Implied domain of h(x) = sqrt[ (x-1) / (x+2) ]?

Question

Can you please answer it slowly, step by step? Thank you.

Here's what I started:

(x-1)/(x+2) > or = 0

Now what? Can you explain the next step?

Answer 1

okay. so when you sketch (x-1)(x+2), notice its x intercepts



as it is a concave up parabola and you want the values for which (x-1)(x+2) is *positive*



pick all the points where the curve is *above* the x-axis.



on the x-axis, draw in a closed circle at the intercepts and following the x-axis, draw a line where the values are positive



therefore, the answer is x< or equal to=-2, x>or equal to 1

Answer 2

(x-1)/(x+2) > or = 0



x = 1 or x = -2 solve for x so that you can test them



pick a number left of -2 like -3 and plug into



(x-1)/(x+2) > or = 0 ==> -4*-1 > or = 0 true so (negative infinity, -2) is part of domain



pick a number between -2 and 1, like 0 and plug into



(x-1)/(x+2) > or = 0 ==> -1*2 > or = 0 is false so (-2, 1)is not your domain



now pick a number right of 1 and check the inequality. Also, you need to include 1 in your domain because number under square root can have zero, but you can't include -2 because it will make denominator of (x-1)/(x+2) zero, not allowed.

Answer 3

The domain of h(x) is the set of all x values where x can be mapped under the function h to a real value.

(Strictly speaking h is a multi-function, since for each x (except x=1), there are two possible values for h(x).)



So, we need to have

(x-1)/(x+2) ≥ 0. --- (*)



Consider the possible cases:



(a) x+2 > 0:

Multiply both sides of (*) by (x+2) to get:

x-1 ≥ 0 (note the inequality sign remains "≥" because we are multiplying by a positive number)

=> x ≥ 1.

We also have to have x+2 > 0, i.e. x > -2,

and so overall we need

x ≥ 1 to satisfy both inequality conditions.



(b) x+2 < 0:

Multiply both sides of (*) by (x+2) to get:

x-1 ≤ 0 (note the inequality sign 'flips over' because we are multiplying by a negative number)

=> x ≤ 1.

We also have to have x+2 < 0, i.e. x < -2,

and so overall we need

x < -2 to satisfy both inequality conditions.



So from (a) and (b), the possible values for x are given by the set {x: x<-2 or x≥ 1}.

Answer 4

Domain is is the set of all real numbers variable x can take such that the expression defining the function is real.

h(x) = sqrt[ (x-1) / (x+2) ]



can take any real number except inside sqrt negative numbers. Sqrt of negative numbers don’t exist. Then

(x-1)/(x+2) >=0. Set of all real numbers except x=-2



Solving:

Between –infinity and-2 Result is – / - = +

Between –2 and 1 Result is – / + = -

Between 1 and infinity Product is + / + = +



Hence the domain in interval notation is given by

(-infinity ,-2) U [1 , +infinity)