Find the domain of h(x)=ln((x-3)/(x+1))?

x>-1 AND x>3

therefore it is the set of all x > 3

You must solve:

( x - 3) / (x + 1) > 0

U

x != -1 (x is different to -1)

For the first part { ( x - 3) / (x + 1) > 0 }, we have:

==>

x - 3 > 0 OR x + 1 > 0

AND

x - 3 < 0 OR x + 1 < 0

==>

x > 3 OR x > -1 ==> x > 3

AND

x < 3 OR x < -1 ==> x < -1

==>

x > 3 AND x < -1

==>

x e R - [ -1, 3 ]

If we finally join to the

AND x != -1, we have:

==>

x e R - [ -1, 3 ]

AND

x != -1

==>

Domain of h(x):

x e R - [ -1, 3 ]

PROOF:

x = -3 ==> h(-3) = ln[ -6 / (-2) ] = ln(3) OK

x = 0 ==> h(0) = ln [ -3 / 1 ] WRONG!!!

x = 8 ==> h(8) = ln ( 5 / 9 ) < 0 OK

:)

We know that ln(x) is defined where x > 0, so we need:

(x - 3) / (x + 1) > 0

We have a quotient of two numbers, x - 3 and x + 1. For the result to be positive, we either need both to be positive or both to be negative, i.e.

1) x - 3 > 0 and x + 1 > 0 ===> x > 3 and x > -1 ===> x > 3, OR

2) x - 3 < 0 and x + 1 < 0 ===> x < 3 and x < -1 ===> x < -1

We also must make sure that the function (x - 3) / (x + 1) is defined in our domain, so we must exclude -1. Fortunately, it's already excluded, so the domain is:

x > 3, or x < -1

or in interval notation:

(-infinity, -1) U (3, infinity)

(x-3)/(x+1) > 0

Case 1: Numerator > 0, Denominator > 0 -----> x > 3, x > -1 -----> x > 3

Case 2: Numerator < 0, Denominator < 0 -----> x < 3, x < -1 -----> x < -1

Domain: x < -1 or x > 3

Mαthmφm