The first one is fraction and the second one is volume.
B. is the answer I believe since most are not functions
Answer: don't know sorry
Step-by-step explanation:
The answer is B) x^3-1 because it's the only one with the 3rd degree
Answer:
x ∈ (-∞, -1) ∪ (1, ∞)
Step-by-step explanation:
To solve this problem we must factor the expression that is shown in the denominator of the inequality.
So, we have:
![x ^ 2-1 = 0\\x ^ 2 = 1](https://tex.z-dn.net/?f=x%20%5E%202-1%20%3D%200%5C%5Cx%20%5E%202%20%3D%201)
So the roots are:
![x = 1\\x = -1](https://tex.z-dn.net/?f=x%20%3D%201%5C%5Cx%20%3D%20-1)
Therefore we can write the expression in the following way:
![x ^ 2-1 = (x-1)(x + 1)](https://tex.z-dn.net/?f=x%20%5E%202-1%20%3D%20%28x-1%29%28x%20%2B%201%29)
Now the expression is as follows:
![\frac{(x-2) ^ 2}{(x-1) (x + 1)}\geq0](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-2%29%20%5E%202%7D%7B%28x-1%29%20%28x%20%2B%201%29%7D%5Cgeq0)
Now we use the study of signs to solve this inequality.
We have 3 roots for the polynomials that make up the expression:
![x = 1\\x = -1\\x = 2](https://tex.z-dn.net/?f=x%20%3D%201%5C%5Cx%20%3D%20-1%5C%5Cx%20%3D%202)
We know that the first two are not allowed because they make the denominator zero.
Observe the attached image.
Note that:
when ![x\geq-1](https://tex.z-dn.net/?f=x%5Cgeq-1)
when ![x\geq1](https://tex.z-dn.net/?f=x%5Cgeq1)
and
is always ![\geq0](https://tex.z-dn.net/?f=%5Cgeq0)
Finally after the study of signs we can reach the conclusion that:
x ∈ (-∞, -1) ∪ (1, 2] ∪ [2, ∞)
This is the same as
x ∈ (-∞, -1) ∪ (1, ∞)