannot be less than zero, otherwise, you will get an imaginary expression.

f(x) = x^2-7x-8 > 0


what happens to the function between the points -1 and 8?

Note that when x=0, f(x) = -8 which is not an allowed value...

So, if you graph the function you will see that x needs to be outside of the interval -1 <=x<=8

Good luck!

Answer 3

Factor first:
1/sqrt(x^2-7x-8) = 1/sqrt((x-8)(x+1)), so x cannot equal 8 or -1 because then the Denomintor would be 0.
x must be less than -1 or greater than 8 or else you get a negative under the square root.

The domain is x < -1 and x > 8

Answer 4

1/(sqrt(x^2 - 7x - 8))
sqrt(x^2 - 7x - 8) = 0
x^2 - 7x - 8 = 0
(x - 8)(x + 1) = 0
Therefore X = -1 or 8

Answer 5

1/(sqrt(x^2 - 7x - 8))

sqrt(x^2 - 7x - 8) >= 0

x^2 - 7x - 8 >= 0

(x - 8)(x + 1) >= 0

Domain
x <= -1
or
x >= 8

Answer 6

7x 8