These are two questions and two answers.
Question 1. For all x in the domain, the function is equivalent to:
Answer:
[tex]J.\frac{1}{x^2-x}[/tex]
Explanation:
First note the the domain of the function is all the real values except those for which the denominator is zero, this is x³ - x = 0.
And x³ - x = 0 ⇒ x (x² - 1) = 0 ⇒ x (x - 1) (x + 1) = 0
⇒ x = 0, x = 1, x = - 1.
∴ The domain is all the real values such that x ≠0, x ≠ - 1, and x ≠ 1.
Now, you can simplify the function following these steps:
[tex]\frac{x+1}{x^3-x} = \frac{x+1}{x(x^2-1)}=\frac{x+1}{x(x+1)(x-1)}= \frac{1}{x(x-1)}= \frac{1}{x^2-x}[/tex]
So, the answer is the option J.
Question 2. What are the real solutions to the equation |x|² + 2|x| - 3 = 0?
Answer: option F. +/- 1
Explanation:
1) Use the fact that |x|² = x²
⇒ x² + 2|x| - 3 = 0
2) Transpose terms to isolate 2|x|:
⇒2|x| = 3 - x²
3) Square both sides:
⇒ [2|x|] ² = (3 - x² )²
⇒ 4|x|² = 9 - 6x² + x⁴
⇒ 4x² = 9 -6x² + x⁴
4) Transpose terms:
⇒ x⁴ - 4x² - 6x² + 9 = 0
⇒ x⁴ - 10x² + 9 = 0
5) Change variable: x² = u
⇒ u² - 10u + 9 = 0
6) Factor:
⇒ (u - 9) (u - 1) = 0
- u - 9 = 0 ⇒ u = 9
- u - 1 = 0 ⇒ u = 1
6) Comeback to the variable x (undo the change of variable):
⇒ x² = u ⇒
- x² = 9 ⇒ x = 3, x = - 3
- x² = 1 ⇒ x = 1, x = - 1
6) Verify extraneous solutions:
|x|² + 2|x| - 3 = 0
⇒ |3|² + 2|3| - 3 = 0
⇒ 9 + 6 - 3 = 0
⇒ 12 = 0 ⇒ FALSE ⇒ extraneous solution ⇒ discarded
⇒ |-3|² + 2|-3| - 3 = 0
⇒ 9 + 6 - 3 = 0
⇒ 12 = 0 ⇒ FALSE ⇒ extraneous solution ⇒ discarded
⇒ |1|² + 2|1| - 3 = 0
⇒ 1 + 2 - 3 = 0
⇒ 0 = 0 ⇒ TRUE ⇒ actual solution
⇒ |-1|² + 2|-1| - 3 = 0
⇒ 1 + 2 - 3 = 0
⇒ 0 = 0 ⇒ TRUE ⇒ actual solution
Conclusion: the solutions are +1 and -1.
