# Math Help - help with Pre-Calc HW questions

1. ## help with Pre-Calc HW questions

1). True or False. The graph r(x) = (2x-1)/(x+2)(x-1) changes signs at x = -2. Justify your answer.

2). Which of the following is the solution to (1)/(x+2)^2 ≥ 0?
a. (-2,∞)
b. All x ≠ -2
c. All x ≠ 2
d. All real numbers
e. There are no solutions

3). Which of the following is the solution to (x^2-1)^2 ≤ 0?
a. {-1,1}
b. {1}
c. [-1,1]
d. [0,1]
e. There are no solutions

I appreciate any help on any or all questions.
Thank you so much!

2. True or False. The graph r(x) = (2x-1)/(x+2)(x-1) changes signs at x = -2. Justify your answer.

complete a sign chart

2). Which of the following is the solution to (1)/(x+2)^2 ≥ 0?
a. (-2,∞)
b. All x ≠ -2
c. All x ≠ 2
d. All real numbers
e. There are no solutions

use some common sense ... what can you say about the sign of 1/(x+2)^2 for all x except x = -2 ???

3). Which of the following is the solution to (x^2-1)^2 ≤ 0?
a. {-1,1}
b. {1}
c. [-1,1]
d. [0,1]
e. There are no solutions

common sense again ... what is the sign of (x^2-1)^2 for all x except x = 1 and x = -1 ???

3. Originally Posted by skeeter
True or False. The graph r(x) = (2x-1)/(x+2)(x-1) changes signs at x = -2. Justify your answer.

complete a sign chart

2). Which of the following is the solution to (1)/(x+2)^2 ≥ 0?
a. (-2,∞)
b. All x ≠ -2
c. All x ≠ 2
d. All real numbers
e. There are no solutions

use some common sense ... what can you say about the sign of 1/(x+2)^2 for all x except x = -2 ???

3). Which of the following is the solution to (x^2-1)^2 ≤ 0?
a. {-1,1}
b. {1}
c. [-1,1]
d. [0,1]
e. There are no solutions

common sense again ... what is the sign of (x^2-1)^2 for all x except x = 1 and x = -1 ???

So.. number 2 is b, and number 3 is c?
Is that correct?

4. Originally Posted by live_laugh_luv27
So.. number 2 is b, and number 3 is c?
Is that correct?
yes ... no

5. Originally Posted by skeeter
yes ... no
Is number 3 b? I don't really understand the notation...

6. you're guessing ... go research the notation from your text.

7. As for #3, the only way when $x^2-1$ be zero happens when $x=1$ or $x=-1.$ Those numbers are unique 'cause $(x^2-1)^2>0,$ hence the solution is $\{-1,1\}.$ ( $\{\cdot\}$ means the numbers you see, that's all; on the other hand $[-1,1]$ means the numbers covered between $-1$ and $1,$ including them.)

8. I know that [ ] is closed interval, quantity, and { } is set, or quantity, but I don't understand how that would relate to the exclusions of x ≠ 1,-1. Would it mean that the set {1,-1} is the solution to the equation? because x ≠ 1,-1?

9. sorry, i meant {-1, 1}