-x^2 + 4 = 2x + 1
-x^2 - 2x + 4 - 1 = 0
-x^2 - 2x + 3 = 0
(x + 3)(-x + 1)= 0
x + 3 = 0 -x + 1 = 0
x = -3 -x = -1
x = 1
so x = -3 and x = 1
Answer:
<u>Domain = R - {3/2}</u>
Step-by-step explanation:
The domain of a rational function consists of all the real numbers x except those for which the denominator is 0
The given function is :
We need to find the zeros of the denominator
∴ 2x - 3 = 0 ⇒ add 3 to both sides
∴ 2x = 3 ⇒ divide both side by 2
∴ x = 3/2
The domain will be all real number except the zeros of the denominator.
So, Domain = R - {3/2}
Also, see the attached figure that represents the given function.
The square root of a a negative integer is imaginary.
It would still be a negative under a square root if you multiplied it by 2, therefor it will still be imaginary, or I’m assuming as your book calls it, undefined.
2•(sqrt-1) = 2sqrt-1
If you add a number to -1 itself, specifically 1 or greater it will become a positive number or 0 assuming you just add 1. In that case it would be defined.
-1 + 1 = 0
-1 + 2 = 1
If you add a number to the entire thing “sqrt-1” it will not be defined.
(sqrt-1) + 1 = 1+ (sqrt-1)
If you subtract a number it will still have a negative under a square root, meaning it would be undefined.
(sqrt-1) + 1 = 1 + (sqrt-1)
however if you subtract a negative number from -1 itself, you end up getting a positive number or zero. (Subtracting a negative number is adding because the negative signs cancel out).
-1 - -1 = 0
-1 - -2 = 1
If you squared it you would get -1, which is defined
sqrt-1 • sqrt-1 = -1
and if you cubed it, you would get a negative under a square root again, therefor it would be undefined.
sqrt-1 • sqrt-1 • sqrt-1 = -1 • sqrt-1 = -1(sqrt-1)
Sorry if this answer is confusing, I don’t have a scientific keyboard, I’ll get one soon.
Integers I believe :) hope this helps
