Question

find the range of: r(x)=(x-2)/((x+1)^2)

Answer 1

I think I saw this question before.

Answer 2

the range is in the y-axis and the entire function becomes undefined if x = -1

Answer 3

I think you may also need to graph this too. like x = 1 2 3 4.. plug them in and you can grab the y values. Then once you see the graph and look at the y-axis we can determine where it is.

Answer 4

I know that but every time I in put the answer into my hw it tells I'm wrong

Answer 5

really? could you provide a screenshot?

Answer 6

hmm... I'm gonna try wolfram alpha and see what's up

Answer 7

this is for the graph on the left right?

Answer 8

the right

Answer 9

oops sorry XD

Answer 10

hmm we could try (-oo,oo) ?

Answer 11

let me try it

Answer 12

didnt work D:

Answer 13

(-10, -3)U(-3, oo) ? maybe? THis is tough D:! It's testing me XDD

Answer 14

wait the function on the left has a small line near -10...does that mean something?

Answer 15

like (-00, -10)? :O

Answer 16

umm.. but the range is in the y-axis

we could try (-oo,-10)U(-10,oo) or something

Answer 17

wait try (-oo,-10) by itself first...

Answer 18

Wouldnt the range just be -infinity?

Answer 19

like all reals.. -oo,oo

Answer 20

Not all reals for sure

Answer 21

:/

Answer 22

would it be (-00,-2)U(-2,00)

Answer 23

(-oo,-10)U(-10,0) ?

Answer 24

no forget the -10, I read the graph wrong lol

Answer 25

i don't remember how to do derivatives...

Answer 26

http://www.wolframalpha.com/input/?i=+r%28x%29%3D%28x-2%29%2F%28%28x%2B1%29^2%29

Answer 27

what does 12r greater then or eqaul to 1 mean? .-. i.e how would that look in interval notation

Answer 28

oops i mean 12r less than or equal to 1

Answer 29

r is a set of real numbers
maybe we could just solve it and be like r is less than or equal to 1/12. not sure. I have never encountered this problem before.

Answer 30

uh I give up, the program doesn't give me many chances to try so I'll just ask my professor, thanks you guys

Answer 31

\[y=\frac{x-2}{(x+1)^2} , x \neq -1 \\ \\ y=\frac{x-2}{(x+1)^2} \\ \text{ solve for } x \\ y(x+1)^2=x-2 \\ y(x^2+2x+1)-x+2=0 \\ y \cdot x^2+(2y-1)x+(y+2)=0 \\ \text{ use quadratic formula } \]

Answer 32

You can consider any restrictions on y from that