Thread: 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!

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 ???

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?

Originally Posted by live_laugh_luv27

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

yes ... no

Originally Posted by skeeter

yes ... no

Is number 3 b? I don't really understand the notation...

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

As for #3, the only way when be zero happens when or Those numbers are unique 'cause hence the solution is ( means the numbers you see, that's all; on the other hand means the numbers covered between and including them.)

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?

sorry, i meant {-1, 1}

Thanks for your help