f(x) = √(x-6)/(x+1)

Question

Find the domain

Answer 1

There are 2 constraints to:

1), Since negatives under the radical are not real numbers, "x - 6" must be either zero or a positive number therefore we can say:

x - 6 > 0 adding 6 to both sides gives:

x > 6

2) Since the denominator of a rational expression cannot be zero, we can say x + 1 ≠ 0 and, subtracting 1 from both sides gives x ≠ -1

Putting these both together we see that if we specify that x > 6 automatically covers the other constraint therefore the domain is x > 6 or, in interval notation, D: [6, ∞)

Answer 2

if you mean f(x) = [sqr(x-6)]/(x+1)

then

x-6 is non-negative. x> or = 6

x+1 cannot be zero, so x cannot =-1, but that condition is redundant

the expression under the square root sign cannot be non-negative

the expression in the denominator cannot be zero, you can't divide by zero.

x can only be in the interval [6, infinity)

the domain is x> 6

BUT more likely you meant the problem as f(x) = sqr[(x-6)/(x+1)]

then again x cannot =-1

and (x-6)/(x+1) cannot be non-negative

set (x-6)/(x+1)>0, equal to or greater than 0, and solve for x

for (x-6)/(x+1)=0

x-6 = 0 and x=6, as that makes the numerator zero

for (x-6)/(x+1)>0,

either both x-6 and x+1 are greater than or equal to zero,

or both are less than zero

either x-6> 0 and x+1>0

or x-6< 0 and x+1<0

when both are greater than zero

x-6>0, x>6

x+1>0, x>-1

net result is x>6

when both are less than zero

x-6<0, x<6

x+1<0, x<-1

net result is x<-1

x is in the intervals (-infinity,-1) U (6, infinity)

the domain is: x<-1 or x>6

x cannot be in the interval [-1,6), -1<x<6 (put a slash through the inequality signs)

check the solution:

test by checking a few simiple integers

let x= 0, then sqr[(x-6)/(x+1)]=sqr(-6/1) which is not a real number

let x=-1 then sqr[(-1-6)/(-1+0)]= undefined

let x=2, then sqr[(2-6)/(2+1)]= sqr(-4/3) which is not a real number

let x=-2, then sqr[-2-6)/(-2+1)] = sqr8 which is a real number

let x=7, then sqr[7-6)/(7+1)]= sqr(1/8) which is a real number

to be in the domain, the square root must be a real number and defined